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Quick Guide for pretreatment sizing of basins
ENSURE YOUR UNITS ARE CONSISTENT AND CORRECT
• Determine, vS, the settling velocity for the particle size targeted (recommend 80 microns for particle size. Determine vs from table on this page)
• Determine FR, the target percent removal (recommend 50-70%)
• Determine A, the area of directly connected impervious draining to the pretreatment practice
• Determine I, the peak rain intensity (0.5 in/hr for a 1-inch event, Type 2 distribution)
• Calculate the area of the pretreatment basin from AS = (FR * I * A * FR) / (vS * (1 - FR))
• Determine the basin depth as the lesser of sqrt(AS) or 6 feet
Quick Guide for pretreatment sizing of filter strips
ENSURE YOUR UNITS ARE CONSISTENT AND CORRECT
• Determine, vS, the settling velocity for the particle size targeted (recommend 80 microns for particle size. Determine vs from table on this page)
• Determine FR, the target percent removal (recommend 50-70%)
• Determine A, the area of directly connected impervious draining to the pretreatment practice
• Determine I, the peak rain intensity (0.5 in/hr for a 1-inch event, Type 2 distribution)
• Calculate the area of the filter strip from LW = - ((c * I * A) / (vS) * ln(1 - FR)) where c = 0.7 for small storms
• Determine the length (L) and the width (W) from the above computation

Initial research was conducted to determine whether an approach for sizing practices was consistently applied across the country. Stormwater programs for every state were reviewed to determine the approaches being used, as well as a consensus in the approach. The results of this review indicated that a consistent approach to stormwater pretreatment across the country was not used, and some states do not require pretreatment of stormwater runoff at all. Because of the lack of consensus and research-backed sizing criteria for pretreatment, stormwater practitioners in Minnesota determined that guidelines were needed to define specific sizing criteria to provide consistent, effective results.

Given the lack of a consistent approach in stormwater pretreatment sizing, analysis was conducted to determine the general guidelines for pretreatment sizing in Minnesota based on the settling velocity of particles in stormwater. Many sizes of suspended solids enter a treatment practice–from clays below 2 μm (2 micrometers or 2000 mm) to silts between 2 μm and 80 μm to sands between 80 μm and 4 mm and beyond. A four order of magnitude difference between the smallest clay particles and largest sand sizes particles entering the practice is noted with a greater order of magnitude difference in settling velocity. A pretreatment practice is designed to settle the largest of these (primarily inorganic) particles in a location where they will be relatively simple to clean out and dispose of. As a result of this pretreatment, required maintenance will be reduced on the primary treatment practice (also called a structural stormwater BMP) because a large portion of the suspended solids will settle in the pretreatment practice, which will prolong the life span and maintain the design pollutant removal efficiency of the primary treatment.

The rationale for pretreatment is, therefore, to reduce maintenance costs and efforts. A pretreatment practice should serve the following three functions.

1. Settle the larger particles (typically sand, which is greater than 80 μm) into an easily cleaned location to capture roughly one-half of the suspended solids.
2. Keep the sediment from scouring out of the device during high flows, which is typically done by providing sufficient depth of water or using a device to calm high flows that come into the basin.
3. Retain floatable trash and large organic particles (e.g., leaves, grass clippings, and sticks) in the pretreatment practice, which is typically done through a submerged outlet so that floatables are retained and can be cleaned out at a later time.
Particle Size Distribution of Suspended Solids in Road Stormwater Runoff [Erickson, 2012]. Click on image to enlarge.

The pretreatment sizing methodology described on this page relies on the settling function of the pretreatment practice (i.e., item No. 1 above) to perform a preliminary sizing of basins and filter strips. Properly designed and sized pretreatment practices will often remove 50 percent of the suspended solids in a typical storm before the water enters the primary treatment practice. However, the amount of sediment removed depends upon the storm intensity. Low-intensity storms will primarily move clay and silt, which do not settle well in a pretreatment device, and pretreatment is expected to remove less than 50 percent. High-intensity storms will move most of the particles lodged on impervious surfaces and will result in a higher removal of suspended solids; greater than 50 percent of the suspended solids will settle. Examples of various solids size distributions are illustrated in the adjacent figure. The x-axis covers four orders of magnitude in particle size, and the median particle diameter (at 50 percent finer than) varies from 50 to 700 μm. Therefore, design of a pretreatment facility to remove 50 percent of suspended solids is difficult, unless the average annual particle size distribution is known. Herein, the design sizing goal will be to remove a given percent of a given particle size of sand and silt, which are the primary components of soil.

## Basins

Turfstone Pavers at the Bottom of Seidel’s Lake Park Forebay and Biofiltration Cell Following Construction (Dakota County Soil and Water Conservation District)
Heavy Leaf and Debris Loading in Pretreatment Manhole Sump [Ramsey Conservation District, 2017].

Basins are stormwater best management practices (BMPs) having an engineered pool that promotes settling of solids (e.g., pond forebay, sump, hydrodynamic separator, and catch basin). These pretreatment facilities are typically small (relative to the primary treatment practice) but can be sized to settle gravel and sands, and possibly silts, at low flow conditions. For preliminary sizing purposes, the small size results in a basin that is close to well-mixed, where the concentration of suspended solids is constant with depth to the bottom of the basin and the particles are settling out because of zero velocity at the wall.

### Development of sizing criteria

Rainfall distribution curves (Types 1, 1A, and 2)

The following assumptions will be made while developing the sizing criteria.

1. A design storm of 1-inch of precipitation.
2. Permeable surfaces in the catchment do not contribute runoff for the 1-inch storm.
3. The design storm distribution over time is Natural Resources Conservation Service (NRCS) Type II with an intensity of 0.5 inches per hour (in/hr) during the most intense portion of the storm.
4. The bottom of the basin does not have sediment scouring.
5. The water column is well-mixed so that the suspended solids concentration is constant over the basin.
6. The watershed time of concentration will be approximated using the simplified 1965 US Federal Aviation Administration (FAA) time of concentration equation, which is meant to be paired with the Rational Method.

Assumption Nos. 1 through 3 are common for rainfall-runoff computations in Minnesota and greatly simplify the design. Assumption No. 4 is correct if the basin is properly cleaned. Assumption No. 5 is applicable with the small size of the pretreatment basin, relative to the discharge and the size of the primary practice and would result in a conservative sizing criteria. This sizing criteria is to be used for preliminary sizing of a pretreatment basin. For a more refined design, a dynamic design criteria similar to SHSAM (Sizing Hydrodynamic Separators And Manholes) is recommended. Assumption No. 6 is a fair assumption with the velocities and depth of the flow over a filter strip and would result in a conservative sizing criteria.

The NRCS Type II distribution of the 24-hour, 1-inch storm was selected be used in these pretreatment sizing calculations. The most intense hour of this design storm occurs around hour 12, when 50% of the storm total falls in one hour [NRCS 2015]. This design intensity, 0.5 inches per hour for that single hour is used in the following design calculations. The design flow into the basin results from the Rational Method by multiplying the directly connected impervious area of the catchment with this intensity, or

Eq. 1 $$Q = cAI$$

where Q is water discharge into the filter strip, A is the directly connected impervious area of the catchment (square feet), I is the peak intensity during the design storm (0.5 in/hr or 1.16x10-5 ft/s), and c is the runoff coefficient equal to 0.7 for smaller storms on impermeable surfaces.

The discharge is also equal to the volume of water in the basin divided by the residence time of the water in the basin, or

Eq. 2 $$Q = A_Sh / t_r$$

where As is the surface area of the water in the pretreatment basin, h is the depth of the water in the basin, and tr is the residence time of the water in the basin.

The residence time of water (tr) can be approximated by assuming that settling occurs in a well-mixed basin (Assumption No. 5) from the relation [Gulliver, 2007]

Eq. 3 $$1 - FR = 1 / (1 + v_S t_r / h)$$

where FR is the fraction of a given settling velocity to be removed and vs is the settling velocity of a given size particle. Equation 3 can be rearranged to give the residence time

Eq. 4 $$t_r = (h / v_S) * (FR / (1 - FR))$$

Substituting to solve for AS gives

Eq. 5 $$A_S = (cAI / v_s) * (FR / (1 - FR))$$

Note that Equation 5 does not depend upon the basin depth. The basin depth should be sufficient so that scour does not occur. Avoiding scour is not considered in these sizing calculations but should be considered in the sump design. To prevent scour, the depth (D) is typically recommended to be equal to the lesser of

Eq. 6 $$D ≤ sqrt (A_S) or 6 feet$$

For , or bioinfiltration, basins that may experience higher flows, a recommended minimum depth to avoid scour is equal to the lesser of

Eq. 7 $$D ≤ sqrt (A_S) or 1.5 feet$$

The fraction of a given settling velocity to be removed by the pretreatment basin will need to be chosen by the designer. The settling velocity of a given size (typical sand shaped) particle has been developed by Ferguson and Church [2004], as

Eq. 8 $$V_S = (gRd^2) / (18 u + (0.75Rd^3)^0.5)$$

Silt and sand Diameter (um) vs (m/s) Header text (ft/s) Settling velocity according to Equation 8 of various equivalent spherical diameter sand and silt particles at 20°C. 10 0.000089 0.00029 30 0.00078 0.0026 80 0.0051 0.017 100 0.0075 0.025 150 0.015 0.049 200 0.023 0.075 500 0.071 0.23 1000 0.13 0.43
Example particle retention curves

where g is the acceleration of gravity, R is the specific gravity of the particle relative to that of water, d is the equivalent spherical diameter of the particle and u is the viscosity of water. Equation 8 combines Stoke's Law at small particle diameters with the turbulent drag coefficient at larger particle diameters and a transition from a dependence upon d2 to one upon √d as the particle size increases. If we assume that the pretreatment basin is being designed to settle silica-based, inorganic particles (i.e., silt and sand) at 20oC, the settling velocities of the particles at various equivalent spherical diameters are given in the adjacent table. Sample results of these computations are given in the adjacent figure.

### Basin sizing examples

Many types of basins can be used as pretreatment practices for various sizes and circumstances of bioretention/biofiltration facilities, including forebays, sumps, proprietary devices, and basins used as pretreatment for bioretention and biofiltration facilities. The examples used herein will use a forebay, sump, and bioretention pretreatment basin. Various-sized retention criteria can be used, which is the user’s choice. Note that these sizing calculations are preliminary. If justified by the cost, it is recommended that these calculations be used as preliminary sizing methodology, to be followed by a more refined dynamic design criteria similar to SHSAM, which uses watershed routing of storms.

#### Example A - forebay pretreatment basin

One of the inlets into a 2-acre pond is draining a directly connected impervious area of 30 acres in a watershed with a 2-percent average slope and a watershed length to the outlet of 2,000 ft. The pretreatment basin for the 2-acre pond will be designed to retain 70 percent of the 80 μm fine-sand particles. Using settling velocities from the above table, the settling velocity (vS) for 80 μm fine-sand particles is 0.017 feet per second (fps). For this case, the values to be input in Equation 5 are

• FR = 0.7
• A = 30 acres
• I = 0.5 in/hr
• vs = 0.017 ft/s
• S = 0.02 ft/ft
• L = 2,000 ft

The rainfall intensity is related to the time of concentration and determining the peak intensity for the storm is essential for sizing pretreatment practices. Using the following equation from the FAA [Chow, et al. 1988], we can estimate the time of concentration (tc in minutes)

Eq. 9 $$t_c = 1.8 (1.1 - c) L^{0.5} / (100S)^{1/3}$$

where c is the Rational Method runoff coefficient of 0.7, S is the average watershed slope (ft/ft), and L is the watershed length (ft) to the outlet. The time of concentration is equal to tc = 26 minutes. This concentration time is within the most intense hour of a NRCS Type II storm [Merkel and Moody, 2015]. Thus, the intensity of the storm, I, is 0.5 inches per hour.

The design discharge is computed from Equation 1

$$Q = 0.7 * 0.5 in/hr * 30 ac = 10.5 ft^3/s$$

Substituting Equation 1 into Equation 5, the equation for the area of the pretreatment basin becomes

Eq. 10 $$A_S = (Q * FR) / (v_S (1 - FR)) = (10.5 ft^3/s * 0.7) / (0.017 ft/s * (1 - 0.7)) = 1442 ft^2$$

Finally, in Equation 6, the forebay depth would be roughly the lesser of (√1,441 = 38 ft) or 6 ft, so the preliminary water depth would be roughly 6 ft. With a large impervious area draining into one outlet into a pond, the designer may want to specify a forebay because of the large pretreatment area that is required.

#### Example B - Sump pretreatment basin

An inlet to a 3,000 ft2 pond will be draining a parking lot with a directly connected impervious area of 3 acres in a watershed with a 1 percent average slope and a watershed length to the outlet of 500 ft. The pretreatment basin for the pond will be designed to retain 50 percent of the 150 μm sand particles during the design storm. For this example, the values to put into Equation 6 are

• FR = 0.5
• A = 3 acres
• I = 0.5 in/hr
• vS = 0.049 ft/s
• S = 0.01 ft/ft
• L = 500 ft

As in the previous example, the peak rainfall intensity should be over a time of concentration, tc, calculated using Equation 9. The time concentration here is equal to 16 minutes. This time of concentration occurs within the most intense hour of an NRCS Type II storm, which accounts for approximately 50 percent of the storm total. Thus, the intensity of the storm is 0.5 in/hr.

The design discharge is computed from Equation 1

$$Q = 0.7 * 0.5 in/h * 3 ac = 1.1 ft^3/s$$

and using an intensity of 0.5 in/hr, AS becomes

$$A_S = (1.1 ft^3/s * 0.5)/(0.049 ft/s * (1-0.5)) = 22 ft^2$$

The sump would have a diameter of 5.3 ft but scaling up to a standard manhole size of 6 ft is recommended as a good practice. Using Equation 6, the water depth in the sump would be the lesser of (√21) = 5 ft or 6 ft, so the preliminary water depth in the sump would be roughly 5 ft.

#### Example C - rain garden pretreatment basin

A directly connected impervious area of 10,000 ft2 with a mean slope of 3 percent and a watershed length to the outlet of 300 ft, drains into a bioretention practice. The area is of sufficient size to install a well-sized pretreatment basin, so the designer chose to settle 70 percent of the 80 μm fine-sand particles in the filter strip during the design storm. For this case, the values to be input for Equation 5 are

• FR = 0.7
• A = 10,000 ft2
• I = 0.5 in/hr
• v_S = 0.017 ft/s
• S = 0.03 ft/ft
• L = 300 ft

As in the previous examples, the peak rainfall intensity should be over a time of concentration, tc, calculated using Equation 9. The time concentration here is equal to 9 minutes. This time of concentration occurs within the most intense hour of an NRCS Type II storm, which accounts for approximately 50 percent of the storm total. The design discharge is computed from Equation 1

$$Q = 0.7 * 0.5 in/hr * 100,000 ft^2 * 1 ac/43560 ft^2 = 0.08 ft^3/s$$

and using Equation 12, AS becomes

$$A_S = (0.08 ft^3/s * 0.7) / (0.017 fr/s * (1-0.7)) = 11 ft^2$$

Using Equation 7 for bioretention facilities, the pretreatment basin depth would be roughly the lesser of (√11) = 3.3 ft or 1.5 ft, so the preliminary depth would be roughly 1.5 ft.

## Filter strips

Vegetated Filter Strip at Curb Cutout (Barr Engineering Company, 2006).

Filter strips are primarily designed to enhance settling; physical filtering of particles in a filter strip is minimal. The filter strip is located at the bottom of a shallow, widely spread, open channel flow to allow for particle settling through the media, which is often grass or rocks. The grass or rocks allow space for the settled particles to collect until cleanout is required.

### Development of sizing criteria

Silt and sand Diameter (um) vs (m/s) Header text (ft/s) Settling velocity according to Equation 8 of various equivalent spherical diameter sand and silt particles at 20°C. 10 0.000089 0.00029 30 0.00078 0.0026 80 0.0051 0.017 100 0.0075 0.025 150 0.015 0.049 200 0.023 0.075 500 0.071 0.23 1000 0.13 0.43

The following assumptions have been made while developing the sizing criteria.

1. A design storm of 1-inch precipitation.
2. The design storm distribution over time is Natural Resources Conservation Service (NRCS) Type II with an intensity of 0.5 inches per hour (in/hr) during the most intense portion of the storm.
3. Permeable surfaces in the catchment do not contribute runoff for the 1-inch storm.
4. There is no scour on the filter strip surface.
5. The infiltration rate through the filter strip surface is minimal and relative to the flow over the filter strip.
6. The water column is well-mixed, such that the concentration of suspended solids is constant, with a loss of solids as the water moves down the filter strip in a plug-flow fashion.

Assumption Nos. 1 through 3 are common for rainfall-runoff computations in Minnesota and greatly simplify the design. Assumption No. 4 is correct if the filter strip does not need maintenance. Assumption No. 5 is made because of the short residence time of the water on the filter strip. Assumption No. 6 is a fair assumption with the velocities and depth of the flow over a filter strip and would result in a conservative sizing criteria.

The highest discharge for the design storm will be during the peak 0.5 in/hr intensity, based upon Assumption Nos. 1 through 3. The flow into the filter strip results by multiplying the directly connected impervious area of the catchment with this intensity, or

Eq. 11 $$Q = cAI$$

where Q is water discharge into the filter strip, c is a Rational Method coefficient equal to 0.7 for small storms, A is the directly connected impervious area of the catchment, and I is the peak intensity during the design storm (i.e., 0.5 in/hr).

The discharge is also equal to the volume of water on the filter strip divided by the residence time of the water on the filter strip, or

Eq 12 $$Q = (LWh) / t_r$$

where L is the length of the filter strip, W is the width of the filter strip, h is the depth of water on the filter strip, and tr is the residence time of the water on the filter strip. Combining Equations 11 and 12 results in a relation for filter strip surface area (LW).

Eq 13 $$LW = (cAI * t_r) / h$$

Rearranging for tr gives the resident time of water on the filter strip.

Eq 14 $$t_r = (hLW) / (cAI)$$

The residence time of water on the filter strip (tr) can also be approximated by assuming a well-mixed water column (Assumption No. 5) with settling and plug flow down the filter strip from the relation [Gulliver, 2007]

Eq 15 $$FR = 1 - exp(-(v_St_r) / h)$$

where FR is the fraction of a given settling velocity to be removed and vS is the settling velocity of a given size particle. Equation 15 can be rearranged to give the residence time

Eq 16 $$t_r = - (h/v_S)ln(1 - FR)$$

Substituting Equation 16 into Equation 13 results in

Eq 17 $$LW = - ((cAI)/v_S) Ln(1 - FR)$$

The natural log of a fraction will result in a negative number, so the area will be a positive value. Note that Equation 17 does not depend upon the flow depth. If the depth is large, the velocity is correspondingly low so the particles are allowed to settle. It is also noteworthy that the surface area of the filter strip (LW) is important, and not just the length of the filter strip. This provides flexibility to designers, as a short filter strip, within reason, can provide adequate pretreatment with a sufficient width to meet the overall surface area requirement. The roughness of the filter strip is unimportant for settling. However, the roughness should have sufficient porosity to hold and avoid sediment scour.

The fraction of a given settling velocity to be removed by the filter strip will need to be chosen by the designer. However, for a limited area of filter strip, Equation 17 can be rearranged to result in the fraction retained for a given filter strip area

Eq 18 $$FR = 1 - exp(-(v_SLW)/(cAI)) = 1 - exp(-(v_SLW)/Q)$$

The settling velocity of a given size (typically sand-shaped) particle was developed by Ferguson and Church [2004], as

Eq 19 $$v_S = (gRd^2) / (18''u'' + (0.75Rd^3)^{0.5})$$

Examples of particle retention for filter strips.

where g is the acceleration of gravity, R is the specific gravity of the particle relative to that of water, d is the equivalent spherical diameter of the particle, and u is the viscosity of water. Equation 19 combines Stoke's Law at small particle diameters with the turbulent drag coefficient at larger particle diameters and a transition in between. The transition is from a dependence upon d2 to one upon √d as the particle size increases.

Assuming the filter strip is being designed to settle silica-based, inorganic particles (i.e., silt and sand) at 20oC, the settling velocity of the particles at various equivalent spherical diameters are presented in the above table specifying settling velocities.

The designer must first choose the fraction of a given silt or sand size that they want to remove through settling into the filter strip. The settling velocity is then chosen from the table or using Equation 19. The design storm intensity has been chosen herein at 0.5 in/hr, and the directly connected impervious area is given by site characteristics. The surface area of the filter strip (LW), can then be computed. If the space is limited for a filter strip, the fraction of the targeted particle size removed can be computed. Sample results of these computations are given in the adjacent figures.

### Filter strip sizing examples

The following section demonstrates the impact that available space has on the sizing of the pretreatment device. The length of the filter strip greatly determines the amount of pretreatment it can provide downstream BMPs. In Example D, an aggressive filter strip is highlighted, which settles 80 percent of the fine particulates that enter the BMP. Example E will demonstrate a filter strip which is fit into a limited space.

#### Example D. Aggressive filter strip settling

A directly connected impervious area of 10,000 ft2 with a mean slope of 3 percent, and a watershed length to the outlet of 300 ft drains into a bioretention practice. The area is a sufficient size to install a well-sized filter strip, so the designer chose to settle 80 percent of the 80 um fine-sand particles in the filter strip during the design storm. For this case, the values to be used are

• FR = 0.8
• A = 10,000 ft2
• I = 0.5 in/hr
• vS = 0.017 ft/s
• c = 0.7
• S = 0.03 ft/ft
• L = 300 ft

As in the previous examples, the peak rainfall intensity should be over a time of concentration, tc, calculated using Equation 9. The time concentration here is equal to 9 minutes and occurs within the most intense hour of an NRCS Type II storm, which accounts for approximately 50 percent of the storm total. Equation 17 then provides the values of LW

$$LW = - ((0.7 * 10000 ft^2 * 0.5 in/hr) / 0.017 ft/s) * ln(1-0.8) = 7.6 ft^2$$

Assuming a 3-ft wide inlet, the filter strip would need to be a minimum of 2.5 ft (L = LW/W = 7.6 ft2/3 ft = 2.5 ft) in length by 2.5-ft. To settle 80 percent of the fine sand particles entering this bioretention practice, the filter strip is recommended to be 3 ft wide by 2.5 ft long.

#### Example E: Limited Space Filter Strip Settling

A directly-connected impervious area of 10,000 ft2 with a mean slope of 3 percent and a watershed length to the outlet of 300 ft drains into a bioretention practice. The space for a filter strip is limited to a width of 2 ft and length of 2 ft. The designer decided to compute the percent of the 100 um fine-sand particles that will settle into the filter strip during the design storm. For this case,

• A=10,000 ft2
• I = 0.5 in/hr
• vs = 0.025 ft/s
• LW = = 2 ft * 2 ft = 4 ft2
• c=0.7
• S=0.03 ft/ft
• L=300 ft

As in the previous examples, the peak rainfall intensity should be over a time of concentration, tc, calculated using Equation 9. The time concentration here is equal to 9 minutes and occurs within the most intense hour of an NRCS Type II storm, which accounts for approximately 50 percent of the storm total.

The value of FR for a 100 um fine-sand particle is

$$FR = 1 - exp( -(0.025ft/s * 4 ft^2)/(0.7 * 10,000 ft^2 * (1ac/43560ft^2/ac) * 0.5 in/hr) = 0.71$$

The filter strip would retain 71 percent of the 100um sand particles during the design storm. This retention for the design storm will still be worth considering because for smaller storms, the retention will be greater. For example, at a 0.25 inch/hour intensity and the other conditions given above, the retention would be 93 percent of the 100 um fine-sand particles. The disadvantage of this limited space filter strip is that more silt and sand will settle downstream in the bioretention practice, requiring more frequent maintenance to maintain the water quality benefits of the BMP.

### Filter strip recommendations

If the site has adequate space, installing a filter strip that would retain 80 percent of the 80 um fine-sand particles during the design storm is recommended. This method will keep the bioretention or bioinfiltration practice from filling up with fine and coarse sand particles too quickly. Both the aggressive treatment and limited space methods are acceptable if the designer is aware of the limitations of the limited space method (i.e., increased frequency of maintenance on the downstream bioretention/bioinfiltration practice). It should be noted that, even though a filter strip will retain more particles per surface area than a basin, filter strips typically require a greater level of effort to clean the pretreatment practice. In deciding between a grass filter strip or a rock filter strip, the cost of pulling and washing rocks versus lowering the level of a grass filter strip to clean out a basin should be compared before a decision is made.

## Literature review of pretreatment sizing practices

Particle Size Distribution of Suspended Solids in Road Stormwater Runoff [Erickson, 2012]. Click on image to enlarge.

We conducted a limited review of pretreatment sizing methods used by other entities.

• New Jersey. Pretreatment is required on all subsurface infiltration basins and must remove 80% of the total suspended solids (TSS) in the runoff from the Water Quality Design Storm. A concern with this approach is the variability in particle size distributions. As the adjacent figure demonstrates, the particle size corresponding with 80 percent of particles ranges from about 2 to nearly 200 microns.
• Massachusetts. Design specifications are provided for a range of devices and structural practices. When used for pretreatment, these devices and practices are assumed to remove 25 percent of TSS.
• Environment Global Stormwater Solutions. Pretreatment is based on three minutes of detention (treatment) time. Sizing therefore focuses on achieving this detention time.
• City of Inver Grove Heights. A flow chart is used to determine the level of pretreatment required. If stormwater runs over an approved filter strip or swale (based on design specifications), no pretreatment is required. Otherwise, pretreatment should remove 85 percent of TSS.
• Vermont. If stormwater is routed to forebay for required pre-treatment, forebay volumetrically sized for 10% of the computed water quality volume. Proprietary devices should be designed to remove 50 percent of TSS. Design specification are provided that are expected to meet these criteria.
• Delaware. Different pretreatment criteria are provided for different post-construction best management practices, ranging from no pretreatment for permeable pavement to 50 percent TSS removal for some infiltration practices.
• New York state. Sizing is expressed in terms of the percentage of the water quality volume that must be pretreated. For example, forebays must treat 10 percent of the and infiltration practices must receive pretreatment for 25 percent of the water quality volume.

## References

• Chow, V.T., D.R. Maidment, and L.W. Mays, 1988. Applied Hydrology, McGraw-Hill, New York, NY.
• Erickson, A.J. 2012. Advanced Stormwater Treatment: Dissolved Pollutants.
• Ferguson, R.I. and Church, M., 2004. “A Simple Universal Equation For Grain Settling Velocity,” Journal Of Sedimentary Research. Vol 74, No. 6. pg 933-937.
• Gulliver, J.S. 2007. Introduction to Chemical Transport in the Environment. Cambridge University Press, Cambridge, England.
• Merkel, W., and H.F. Moody. 2015. NOAA Atlas 14 Rainfall for Midwest and Southeast States, prepared by the Natural Resources Conservation Service, Washington, D.C.

## Related pages

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text:
NZ Level 8 (NZC) Level 3 (NCEA) [In development]
Mixed problem

## Interactive practice questions

If you can choose one item from a group of $a$a items and a second item from a group of $b$b items, what is the total number of two-item choices?

Easy
Approx a minute

The number of possible permutations of $r$r objects taken from $n$n items is $\nPr{n}{r}$nPr. Express this in factorial form.

The number of possible combinations of $r$r objects taken from $n$n items is $\nCr{n}{r}$nCr. Express this in factorial form.

Evaluate $\frac{\nPr{8}{3}}{3!}-\nCr{8}{3}$8P33!8C3.

### Outcomes

#### M8-3

Use permutations and combinations

format: plain


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Grade 11 - Physics - Gravitation - Concept of Gravitation

Acceleration due to gravity

The acceleration with which a body falls towards the earth is known as acceleration due to gravity.
The acceleration due to gravity g near the surface of the earth is .

Case 1 :

Let the body of mass m be at a height h from the surface of the earth.If the radius of the earth is R, then as its distance r from the center of the earth is r=(R+h), the downward force on the body is given by,

Where M is the mass of the earth. If the body happens to be very close to the surface, h=0, and the gravitational force on it is-

The acceleration generated due to this gravitational force is, therefore, given by-

The above relation is true for all planets and other heavenly bodies.

(i) Mass of the earth

It is possible to calculate the mass of the earth using the above relation. As radius of the earth ${}^{}$m and, putting weget,

Acceleration due to gravity
(ii) Mean density of the earth
Suppose the mean density of the earth is ,then as

Acceleration due to gravity
Above the surface
The gravitational force acting on a body of mass m when at a height h from the surface of the earth is ,

hence, the body is subjected to a downward acceleration given by,

As ,therefore Where r=(R+h)

NOTE: The above expression shows that the value of acceleration due to gravity decreases with height. It falls inversely with the square of the distance provided the distances are measured from the center of the earth.

Acceleration due to gravity
When (h is negligible to R)
In that condition the relation can be further simplified by using binomial theorem. Accordingly,


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# Efficiently determine if convex hull contains the unit ball

Given a set of $n$ points in $\mathbb{R}^d$, is there an algorithm to determine if the convex hull contains the unit ball centered at the origin in polynomial time? The convex hull itself might have an exponential number of facets so we cannot afford explicitly to compute it.

My main interest is not in computer precision so we can make whatever assumptions help avoid that in relation to the points themselves (for example they only have integer coordinates).

• Polynomial time in what? Are you fixing $d$ and varying $n,$ or are you varying both? – Igor Rivin Sep 10 '13 at 16:38
• @IgorRivin I should have said. Polynomial in both $n$ and $d$. You can, for example, determine if an arbitrary fixed point is in the convex hull in polynomial time by a reasonably straightforward application of linear programming. – octonots Sep 10 '13 at 16:45
• how is the set of points given? They could be either in random order or, be sorted in some way (e.g. lexicographical). Further, you should define what is meant by "the unit ball": is it centered at the origin or did you mean "a unit ball", which could be centered anywhere? – Manfred Weis Sep 11 '13 at 5:26
• In your answer to Igor Rivin you mentioned linear programming for checking whether a point is inside the convex hull; in linear programming the points of the convex hull are not explicitly given, but rather implicitly via the intersection of half-spaces. Maybe you should rethink, how your convex hull is given, via points or via the intersection of half-spaces. In the latter case the problem can be solved efficiently via linear programming – Manfred Weis Sep 11 '13 at 5:49
• Yet another question: are the $n$ points in convex configuration or can some of the points be inside the convex hull? Is it assumed that the convex hull is $d$-dimensional? – Manfred Weis Sep 11 '13 at 6:13

The problem is NP hard. Here is a proof sketch.

The problem is to determine if there is a point $y$ with $\|y\|=1$ outside of the convex hull of given points $x_1,\dots, x_n$. Note that such point exists if and only if there is hyperplane at distance less than $1$ from the origin such that all points $x_1, \dots, x_n$ and $0$ lie on one side of the hyperplane (consider a hyperplane $\pi$ separating $x_1,\dots, x_n, 0$ and $y$).

So the problem can be stated as follows: is there $a\in {\mathbb R}^d$ s.t.

1. $\langle a, x_i\rangle \leq 1$ (that is, all $x_i$ lie in the half-space $\{x: \langle a, x\rangle\leq 1\}$);
2. $\|a\| > 1$.

Note that this problem is equivalent to the following well-known problem:

We are given a convex polytope $\cal P$, a positive definite matrix $A$ and a number $t$, find $x\in \cal P$ such that $x^T A x > t$. ($\cal P$ is described by a system of linear equations).

The optimization version of this problem is:

Quadratic Programming (Non-convex Linearly Constrained Quadratic Programming with Positive Definite Matrix). We are given a convex polytope $\cal P$ and and a positive definite matrix $A$, find $x\in \cal P$ that maximizes $x^T A x$.

This problem is known to be NP hard.

It is even NP-hard to optimize $x^T A x$ when $\cal P$ is the unit cube $\{(b_1, \dots, b_d): -1\leq b_i \leq 1\}$ (this problem is known as Integer Quadratic Programming with Positive Definite Matrix). In particular, the MAX CUT problem is a special case of this problem. Let $G$ be a graph on $n$ vertices and $L$ be its Laplacian. Then $\max_{x\in\{\pm 1\}^n} x^T L x$ is equal to the size of the maximum cut in $G$ ($L$ is positive semi-definite, not positive definite, but this difference is not important; e.g. we can consider $L'=L+\varepsilon I$ with very small $\varepsilon$). The NP-hardness of MAX CUT was proved by Karp in

Richard M. Karp (1972). Reducibility Among Combinatorial Problems. In R. E. Miller and J. W. Thatcher (editors). Complexity of Computer Computations. New York: Plenum. pp. 85–103.

• Thank you. Does this well-known problem have a standard name so I can look it and its NP-hardness proof up? – octonots Sep 12 '13 at 10:17
• I added some references. – Yury Sep 12 '13 at 12:10

Under the assumption, that $n$ points in random order are given, the best algorithm seems to be to construct the generalization of the Delaunay Triangulation to $d$-dimensional Euclidean space; that yields a collection of empty hyper-balls that are defined via $d+1$ of the points; the number of those hyper-balls is $O(n^{\lceil d/2 \rceil})$.
The bound on the number of empty hyper-balls proves that the convex hull of $n$ points can't have an exponential number of faces like e.g. $O(2^n)$.

From that collection of hyperballs the ones, whose center is outside the convex hull of their defining $k>=d+1$ points, are not inside the convex hull of the $n$ points and are not considered further.

Then one has to check, whether the radius of any of the remaining hyper-balls is at least $1$.

From the efficient construction of the Delaunay Triangulation in higher dimensions, it follows, that the answer to the question is yes.

The situation doesn't change, if only the points on the convex hull shall be taken into account; this is so, because the points on the convex hull can also be efficiently determined via a generalized gift-wrapping algorithm and then the generalized Delaunay Triangulation can be constructed for those points and finally the largest empty hyper-ball can be determined as described before.

If the unit ball is centered at the origin, then one has to check, whether distance of the the faces of the convex hull that were reported by the gift-wrapping algorithm, is not less than $1$.

• Thank you. I consider $n^{\lceil d/2 \rceil}$ to be exponential in $d$ however. My question is whether there is an algorithm that is polynomial in both $n$ and $d$. – octonots Sep 11 '13 at 9:56
• @octonots as the construction of the convex hull is exponential in $d$, the answer to your question is no, because if the unit sphere is inside the convex hull, it can come arbitrarily close to a face, while the distance to the vertices may stay above some fixed value; so one has to construct the hyperplanes through the faces of the convex hull in order to calculate their distance from the origin. – Manfred Weis Sep 11 '13 at 10:08
• I suppose an interesting subquestion is whether it is even in NP. However I am not sure one can rule out a polynomial time algorithm so easily. – octonots Sep 11 '13 at 10:44
• A first step to prove that it could be polynomial, would be to prove that it is not necessary to check $O(n^{\lceil d/2 \rceil})$ faces of the convex hull; I doubt that that is possible. Suppose you have a convex hull that contains the unit ball; intersect that with a halfspace that contains the origin and intersects the unit ball hoping to be able to detect efficiently from the resulting set of points the ones that define the hyperplane that cuts the unit-ball is very optimistic. – Manfred Weis Sep 11 '13 at 11:22


text:
## Formal charge on central atom

$FC=V-(L+\frac{S}{2})$

marcus lin 1E
Posts: 54
Joined: Fri Sep 28, 2018 12:28 am

### Formal charge on central atom

Why does the formal charge on the central atom have to be as close to zero as possible?

Diego Gonzalez 3F
Posts: 32
Joined: Fri Sep 28, 2018 12:26 am

### Re: Formal charge on central atom

the closer to zero the charge is, the more stable the entire compound will be. Since its the central atom its especially important since its bonding with every other atom in the compound.

Angela Cong 3C
Posts: 69
Joined: Fri Sep 28, 2018 12:25 am

### Re: Formal charge on central atom

additionally, the central atom is alway the least electronegative so the surrounding atoms bonded would much rather have the more negative charge

Jeremy Feng 1A
Posts: 30
Joined: Fri Sep 28, 2018 12:18 am

### Re: Formal charge on central atom

In nature, multiple structures exist, as do resonance structures. To my understanding, by drawing the most stable structure, we are illustrating the patterns most likely to exist at any given time.

Jayde Felix 4H
Posts: 59
Joined: Fri Sep 28, 2018 12:28 am

### Re: Formal charge on central atom

Closer to zero means more stable than other structures with different charges.

uhedlund
Posts: 38
Joined: Wed Nov 15, 2017 3:03 am

### Re: Formal charge on central atom

A formal charge closer to zero is more stable than other higher charges.

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text:
VGAM (version 1.0-4)

# gamma2: 2-parameter Gamma Distribution

## Description

Estimates the 2-parameter gamma distribution by maximum likelihood estimation.

## Usage

gamma2(lmu = "loge", lshape = "loge",
imethod = 1,  ishape = NULL,
parallel = FALSE, deviance.arg = FALSE, zero = "shape")

## Arguments

lmu, lshape

Link functions applied to the (positive) mu and shape parameters (called $$\mu$$ and $$a$$ respectively). See Links for more choices.

ishape

Optional initial value for shape. A NULL means a value is computed internally. If a failure to converge occurs, try using this argument. This argument is ignored if used within cqo; see the iShape argument of qrrvglm.control instead.

imethod

An integer with value 1 or 2 which specifies the initialization method for the $$\mu$$ parameter. If failure to converge occurs try another value (and/or specify a value for ishape).

deviance.arg

Logical. If TRUE, the deviance function is attached to the object. Under ordinary circumstances, it should be left alone because it really assumes the shape parameter is at the maximum likelihood estimate. Consequently, one cannot use that criterion to minimize within the IRLS algorithm. It should be set TRUE only when used with cqo under the fast algorithm.

zero

See CommonVGAMffArguments for information.

parallel

Details at CommonVGAMffArguments. If parallel = TRUE then the constraint is not applied to the intercept.

## Value

An object of class "vglmff" (see vglmff-class). The object is used by modelling functions such as vglm and vgam.

## Details

This distribution can model continuous skewed responses. The density function is given by $$f(y;\mu,a) = \frac{\exp(-a y / \mu) \times (a y / \mu)^{a-1} \times a}{ \mu \times \Gamma(a)}$$ for $$\mu > 0$$, $$a > 0$$ and $$y > 0$$. Here, $$\Gamma(\cdot)$$ is the gamma function, as in gamma. The mean of Y is $$\mu=\mu$$ (returned as the fitted values) with variance $$\sigma^2 = \mu^2 / a$$. If $$0<a<1$$ then the density has a pole at the origin and decreases monotonically as $$y$$ increases. If $$a=1$$ then this corresponds to the exponential distribution. If $$a>1$$ then the density is zero at the origin and is unimodal with mode at $$y = \mu - \mu / a$$; this can be achieved with lshape="loglog".

By default, the two linear/additive predictors are $$\eta_1=\log(\mu)$$ and $$\eta_2=\log(a)$$. This family function implements Fisher scoring and the working weight matrices are diagonal.

This VGAM family function handles multivariate responses, so that a matrix can be used as the response. The number of columns is the number of species, say, and zero=-2 means that all species have a shape parameter equalling a (different) intercept only.

## References

The parameterization of this VGAM family function is the 2-parameter gamma distribution described in the monograph

McCullagh, P. and Nelder, J. A. (1989) Generalized Linear Models, 2nd ed. London: Chapman & Hall.

gamma1 for the 1-parameter gamma distribution, gammaR for another parameterization of the 2-parameter gamma distribution that is directly matched with rgamma, bigamma.mckay for a bivariate gamma distribution, expexpff, GammaDist, golf, CommonVGAMffArguments, simulate.vlm, negloge.

## Examples

Run this code
# NOT RUN {
# Essentially a 1-parameter gamma
gdata <- data.frame(y = rgamma(n = 100, shape = exp(1)))
fit1 <- vglm(y ~ 1, gamma1, data = gdata)
fit2 <- vglm(y ~ 1, gamma2, data = gdata, trace = TRUE, crit = "coef")
coef(fit2, matrix = TRUE)
c(Coef(fit2), colMeans(gdata))

# Essentially a 2-parameter gamma
gdata <- data.frame(y = rgamma(n = 500, rate = exp(-1), shape = exp(2)))
fit2 <- vglm(y ~ 1, gamma2, data = gdata, trace = TRUE, crit = "coef")
coef(fit2, matrix = TRUE)
c(Coef(fit2), colMeans(gdata))
summary(fit2)
# }

Run the code above in your browser using DataCamp Workspace


text:
## Algebra: A Combined Approach (4th Edition)

When $t=2$, $528.1$ feet. When $t=4$, $336.1$ feet.
$-16t^2 + 592.1$ $=-16(2)^2 + 592.1$ $=-16(4)+592.1$ $=-64+592.1$ $=528.1$ feet $-16t^2 + 592.1$ $=-16(4)^2 + 592.1$ $=-16(16)+592.1$ $=-256+592.1$ $=336.1$ feet


text:
## Notre Dame Journal of Formal Logic

### Russell, Presupposition, and the Vicious-Circle Principle

Darryl Jung

#### Abstract

Prompted by Poincaré, Russell put forward his celebrated vicious-circle principle (vcp) as the solution to the modern paradoxes. Ramsey, Gödel, and Quine, among others, have raised two salient objections against Russell's vcp. First, Gödel has claimed that Russell's various renderings of the vcp really express distinct principles and thus, distinct solutions to the paradoxes, a claim that gainsays one of Russell's positions on the nature of the solution to the paradoxes, namely, that such a solution be uniform. Secondly, Ramsey, Gödel, and Quine have protested that the vcp's proscription against impredicative specification is incompatible with a realistic conception of the domain of quantification, a conception that Russell certainly held. I examine Russell's vcp and defend it against these objections.

#### Article information

Source
Notre Dame J. Formal Logic Volume 40, Number 1 (1999), 55-80.

Dates
First available in Project Euclid: 5 December 2002

http://projecteuclid.org/euclid.ndjfl/1039096305

Digital Object Identifier
doi:10.1305/ndjfl/1039096305

Mathematical Reviews number (MathSciNet)
MR1811203

Zentralblatt MATH identifier
0968.03012

#### Citation

Jung, Darryl. Russell, Presupposition, and the Vicious-Circle Principle. Notre Dame J. Formal Logic 40 (1999), no. 1, 55--80. doi:10.1305/ndjfl/1039096305. http://projecteuclid.org/euclid.ndjfl/1039096305.

#### References

• [1] Boolos, G., To be is to be the value of a variable (or some values of some variables),'' Journal of Philosophy, vol. 81 (1984), pp. 430--50.
• [2] Boolos, G., Nominalist Platonism,'' Philosophical Review, vol. 94 (1985), pp. 327--44.
• [3] Carnap, R., The Logical Syntax of Language, translated by A. Smeaton, Harcourt Brace, London, 1937.
• [4] Chihara, C., Ontology and the Vicious-Circle Principle, Cornell University Press, Ithaca, 1973.
• [5] Copi, I., The Theory of Logical Types, Routledge and Kegan Paul, London, 1971.
• [6] de Rouilhan, P., Russell et le cercle des paradoxes, Presses Universitaires de France, Paris, 1996.
• [7] Fraenkel, A., Y. Bar-Hillel, and A. Levy, Foundations of Set Theory, North-Holland, Amsterdam, 1984.
• [8] Gödel, K., Über formal unentscheidbare Sätze der Principia Mathematica und verwandter Systeme, 1,'' Monatshefte für Mathematik und Physik, vol. 38 (1931), pp. 173--98. Translated as On formally undecidable propositions of Principia Mathematica and related systems, 1,'' pp. 145--205 in Kurt Gödel: Collected Works, vol. 1, edited by S. Feferman et al., Oxford University Press, Oxford, 1986, and pp. 596--616 in J. van Heijenoort, From Frege to Gödel, Harvard University Press, Cambridge, 1967.
• [9] Gödel, K., Russell's mathematical logic,'' pp. 119--41 in Kurt Gödel: Collected Works, vol. 2., edited by S. Feferman et al., Oxford University Press, Oxford, 1990.
• [10] Goldfarb, W., Russell's reasons for ramification,'' pp. 24--40 in Rereading Russell: Essays on Bertrand Russell's Metaphysics and Epistemology, Minnesota Studies in the Philosophy of Science, edited by C.W. Savage and C. A. Anderson, University of Minnesota Press, Minneapolis, 1989.
• [11] Hallett, M., Cantorian Set Theory and Limitation of Size, Clarendon Press, Oxford, 1984.
• [12] Hylton, P., Russell, Idealism, and the Emergence of Analytic Philosophy, Oxford University Press, Oxford, 1990.
• [13] Hylton, P., The Vicious-Circle Principle: Comments on Philippe de Rouilhan,'' Philosophical Studies, vol. 65 (1992), pp. 183--91.
• [14] Kant, I., The Critique of Pure Reason, translated by N. K. Smith, MacMillan Press, London, 1982.
• [15] Kremer, M., The argument of On Denoting','' The Philosophical Review, vol. 103 (1994), pp. 249--97.
• [16] Landini, G., Russell's substitutional theory of classes and relations,'' History and Philosophy of Logic, vol. 8 (1987), pp. 171--200.
• [17] Landini, G., New evidence concerning Russell's substitutional theory of classes,'' Russell, n.s., vol. 9 (1989), pp. 26--42.
• [18] Landini, G., Reconciling PM's ramified type theory with the doctrine of the unrestricted variable of the Principles,'' pp. 361--94 in Russell and Analytic Philosophy, edited by A. D. Irvine and G. A. Wedeking, University of Toronto Press, Toronto, 1993.
• [19] Levy, A., Basic Set Theory, Springer-Verlag, Berlin, 1979.
• [20] Linsky, B., Propositional functions and universals in Principia Mathematica,'' Australasian Journal of Philosophy, vol. 66, (1988), pp. 447--60.
• [21] Linsky, B., Why Russell abandoned Russellian propositions,'' pp. 193--209 in Russell and Analytic Philosophy, edited by A. D. Irvine and G. A. Wedeking, University of Toronto Press, Toronto, 1993.
• [22] Linsky, B., Russell's Metaphysical Logic, CSLI Publications, Stanford, 1999.
• [23] Pelham, J., and A. Urquhart, Russellian propositions,'' pp. 307--26 in Logic, Methodology, and Philosophy of Science, vol. 9, edited by D. Prawitz, B. Skyrms, and D. Westerstå hl, Elsevier Science, Amsterdam, 1994.
• [24] Poincaré, H., Les mathématiques et la logique,'' Revue de Métaphysique et de Morale, vol. 13 (1905), pp. 815--35; vol. 14 (1906), pp. 17--34 and 294--317.
• [25] Poincaré, H., The latest efforts of the logicians,'' pp. 472--85 in The Foundations of Science: Science and Hypothesis, the Value of Science, Science and Method, translated by G. B. Halstead, The Science Press, New York, 1929.
• [26] Quine, W. V., New foundations for mathematical logic,'' American Mathematical Monthly, vol. 44 (1937), pp. 70--80.
• [27] Quine, W. V., Set Theory and Its Logic, Harvard University Press, Cambridge, 1969.
• [28] Ramsey, F. P., Foundations of mathematics,'' pp. 152--212 in Foundations, edited by D. H. Mellor, Humanities Press, Atlantic Highlands, 1978.
• [29] Russell, B., Paper 1'' (1899--1900 unpublished draft of Principles), pp. 13--180 in The Collected Papers of Bertrand Russell: Toward the Principles of Mathematics'' 1900--02, vol. 3, edited by G. H. Moore, Routledge, London, 1993.
• [30] Russell, B., Paper 2'' (1901 unpublished draft of Principles), pp. 184--208 in The Collected Papers of Bertrand Russell: Toward the Principles of Mathematics'', 1900--02, vol. 3, edited by G. H. Moore, Routledge, London, 1993.
• [31] Russell, B., The Principles of Mathematics (Principles), 2d edition, Norton, New York, 1938.
• [32] Russell, B., On denoting,'' pp. 103--19 in Essays in Analysis, edited by D. Lackey, George Braziller, New York, 1973.
• [33] Russell, B., On some difficulties in the theory of transfinite numbers and order types,'' pp. 135--64 in Essays in Analysis, edited by D. Lackey, George Braziller, New York, 1973.
• [34] Russell, B., Les paradoxes de la logique,'' translated as On Insolubilia' and their solution in symbolic logic'' pp. 190--214 in Essays in Analysis, edited by D. Lackey, George Braziller, New York, 1973.
• [35] Russell, B., Mathematical logic as based on the theory of types,'' pp. 59--102 in Logic and Knowledge, edited by R. C. Marsh, George Allen and Unwin, London, 1956.
• [36] Tarski, A., The concept of truth in formalized languages'' (Wahrheitsbegriff''), pp. 152--278 in Logic, Semantics, and Metamathematics, Hackett, Indianapolis, 1983.
• [37] Whitehead, A. N., and B. Russell, Principia Mathematica (Principia), 2d edition, vol. 1, Cambridge University Press, Cambridge, 1927.


text:
# Time dilation ll

by cos
Tags: dilation, time
P: 645
 Quote by jtbell So, does this thread basically boil down to a disagreement over the meanings of words like "physical" and "reality"? If so, then this properly belongs in the philosophy forum, IMO.
I think cos is moving towards something like this:

"if the slowing down of clocks is supposed to real or physical, then something is amiss, and if results (like the twin paradox) suggest that only one clock ran slow, then the mathematics should be suspected, rather than the reality or physical results"

cos,

I am not saying categorically that that is your contention, it is merely what I think you are saying.

However, I should point out that you have to take into account everything in a situation like the twin's paradox. Specifically, you need to consider simultaneity (I say this even though this is not my favoured approach). Both twins will calculate that the other twin's clock ran slow during the inertial phases. What the twins will not agree on is how long those inertial phases lasted.

So the results will be (time on a slow clock running for a longer time) and (time on a slow clock running for a shorter time). The change in direction was an event that was colocal with the acelerated twin, but not colocal with the twin who was inertial throughout, which means that - taking into account simultaneity - the stationary twin will calculate that the travelling twin turned around later than travelling twin calculated. This is totally in agreement with the fact that the travelling twin's clock ran slow compared to the stationary twin.

So if you like, both clocks really ran slow, the travelling twin really turned around later than was said on his clock and at the end of the journey one clock will really show more time elapsed than shown on the other - and both clocks are showing the real time elapsed for that clock.

cheers,

neopolitan
P: 212
 Quote by neopolitan cos, It appears that you have had some misfortune, but it is not relevant to the discussion.
It is relevant to the discussion inthat, in the discussion you made what I considered to be snide and belittling comments which were themselves not relevant to the discussion.

 "I am of the opinion that Einstein obviously did not reject maths but that he insisted that it does not refer to reality."
 Quote by neopolitan This is central. You are taking one quote and misrepresenting it terribly. You should try to take Einstein's comments in context. In 1917 he found that his equations showed that the universe is expanding. He then spent quite a few years trying to fit a cosmological constant in order to make the universe static. In other words, the mathematics were telling him that the universe is expanding and "reality" was telling him it isn't. Edwin Hubble came to the rescue with observations which showed that the universe is in fact expanding and Einstein's maths were correct, not his perception of what must be real.
It is my understanding that Hubble argued that the greater redshift of the more distant galaxies was not an indication that the universe is expanding thus he would not have accepted the veracity of Einstein's calculations that it is!

 Quote by neopolitan So, I put it to you again, try the maths. The maths worked for Einstein......
Einstein's maths indicated the amount by which, in his opinion, clock A lags behind B and, as Einstein pointed out, because A lags behind B it must have 'gone more slowly' (i.e. ticked over at a slower rate) than B whilst A was moving.

According to Einstein's section 4 maths, clock A is ticking over at a slower rate than clock B but according to the maths employed by the observer accompanying clock A it is B that is ticking over at a slower rate than his clock.

Having calculated that B 'is' ticking over at a slower rate than his own clock, observer A 'determines' or 'predicts' that when he arrives at B's location he will find that it lags behind his clock yet he learns that it does NOT!

HE has 'done the math' yet finds that it gave an erroneous answer. Of what value his math?

You are, I believe, confusing Einstein's depiction of a non-inertial observer with his previously depicted inertial observer.

 Quote by neopolitan You would probably be better off if you search for one of his papers or essays written after 1929.
I assume that you have read his papers and essays written after 1929 and to save time not only for myself but also for others that may be following this thread perhaps you would be so kind as to nominate just one of those papers showing that he recanted or amended his section 4 STR comments.

Albert Einstein's 1905 article 'On the Electrodynamics of Moving Bodies' is said to be the foundation of modern-day physics and in that article he indicates that inertial observers that are moving relative to each other will both determine that the other person's clock will be running slower than their own however in section 4 of that same article he shows that an observer who has accelerated will not find that the other clock is ticking over at a slower rate than their own clock but at a faster rate!
Mentor
P: 16,951
 Quote by cos According to Einstein's section 4 maths, clock A is ticking over at a slower rate than clock B but according to the maths employed by the observer accompanying clock A it is B that is ticking over at a slower rate than his clock. Having calculated that B 'is' ticking over at a slower rate than his own clock, observer A 'determines' or 'predicts' that when he arrives at B's location he will find that it lags behind his clock yet he learns that it does NOT! HE has 'done the math' yet finds that it gave an erroneous answer. Of what value his math?
His math is fine and does not give an erroneous answer, you just made a mistake. Clock A is non-inertial, so the acompanying observer correctly determines that the other clock "physically" runs faster overall. In fact, all reference frames (including both inertial and non-inertial frames) will agree that clock B runs faster on average.
 P: 1,060 Hello cos. Probably of no consequence but in an early traslation of about 1920 part of your quoted text reads:- -----From this, we conclude that a clock placed at the equator must be slower by a very small amount than a similarly constructed clock which is placed at the pole, all other conditions being identical.----- The words "be slower" are used rather than "gone more slowly". The words "balance clock" do not appear. This may of course be due to the transators M.N. Saha and S.N. Bose. I am not drawing any conclusions from this, it is just a point of interest. Mateinste
P: 212
 Quote by DaleSpam His math is fine and does not give an erroneous answer, you just made a mistake. Clock A is non-inertial, so the acompanying observer correctly determines that the other clock "physically" runs faster overall. In fact, all reference frames (including both inertial and non-inertial frames) will agree that clock B runs faster on average.
On the assumption that "..the accompanying observer correctly determines that the other clock "physically" runs faster overall." and on the assumption that he is an enquiring scientist is it not possible that he might ask himself what force has made clock B PHYSICALLY run faster than it did before he started moving?

He is an astronaut returning to the planet following turn-around; he 'sees' Earth clocks ticking over at a faster rate than they did before he started accelerating hence he 'sees' shorter Earth seconds than he did before he started moving thus he must also 'see' Earth minutes, hours and days to also 'be' ticking over at a faster rate than they were before he started moving ergo he 'sees' the Earth spinning on its axis and orbiting the sun at a considerably faster rate than it did before he started moving.

Is he not likely to ask himself what physical force has made the planet spin faster on its axis and orbit the sun at a much faster rate than it did before he started his return journey?

Although he 'determines' that the Earth clock 'is' running faster than it did before he started moving and that the planet 'is' spinning faster on its axis than they did before he started accelerating is he not likely to realize that this is nothing more than a (mathematically generated) illusion created by his non-inertial motion?

In section 4 STR Einstein pointed out, in effect, that clock A, having moved to clock B's location will lag behind clock B due to the fact that clock A 'goes more slowly' (i.e. ticks over at a slower rate) than clock B not that clock B would leap ahead of clock A thus that clock B would incur time contraction which I believe was for Einstein an anathema.

You wrote "... all reference frames (including both inertial and non-inertial frames) will agree that clock B runs faster on average." and i agree with that comment; clock B does 'run faster' than A due to the fact that, as Einstein pointed out, clock A runs slower than B however for them to be of the opinion that clock B runs faster than it did before A started moving is erroneous - according to Einstein's section 4 STR.

According to Einstein - clock A ticks over at a slower rate than it did before it started moving NOT that B starts ticking over at a faster rate. Clock A accelerates and it is, according to Einstein, this factor that physically causes it to tick over at the slower rate. There is no force, no action on it's behalf which causes clock B to physically tick over at a faster rate than it did before A started moving.
P: 212
 Quote by jtbell So, does this thread basically boil down to a disagreement over the meanings of words like "physical" and "reality"? If so, then this properly belongs in the philosophy forum, IMO.
It is not simply a disagreement over the words "physical" and "reality" but a disagreement over those words as they apply to Einsteins section 4 STR depictions.
P: 212
 Quote by matheinste Hello cos. Probably of no consequence but in an early traslation of about 1920 part of your quoted text reads:- -----From this, we conclude that a clock placed at the equator must be slower by a very small amount than a similarly constructed clock which is placed at the pole, all other conditions being identical.----- The words "be slower" are used rather than "gone more slowly". The words "balance clock" do not appear. This may of course be due to the transators M.N. Saha and S.N. Bose. I am not drawing any conclusions from this, it is just a point of interest. Mateinste
My reference was specifically in relation to Einstein's 1905 paper not any other translation.

In section 4 he wrote -

"Thence we conclude that a balance-clock at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions."
P: 645
cos,

I assume that you are refering to the section titled "Physical Meaning of the Equations Obtained in Respect to Moving Rigid Bodies and Moving Clocks" since that section ends with:

 Thence we conclude that a balance-clock at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions.
Shortly before, in that same section:

 From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by 1/2 tv2/c2 (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B.
Now it should be remembered that this is Einstein. If I wrote something similar on this forum, I would probably be appropriatedly chastised for an inaccuracy - perhaps not for being incorrect, but for not clearly stating something.

The time t is the time occupied in the journey from A to B, in the frame of the clock which remains stationary. According to the stationary clock, the clock which was moved travelled a distance of x = v.t.

I'd say that the time t' on the moving clock at the end of that journey would be:

$$\gamma ( t - x.v / c^{2} ) = \gamma ( t - t.v^{2} / c^{2} ) = \gamma t (1 - v^{2} / c^{2} ) = t / \gamma$$

The difference is therefore $$\Delta = t - t' = t ( 1 - 1 / \gamma) \approx 1/2t . v^{2} / c^{2}$$.

When that moving clock stops, there are a number of ticks from the "stationary" clock still travelling to catch up, x/(c-v) = vt/(c-v) worth. Those ticks in transit will, when added to the ticks already received, show that the more time has elapsed has elapsed on the stationary clock, even though that time elapsed at a slower rate.

So Einstein's answer, while possibly not immediately intuitive, is not erroneous.

cheers,

neopolitan
P: 645
 Quote by cos My reference was specifically in relation to Einstein's 1905 paper not any other translation. In section 4 he wrote - "Thence we conclude that a balance-clock at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions."
I have to admit that I laughed out aloud at this.

Einstein's paper was written in German. The original text can be seen here.

The 1920 translation is of the same paper. A translation today (using more modern language) would be a translation of the same paper.

cheers,

neopolitan
Mentor
P: 16,951
 Quote by cos On the assumption that "..the accompanying observer correctly determines that the other clock "physically" runs faster overall." and on the assumption that he is an enquiring scientist is it not possible that he might ask himself what force has made clock B PHYSICALLY run faster than it did before he started moving? He is an astronaut returning to the planet following turn-around; he 'sees' Earth clocks ticking over at a faster rate than they did before he started accelerating hence he 'sees' shorter Earth seconds than he did before he started moving thus he must also 'see' Earth minutes, hours and days to also 'be' ticking over at a faster rate than they were before he started moving ergo he 'sees' the Earth spinning on its axis and orbiting the sun at a considerably faster rate than it did before he started moving. Is he not likely to ask himself what physical force has made the planet spin faster on its axis and orbit the sun at a much faster rate than it did before he started his return journey?
Certainly. And the answer would be the fictitious forces present in his non-inertial frame.

I don't mind how you use the word "physical", but you need to be self-consistent. You cannot claim that frame-variant quantities, like the rate of a clock, are "physical" and then exclude the frame-variant fictitious forces from being "physical" also. In a non-inertial reference frame fictitious forces can do work, can have potential energy, can cause mechanical stress and strain, and have many other measurable effects.

 Quote by cos Although he 'determines' that the Earth clock 'is' running faster than it did before he started moving and that the planet 'is' spinning faster on its axis than they did before he started accelerating is he not likely to realize that this is nothing more than a (mathematically generated) illusion created by his non-inertial motion?
Yes, that is why they are called fictitious forces. That is also why he is not likely to try to do the analysis in his non-inertial rest frame, but is more likely to do the analysis in some inertial frame.

 Quote by cos You wrote "... all reference frames (including both inertial and non-inertial frames) will agree that clock B runs faster on average." and i agree with that comment; clock B does 'run faster' than A due to the fact that, as Einstein pointed out, clock A runs slower than B however for them to be of the opinion that clock B runs faster than it did before A started moving is erroneous - according to Einstein's section 4 STR. According to Einstein - clock A ticks over at a slower rate than it did before it started moving NOT that B starts ticking over at a faster rate. Clock A accelerates and it is, according to Einstein, this factor that physically causes it to tick over at the slower rate. There is no force, no action on it's behalf which causes clock B to physically tick over at a faster rate than it did before A started moving.
Your explanation is correct only in the inertial reference frame where B is at rest. In other reference frames there will be other explanations. But all reference frames will agree on the conclusion.
P: 212
 Quote by neopolitan cos, I assume that you are refering to the section titled "Physical Meaning of the Equations Obtained in Respect to Moving Rigid Bodies and Moving Clocks" since that section ends with: "Thence we conclude that a balance-clock at the equator must go more slowly, by a very small amount, than a precisely similar clock situated at one of the poles under otherwise identical conditions."
On the basis that this precisely the comment to which I refer I fail to see how you could possibly be of the opinion that I may not have been '...referring to the section titled "Physical Meaning of the Equations Obtained in Respect to Moving Rigid Bodies and Moving Clocks."'

It would be very much appreciated if we could adhere to relevant matters.

 Quote by neopolitan Shortly before, in that same section: "From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by 1/2 tv2/c2 (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B." Now it should be remembered that this is Einstein. If I wrote something similar on this forum, I would probably be appropriatedly chastised for an inaccuracy - perhaps not for being incorrect, but for not clearly stating something.
You may not have noticed but Einstein's comments have been criticized in this forum! Einstein has, somewhat belatedly, been chastised!

 Quote by neopolitan The time t is the time occupied in the journey from A to B, in the frame of the clock which remains stationary. According to the stationary clock, the clock which was moved travelled a distance of x = v.t.
And according to the traveler the distance he travels is less than the distance from A to B as measured by the stationary observer.

 Quote by neopolitan When that moving clock stops, there are a number of ticks from the "stationary" clock still travelling to catch up, x/(c-v) = vt/(c-v) worth. Those ticks in transit will, when added to the ticks already received, show that the more time has elapsed has elapsed on the stationary clock, even though that time elapsed at a slower rate.
What on Earth does "...there are a number of ticks from the 'stationary' clock still travelling to catch up." mean?

When the moving clock stops, A and B could press switches whereupon both clocks stop ticking so any 'ticks in transit' would be out of luck. Their time will have expired and they can no affect on the times registered by clocks A and B.

So Einstein's answer, while possibly not immediately intuitive, is not erroneous.

P: 212
 Quote by DaleSpam Certainly. And the answer would be the fictitious forces present in his non-inertial frame.
My references are to reality and I am of the opinion that fictitious forces do not come under that category.

 Quote by DaleSpam I don't mind how you use the word "physical", but you need to be self-consistent. You cannot claim that frame-variant quantities, like the rate of a clock, are "physical" and then exclude the frame-variant fictitious forces from being "physical" also. In a non-inertial reference frame fictitious forces can do work, can have potential energy, can cause mechanical stress and strain, and have many other measurable effects. Yes, that is why they are called fictitious forces. That is also why he is not likely to try to do the analysis in his non-inertial rest frame, but is more likely to do the analysis in some inertial frame.
So if something cannot be logically identified or physically determined it comes under the heading of a 'fictitious force'? It was a fictitious force that many years ago exchanged my tooth for a dime.

The concept of a 'fictitious force' is in my opinion a desperate grasping at straws analogous to the 'parallel universes' escape-clause, suitably impossible-to-disprove, concept.

People who believe in God are criticized by others for their faith in a 'fictitious force' yet apparently some people are apparently of the opinion that a non-material 'force' can result in an equal and opposite reaction provided the results supply the solution they seek.

 Quote by DaleSpam Your explanation is correct only in the inertial reference frame where B is at rest. In other reference frames there will be other explanations. But all reference frames will agree on the conclusion.
Einstein indicated that clock A will lag behind B due to the fact that, whilst it is moving, clock A 'goes more slowly' (ticks over at a slower rate) than clock B.

It has been pointed out in relation to my previous thread in this forum that there could be third observer, C, relative to whom A and B were initially moving at v. When A accelerates he, from C's point of view, decelerates and comes to a stop in C's reference frame (thus ticks over at the same rate as C's clock) whereas B keeps moving relative to C at v thus from C's point of view clock B is ticking over at a slower rate than his own clock ergo also at a slower rate than clock A so when A 'accelerates' back to B's location (in B's reference frame, decelerates and comes to a stop alongside B) it is, in C's opinion, clock B that will lag behind A.

C moves to B's location and comes to a stop alongside A and B and finds, much to his consternation, that B does not lag behind A as indicated by his 'calculations' or 'determinations' or 'predictions' but that A lags behind B!
P: 645
 Quote by cos I made no suggestion whatsoever that Einstein's answer is erroneous!
 Quote by cos Einstein's maths indicated the amount by which, in his opinion, clock A lags behind B and, as Einstein pointed out, because A lags behind B it must have 'gone more slowly' (i.e. ticked over at a slower rate) than B whilst A was moving. According to Einstein's section 4 maths, clock A is ticking over at a slower rate than clock B but according to the maths employed by the observer accompanying clock A it is B that is ticking over at a slower rate than his clock. Having calculated that B 'is' ticking over at a slower rate than his own clock, observer A 'determines' or 'predicts' that when he arrives at B's location he will find that it lags behind his clock yet he learns that it does NOT! HE has 'done the math' yet finds that it gave an erroneous answer. Of what value his math?
????

 Quote by cos What on Earth does "...there are a number of ticks from the 'stationary' clock still travelling to catch up." mean?
When the clocks are not colocated it takes time for the information from one clock to reach the other. The information travels at the speed of light. If the moving clock "looked" back at the stationary clock (as per Einstein's scenario), just before stopping, it would see only the time on the stationary clock that happened, in the moving clock's frame, x'/c ago (where x' is the separation that the moving clock thinks that it has from the stationary clock based on the travelling time). There will be more information still in transit.

Given that we can't agree as to whether you are saying Einstein's maths was erroneous or not, or that what he said matches with his maths, I don't feel this is going anywhere.

If you have a go at the maths, you will see that it matches the "reality" of what Einstein said (at least wrt to the 1905 paper). Until you do that, I really think I have to agree with Jtbell, at least in part, this is not a physics discussion. I just don't think it qualifies as philosophy either.

cheers,

neopolitan
P: 1,886
 C moves to B's location and comes to a stop alongside A and B and finds, much to his consternation, that B does not lag behind A as indicated by his 'calculations' or 'determinations' or 'predictions'
Then why don't you help C with his predictions? Let the movement be sinusoidal, and let the "clock rate" be adjusted by -1/2 v². Sometimes a decent calculation saves many lines of philosophical debate.
P: 801
 Quote by cos "...on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other ." In other words, according to Einstein, something physically happens to the moving clock. It is no longer synchronized with the inertial clock.
No, that's not why they're no longer in synch.
 Having, during his trip, 'determined' that B is, as you say, running slow compared to his clock that is at rest in his frame he arrives at B's location to find that B does not lag behind (having 'run slower' than) his clock but that his clock lags behind B.
No. The ship's twin knows 2 things during the outbound inertial part of the trip:
1. Earth's clock runs slower than the ship's clock in the ship's frame.
2. The ship's clock runs slower than earth's clock in earth's frame.

When he stops at the turnaround, the ship's twin doesn't "find" that his clock ran slower than earth's clock, he knew all along that the ship's clock ran slower than earth's clock in earth's frame.
 The idea (during the astronaut's period of acceleration following turn around) that the stationary clock incurs time contraction (i.e. 'is' ticking over at a faster rate than it was before he accelerated) was, for Einstein, an anathema and it it is his depictions to which I refer not interpretations arrived at by anybody else.
The relative ticking rate of clocks is frame-dependent, not absolute. Nothing changed with earth's clock during the turnaround, the relative speed of the ship changed. Earth's clock didn't change its ticking rate, we changed which frame we're referring to, and the ticking rate of a clock is frame dependent.

It seems that you're ignoring Einstein's most important contribution to modern physics, that the rate that any clock runs is dependent on the relative speed of the observer. That means that if someone changes his speed relative to a given clock, the rate of that clock will be different. Not because the clock changed, but because the reference frame of the observer changed.
Mentor
P: 16,951
 Quote by cos My references are to reality and I am of the opinion that fictitious forces do not come under that category.
That is fine by me, but then neither should coordinate time (and therefore the rate of a clock wrt coordinate time). Again, I don't care how you use the words "physical" or "real" but you need to be consistent.

 Quote by cos The concept of a 'fictitious force' is in my opinion a desperate grasping at straws analogous to the 'parallel universes' escape-clause, suitably impossible-to-disprove, concept.
This is also fine, but if you do not like fictitious forces then you cannot do any analysis in any non-inertial frame. You must stick exclusively to inertial frames. As I mentioned previously, all inertial frames agree on the results also.

 Quote by cos It has been pointed out in relation to my previous thread in this forum that there could be third observer, C, relative to whom A and B were initially moving at v. When A accelerates he, from C's point of view, decelerates and comes to a stop in C's reference frame (thus ticks over at the same rate as C's clock) whereas B keeps moving relative to C at v thus from C's point of view clock B is ticking over at a slower rate than his own clock ergo also at a slower rate than clock A so when A 'accelerates' back to B's location (in B's reference frame, decelerates and comes to a stop alongside B) it is, in C's opinion, clock B that will lag behind A.
No, you have forgotten that A is moving at 2v/(1+v²/c²) on the second leg of the trip. Since A travels at a faster speed than B, A experiences more time dilation than B. Also, that second leg lasts for a longer coordinate time in C's frame. Because of that, C is of the (correct) opinion that A will lag behind B when they meet and that A was "physically" slower than B on average. Again, all frames agree on this result.
P: 801
 Quote by cos Having calculated that B 'is' ticking over at a slower rate than his own clock, observer A 'determines' or 'predicts' that when he arrives at B's location he will find that it lags behind his clock yet he learns that it does NOT!
This is not true. Observer A makes no such prediction. Each twin will make the same prediction. That is that when they reunite, A's clock will read less elapsed time than B's clock.

What might be a source of confusion is that the fact that time dilation is symmetrical, but proper elapsed time is not. The reason that the ship's clock readings coincide with the end result of elapsed time is because the ship's clock is local to every relevant event and therefore represents the proper elapsed time for the ship for every event in every frame. That is not true of earth's clock. Earth's clock is local to the departure and return only, so only those readings represent proper time for the earth twin in a different frame. For events not local to earth, the earth clock represents proper time in earth's frame only. Clocks in relative motion only read proper time locally, not at a distance. So an earth clock reading in the ship's frame does not represent the proper time in earth's frame for any event not local to earth, like the ship's turnaround.

If that's what you mean by not being "reality", then just say so and I think everyone would agree.
 HE has 'done the math' yet finds that it gave an erroneous answer. Of what value his math
He got no erroneous answer. All of his math gave correct answers. And it's not contradictory because he didn't confuse reciprocal time dilation between the twins' clocks with the non-symmetrical elapsed time of the clocks.
P: 801
 Quote by neopolitan You would probably be better off if you search for one of his papers or essays written after 1929.
Hi, neopolitan,

I think Einstein's best work was prior to 1929, despite not being perfect.

His 1918 paper on the twins paradox is nothing like the way cos is interpreting it. The only difference between it and standard resolutions is that instead of using an instantaneous turnaround with earth's clock "jumping" ahead suddenly, he uses a realistic acceleration turnaround with earth's clock "running fast" in the ship's frame during the turnaround. It's essentially the same as the standard resolutions, at least as far as this topic is concerned.

 Related Discussions Special & General Relativity 17 Introductory Physics Homework 5 Special & General Relativity 5 Special & General Relativity 3 Introductory Physics Homework 1


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# Ralf Schindler - Talk 1 on Logic Summer School of Fudan University, 2020

Content:

• Stationary sets;
• Forcing revisited;
• Forcing Axioms: $\mathbf{MA}$;
• Proper forcing; semi-proper forcing; stationary set preserved forcing;
• $\mathbf{PFA}$, $\mathbf{SPFA}$, $\mathbf{MM}$.

# Ralf Schindler - Talk 2 on Logic Summer School of Fudan University, 2020

Content:

• Restate $\mathbf{PFA}$, $\mathbf{SPFA}$, $\mathbf{MM}$ as well as $\mathbf{PFA}^{++}$, $\mathbf{SPFA}^{++}$, $\mathbf{MM}^{++}$;
• A few words on iterated forcing
• Supercompact Cardinals, Laver functions;
• Forcing $\mathbf{SPFA}^{(++)}$
• Weak reflection principle;
• $\mathbf{MM}\Rightarrow2^{\aleph_1} = \aleph_2$.

# Course Notes of Axiomatic Set Theory

This is a collective course note taken in Prof. Simon Thomas’ course Axiomatic Set Theory of Rutgers University, which is held on Spring semester, 2019. The main topic of this course is forcing, forcing axioms such as $\mathbf{MA}$, Open Coloring Axiom, Axiom A of Baumgartner and the Proper Forcing Axiom of Shelah. Also, this course discussed the relation among themselves and basic independent statements like $\mathbf{CH}$. The main reference would be Kunen’s book and Jech’s book. If there is any mistakes or comments, please feel free to contact me.

# 《先天性》：阅读笔记

《先天性》，节选自阿尔弗雷德·朱尔斯·艾耶尔《语言，真理和逻辑》

1. 经验论的数学哲学：不存在关于现实的先天知识。数学无关乎现实世界，数学命题是分析命题。在现实世界的（“出乎意料的”）有效性是纯粹偶然的。对于更一般的科学来说，所有科学命题都是“大概的假设”。
2. 数学是有意义的，但数学命题的意义并非在于其在物理世界的映照，而是在于论证分析命题在我们的世界（一个低智商人群组成的世界）是合理的。“一个具有无限智力的人对逻辑和数学毫无兴趣。”
3. 批驳康德主义唯理论：“5+7=12”不是综合命题，是分析命题，错误的根源在于康德使用了非逻辑的论据；几何学是分析的且无关乎人类空间直觉的，因为几何学也不描述物理世界：康德受到了时代的限制。（康德的这个错误常常使得现代数学家忍俊不禁。）


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# Turnigy RotoMax 150cc Size Brushless Outrunner Motor

### Help Support Homebuilt Aircraft & Kit Plane Forum:

Log Member
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#### Vigilant1

##### Well-Known Member
I think you might have typo there, the HobbyKing web site says the motor is rated at 190 amps, or 9800 watts (roughly 13HP). If somebody is putting 360 amps through it at the same voltage, then it should be producing roughly 26 HP (for however long it lasts).

#### pictsidhe

##### Well-Known Member
I'm still pondering power options. If you buy two, the second could go on ebay for profit... I might get one to test ducted fans, though. Ducted fan math is horrible, btw...
The 120100 have a 50% longer stator than the '150cc' 12090 motors.

#### Sockmonkey

##### Well-Known Member
Remember the rule for electric and/or hybrid planes. If you don't want it to suck, you need to design the plane from the ground up to take advantage of the flexibility electric and hybrids give you.

#### pictsidhe

##### Well-Known Member
Remember the rule for electric and/or hybrid planes. If you don't want it to suck, you need to design the plane from the ground up to take advantage of the flexibility electric and hybrids give you.
And to avoid their bad points...

#### henryk

##### Well-Known Member
I'm still pondering power options. If you buy two, the second could go on ebay for profit... I might get one to test ducted fans, though. Ducted fan math is horrible, btw...
The 120100 have a 50% longer stator than the '150cc' 12090 motors.
-see FREERCHOBBY motors...(big power!)

#### proppastie

##### Well-Known Member
Log Member
Any good charts,...spread sheets,...on batteries required and running time......also different controllers? Seems like there are lots of variables and every installation I see is different.

Saw a reference to "junk yard Nissan Leaf".....is that an option for batteries or are those batteries too big....and are there really any "junk yard" stuff like that yet?

Lots of questions

#### pictsidhe

##### Well-Known Member
I'd favour 18650 cells rather than hobby ones myself.
Capacity is a tradeoff with max current.
The not-latest Tesla uses panasonic NCR cells, you may find someone splitting a pack on ebay. DeWalt 1.5Ah batts use LG HB4 cells.
Pick cells with a higher max current than you will use. Some on the web have tested many of these cells. Beware fakes! 18650 industry is littered with them.
This site has a big selection: www.imrbatteries.com

The math is pretty easy:
Wh required = A x V x hours run time.
750W = 1hp. The motor will be up to 85% efficient. So power draw will be around 900W per hp. Then you've got losses in the controller, lets say, to be safe and easy, a round 1kW from the batteries per prop hp.

Battery energy = Ah x 3.6V
A 3Ah battery will be 9.8Wh. Batteries will last much longer if you avoid full charge and discharge. Using only 2/3 or 3/4 the capacity will greatly increase their life.
So, lets say 20hp for 15 minutes.
20kW x 15 / 60 = 5kWh
Using 3Ah cells. 5000/9.8 = 510 cells. That doesn't include extra capacity so you don't fully charge and discharge, 510 / 0.75 = 680 cells. Yeah, that's going to be pricey at ~\$5 a cell. Time to see what is on ebay...

#### proppastie

##### Well-Known Member
Log Member
thanks much...... all in all great post.....

680 of the 18650 = 680 x Weight: 47.9g =31960g=71.8 Lb.

5 min = .33 x 71.8 = 23.7 lb.......not so unreasonable, especially if perhaps I only need 15-16 hp to climb and can throttle back after liftoff.

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#### pictsidhe

##### Well-Known Member
Most efficient climb will be at max power. You'll be putting a larger proportion of power into climb. For only 5 min, you need to be careful not to exceed the max current of your cells. If you only use 3/4 of the pack capacity, you'll need cells that run at 60/5 * 3/4 = 9C, where C is the cell Ah. A 3Ah cell would need to be good for 9x3 = 27A. They generally aren't, so you'll need to look at higher current, lower capacity cells. Such as 2Ah, 25A cells. Then you need 50% more cells to get the capacity. They are likely to be a littke cheaper each, though...


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# First passage time of Brownian motion

I am currently trying to understand why $P(\tau_b < \infty)$ for $b \in \mathbb{R}$, where $\tau_b := \inf\{t \geq 0: B_t = b\}$. There was a little step in the argument below which I did not understand.

Let $(B_t)_{t \geq 0}$ be a one-dimensional Brownian motion. Let us write $$\sup_{t \geq 0}B_t \geq \sup_{n \in \mathbb{N}}B_n = \sup_{n \in \mathbb{N}} (\xi_1 +...+\xi_n)$$ where $\xi_j=B_j-B_{j-1}$ are i.i.d. standard normal random variables. If we further write $$S:=\sup_{n\in \mathbb{N}}(\xi_1 +...+\xi_n) = \xi_1 + \sup_{n \in \mathbb{N},n \geq 2}(\xi_2 +...+\xi_n) =: \xi_1 + S'$$ then $S$ and $S'$ have the same distribution and $\xi_1$ and $S'$ are independent. I need to somehow conclude, since $\xi_1$ is standard normal distributed, that then $S=\infty$ a.s. How do I see this?

Thanks a lot in advance!

By 0-1 law for exchangeable random variables $P\{|S|<\infty\}$ is $0$ or $1$. If it is $1$ then $Ee^{itS}Ee^{it\xi}=Ee^{itS}$ for all $t$ and $Ee^{itS}=Ee^{itS}$ because $S$ and $S'$ have the same distribution. This gives $Ee^{it\xi}=1$ at least for $|t|$ sufficiently small, which is a contradiction because$Ee^{it\xi}=e^{-t^{2}/2}$.Hence $P\{|S|=\infty\}=1$. By definition $S>-\infty$ almost surely so we get $S=\infty$ almost surely.
• Doesn't it hold for arbitrary $t$ instead of sufficiently small $|t|$? – vaoy Aug 30 '18 at 9:26
• I cannot cancel $Ee^{itS}$ unless I know it is not zero. For $|t|$ small it is close to $1$ and hence non-zero. – Kavi Rama Murthy Aug 30 '18 at 9:29


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# Singular values of the sum of A and A^T

As a part of my research, I need to achieve a lower bound to the smallest singular value, $\sigma_{n}(A+A^{T})$ for a stochastic $A$ (as a function of the singular values of $A$), which is generally not Hermitian.

I am aware of the upper bounds (due to Weyl and Fan) and of the fact that for general $\sigma_{i}(A+B)$ no lower bound is known. Do you see a way?

Thank you.

Edit: A can be considered a power of a lazy row stochastic matrix. I.e., $A=P^k$ for some strongly diagonally dominant row stochastic $P$.

-
See my answer mathoverflow.net/questions/97746 to a similar MO question. However, the assumption that $A$ is stochastic might change the answer. – Denis Serre Dec 30 '12 at 7:29
I came to a relaxation to my problem, wherein A can be a power of a lazy row stochastic matrix. I.e., A=P^k for some strongly diagonally dominant row stochastic P. Hence, P is positive definite, however A is generally not. Do you see a way that it simplifies the problem? – Daniel86 Jan 3 '13 at 6:22

It does not seem like any obvious bound on $\sigma_n(A+A^T)$ is possible in terms of the singular values of $A$. Indeed, consider $$A = \left( \matrix{ 1/3 & 1/2 & 1/6 \cr 1/6 & 1/3 & 1/2 \cr 1/2 & 1/6 & 1/3} \right).$$ Clearly, $A$ is stochastic and a computation reveals that it is nonsingular so that all of its singular values are positive. On the other hand, $A+A^T$ is a multiple of the all-ones matrix, so $\sigma_3(A+A^T)=0$.
Thankyou, it is a nice example, but can this situation occur where $A$ is an integer power of a positive definite matrix? I edited my original post. Thanks. – Daniel86 Jan 3 '13 at 6:31
@Daniel86 - under standard definitions, a real positive definite matrix is automatically symmetric. When you use the words "positive definite" I am guessing you mean that $x^T A x > 0$ for any $x \in R^n$ but $A$ is not necessarily symmetric. Is my understanding correct? – alex o. Jan 4 '13 at 21:24
@alex o. - If $P$ is diagonally dominant but not necessarily symmetric, $P+P^{T}$ is diagonally dominant and symmetric and hence positive definite. However, for $A=P^{k}$ this is not always the case. For such a case, I want to lower bound the smallest singular value of $A+A^{T}$ in terms of the singular values of $P$. – Daniel86 Jan 4 '13 at 22:30


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# Problem #1134

 1134 Three circles of radius $1$ are externally tangent to each other and internally tangent to a larger circle. What is the radius of the large circle? $\mathrm{(A) \ } \frac{2 + \sqrt{6}}{3} \qquad \mathrm{(B) \ } 2 \qquad \mathrm{(C) \ } \frac{2 + 3\sqrt{2}}{2} \qquad \mathrm{(D) \ } \frac{3 + 2\sqrt{3}}{3} \qquad \mathrm{(E) \ } \frac{3 + \sqrt{3}}{2}$ This problem is copyrighted by the American Mathematics Competitions.
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