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# How to drive high powered LEDs as efficiently as possible

I have come across some very high powered LEDs.

My question is: for these high powered LEDs how can I create a driver circuit that will be energy effiecent or is there a driver for high powered LEDs that I'm just not finding.

The LEDs will eventually be used in a grow light for an aquaponics system so effiecency is essential.

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I seriously doubt one LED operates at 40 V. –  Olin Lathrop Jun 28 '12 at 23:10
Maybe he meant 4.0V... –  Chris Harris Jun 28 '12 at 23:43
... or, may be, it's an assembly with multiple LEDs in series. –  Nick Alexeev Jun 29 '12 at 3:10
Do you mean 0.8A or 0.08A? –  jippie Jun 29 '12 at 7:12
From what I'm told, it operates at 40V .8A and they were very very expensive to buy –  Mike Jul 2 '12 at 1:36
show 2 more comments

For best overall LED efficiency (which is your stated aim) you want high efficiency LEDs plus high efficiency drivers. There are may high power LED modules available but they in many cases use LEDs that are less efficient than the best available. No amount of efficient driving of a very low efficiency LED will make up for its low efficiency.

You mention 0.8A x 40V (which will use multiple LED dies or separate LEDs) = 32 Watts. This input power could be obtained with about 5 to 8 of the top available LEDs and it will often be better to use the best available if top efficiency is wanted.

It will probably be most cost-effective to buy a commercially available LED driver rather than making your own, looking for good efficiency. ebay lists a vast quantity of possible candidates

You also need to specify your energy source. Is this 110 VAC mains, or a 12 Volt battery or?

The standard way of driving LEDs correctly and at maximum efficiency is straightforward (fortunately).

• Determine required LED operating current.

• From data sheets, determine maximum voltage drop that LED will have at desired current.

• Provide a switched mode constant current source that will produce the required current at at least the maximum possible voltage.

• Extra points:

• Minimise LED temperature with cooling

• Operating an LED at below its maximum current ratings will result in somewhat increased efficiency - perhaps 10% to 20% more light out per Watt in at 20% of full ratings compared to full power.

• LED lifetimes increase with decreasing temperature and with decreasing current. While more current will usually result in more heat produced, These effects are independent of each other.

To get an LED driver of essentially any desired rated current. use of an external MOSFET of suitable rating, driven by a controller that sits, will allow whatever current the MOSFET can handle. This can be very large if desired.

For very best efficiencies use "synchronous rectification" - this replaces rectifier diodes with switched MOSFETs and gives lower losses.

Efficiencies around 85% from a low voltage DC source to an LED string will be achievable.Over 90% in selected cases and with care.

Comment on the above and we can provide better information.

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The cost of consumer commodity PSU's for PC's range from cheap and dirty $50/500W to 1200W @$300 Corsair Pro, so ranging from $100/kW to$250/kW. But LED approved drivers & PSU's tend to cost much more than this.

# My suggestion is to use a good consumer 85W "universal laptop charger" (ULC)

• They range in price and this year I've purchased ones for $35 @ 85W, or$0.41/W qty=1.
• The key point here is LED drivers, supplies etc tend to cost as much or more than the LED Engines where here, I propose a cheap consumer solution.
• Power LEDs range $1~$1.5/W and higher for improved quality
• Expect these costs to shrink < $0.10 to$0.50/W in volume buys.
• I have several installed running non-stop for a few years now.
• Heatsink costs have a wide variation due to choices of large material costs with massive fins or moderate size with force air DC fans.

If you find a cheaper complete power solution for LED Engines, let me know.

• Some non-OEM chargers have all std DC voltage settings & plugs for laptops.
• I use the 4.5A @ 19V ULC's for 6x1W white or blue on aluminum clad PWB's
• Shown below with 6x3 serial string and 6x1 string on aluminum boards
• I mount these under aluminum 0.5"x1" extruded end strips uses for tiles for elimination of LED glare and additional heat-spreading and the mount under exterior hand or deck rails.

• Moisture protection is added and essential for external use.

• Used 13 strings of 6 x 1W LED MCPCB's (78 LEDs) in parallel on 19V @ 4.5A (85W)
• Result was cool LED's, cool PS, but almost like daylight, so it was dimmed to 50% with 5Ω, 25W added. Efficiency was not a concern. Cost \$1.5
• 1W per linear foot under handrail or one strip every 5~6 feet
• 400~500 Lumen per 6 x 1W strip each hidden under the handrail.
• Results exceed user's expectations, pc's below prior to modification/ installation.

To use this for powering LED's depends on your expertise in matching an array of LEDs to the V-I operating points of the ULC. If you can buy from LEDs from the same batch, they V-I curve will be matched, but Disti's refuse to do this, so it takes purchasing skill, otherwise you will have issues.

Basically, it's all about having good low thermal resistance and raising circuit ESR slightly above nominal ESR characteristics with same batch sourcing on LEDs to achieve low cost reliable operation. You need to understand open circuit voltage a rise is expected with no load by the load regulation or ESR of the charger. 2% typ. or 100mΩ for 100W PS on long cables.

# Ticky Tackie Technical secrets for Powering any LED Engine,

Now let's examine @Mike's desire to use Cree's 45W Xlamp cXa2011 LED

All PN diodes have a negative temperature coefficient, k (Shockley Effect) or voltage drop due to temperature rise. The fear with designers is with CV, is "thermal runaway" so they only consider constant current drivers whereas if you have tight control of a batch and sufficient ESR with low Tja, it is stable. When ambient & junction gets hotter, the LED's internal junction voltage drops demanding more current from the voltage difference across the ESR inside.

I'll show how to overcome this effect on the XLamp and in another article show how to measure junction temperature using the LED itself as a thermometer with the V-T curve which is standard in hospitals now.

• from curves on p.5, using conservative specs.
• 40.0V TYP @ 270mA ∂V/∂I = ESR = 7Ω at 10.4W TYP. (p.5 spec)
• Thermal resistance jcn-case = 0.4 °C/W TYP.
• Junction thermal voltage, k = -35 mV/°C. TYP. (p.2 spec)
• We know that heatsink design is crucial and Rja is sum of all thermal resistance from junctions to ambient.
• Let's choose 2 °C/W as heatsink + interface starting point, so Rja = 0.4+2 = 2.4°C/W

- Note: that a decent CPU heatsink is ~ 0.2°C/W for a forced air CPU heatsink ) and an excellent one is <<0.1 °C/W.

• V forward = 40V typ, 48V max @ 270mA this is a wide range from batch to batch - So you need a good source LED engines from same batch or ask me.

• Note this a string of 12 chips in one hence 40V or 3.3V/chip

Thermal equilibrium exists where there is an intersect between the linear temperature with thermal resistance and the quadratic effects of If^2*Rs) If ambient is too high with high thermal resistance, runaway can occur. It can be prevented with low Thermal resistance and adding ESR in the cable.

-


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## Quiet Jet Engines

A number of deceptively simple, but hopefully interesting questions.

1. Why are jet engines so noisy?
2. Is there a way to make them quieter, or otherwise register less on the human ear?
 PhysOrg.com science news on PhysOrg.com >> City-life changes blackbird personalities, study shows>> Origins of 'The Hoff' crab revealed (w/ Video)>> Older males make better fathers: Mature male beetles work harder, care less about female infidelity
 Recognitions: Science Advisor I've seen Mach diamonds in rocket engine exhaust, and I think also in military jet engine exhaust. Regardless of how much noise is coming from the engine proper, that sort of high-speed exhaust is going to interact with the atmosphere in a very noisy way, I would think.
 Recognitions: Gold Member Science Advisor Staff Emeritus All a fanjet engine is, is a giant turbine. You're also pushing out air at really high velocity. You can definately hear wind at 30mph... you'll really hear it at 500.

Mentor
Blog Entries: 9

## Quiet Jet Engines

While I remember well, the screech of a F4 Phantoms afterburners, a question perhaps related to this one, and much more important to my current life...

How about a quite vacuum cleaner?
 Recognitions: Science Advisor There are a few quiet vacum cleaners. Go down to Sears and look at their best models, for $300 its got great suction and the canister is easier then the uprights IMO. The upright featured on the TV show Monk is also quiet, but a friend paid like close to$500 for his verison of that company's product. Low 70 dbs would be my guess, I could meter it on my Audiocontrol and tell you if you really want to know. :) How about the jet engine diffusors on the stealth planes? Or the other jets that hide the sound behind the 'cone' of air as the plane approaches mach? Still not going to be as quiet as a luxury car. Cliff
 Noise reduction for jet engines is a major design problem, especially since federal noise regulations for airports have become tougher. You have to fight the fact that you are dealing with pressurized air (and sound is a pressure disturbance) the mechanical noise generated by the engine (friction, spinning shafts, bearings, etc.) and the noise of high speed flow. You can see some of the design elements that help reduce noise next time you are at an airport. Look for a Boeing 727, a two engine domestic aircraft. Check out the shape of the cowl of the engines, that is to say, the front part of the engine. You will notice that it looks like a flexible hoop being smooshed against the ground, with a flatter part at the bottom. This design reduces the noise produced by the engine while having a minimal impact on engine performance.
 Recognitions: Gold Member Science Advisor The sounds so simple that I'm sure someone has already thought it, but; Am sure you all seemed noise-reducing headphones, right? Perhaps a similar trick could be applied to get exhaust. What about a secondary cowling inside the primary? This inner cowling could be perforated to allow one-half of the sound waves that reach it to pass through, while the other half are deflected out the rear of the engine. The one-half that pass through this barrier strike the outer cowling and are deflected out the rear also. By properly setting the distance between the primary and secondary cowlings the apparatus could be engineered in such a way that the two sets of sound waves was less emitted are of equal amplitude and frequency, with a one-half phase variance. These two sets of sound waves might then partially cancel one another out.
 Mentor Noise reduction in airflow is actually quite simple: lower velocity. I deal with the issue on a daily basis designing air conditioning systems. This is the reason turbofan engines are quieter than turbojets: massive fans and high bypass ratios mean lower airflow velocities at low speed. Unfortunately, this also means the only way to reduce noise further is make the engines so big you could park a semi inside.

Recognitions:
Gold Member

format: plain


text:
Power in Circuits Video Lessons

Concept

# Problem: A real battery is not just an emf. We can model a real 1.5 V battery as a 1.5 V emf in series with a resistor known as the "internal resistance", as shown in the figure. A typical battery has 1.0 Ω internal resistance due to imperfections that limit current through the battery. When there's no current through the battery, and thus no voltage drop across the internal resistance, the potential difference between its terminals is 1.5 V, the value of the emf. Suppose the terminals of this battery are connected to a 2.4 Ω resistor.What fraction of the battery's power is dissipated by the internal resistance? ΔP/P= %

###### FREE Expert Solution

Power in circuits:

$\overline{){\mathbf{P}}{\mathbf{=}}{\mathbf{V}}{\mathbf{i}}{\mathbf{=}}{{\mathbf{i}}}^{{\mathbf{2}}}{\mathbf{R}}{\mathbf{=}}\frac{{\mathbf{V}}^{\mathbf{2}}}{\mathbf{R}}}$

###### Problem Details

A real battery is not just an emf. We can model a real 1.5 V battery as a 1.5 V emf in series with a resistor known as the "internal resistance", as shown in the figure. A typical battery has 1.0 Ω internal resistance due to imperfections that limit current through the battery. When there's no current through the battery, and thus no voltage drop across the internal resistance, the potential difference between its terminals is 1.5 V, the value of the emf. Suppose the terminals of this battery are connected to a 2.4 Ω resistor.

What fraction of the battery's power is dissipated by the internal resistance?

ΔP/P= %


text:
• February 27, 2021, 10:03:03 PM

### Author Topic: (Question) iterating corners to find period  (Read 1314 times)

0 Members and 1 Guest are viewing this topic.

#### quaz0r

• Fractal Feline
• Posts: 152
##### (Question)iterating corners to find period
« on: October 09, 2020, 10:28:18 AM »
http://www.mrob.com/pub/muency/period.html

im trying to figure out what the proper way to implement this is.  it sounds like we are checking to see when the points surround the origin.  first i tried just checking if each of four points occupy each quadrant, but that fails immediately.  he makes some weird statement about an odd number of vertices crossing one half of either axis, but i cant picture how 3 of 4 vertices can cross one half of one axis and surround the origin at the same time.  unless the vertices are allowed to criss-cross each other, is that supposed to count?   he goes on to say that you also get a rough estimate for the location of the root by the location of the origin with respect to the points when they surround it, which makes it sound like for example if you start with a square you more or less should have retained that shape.

#### claude

• 3f
• Posts: 1784
##### Re: iterating corners to find period
« Reply #1 on: October 09, 2020, 11:15:40 AM »
the shape only gets folded by squaring if it surrounded the origin, but we stop then so don't have to worry about that
so the shape (probably, maybe, usually?) remains a loop that doesn't cross itself
if only one edge crosses the positive real axis (or any other ray from the origin to infinity), the other 3 edges must be wrapped around the origin
if three edges cross the positive real axis, and the 4th edge doesn't, again the shape must enclose the origin
the other 3 cases don't surround the origin (maybe try drawing some diagrams)

see also: https://en.wikipedia.org/wiki/Even%E2%80%93odd_rule "On a simple curve, the even–odd rule reduces to a decision algorithm for the point in polygon problem. "

for analytic functions like Mandelbrot iterations, ball arithmetic is usually superior than this method; and ball method can be extended to non-analytic functions using 2x2 Jacobian matrices

#### marcm200

• 3c
• Posts: 925
##### Re: iterating corners to find period
« Reply #2 on: October 09, 2020, 11:33:27 AM »
I tried this with period-2 (around -1+0i) and period-10 (value in image) and used a 4-polygon (lines). At the correct iteration the origin (cyan point) is enclosed. Quite interesting!

#### quaz0r

• Fractal Feline
• Posts: 152
##### Re: iterating corners to find period
« Reply #3 on: October 09, 2020, 12:51:03 PM »
thanks for the replies.  is there a post on the forums that explains how to implement the ball method?  i remember implementing knighty's ball method for calculating an error estimate on the series approximation, but im not sure how to apply it to this.

#### claude

• 3f
• Posts: 1784
##### Re: iterating corners to find period
« Reply #4 on: October 09, 2020, 02:50:59 PM »
The derivative is basically a scale factor applied to the ball radius (which starts at your desired region of interest), then you just check whether the distance to the origin is less than the radius of the scaled ball.  But I think knighty might have extended it with second derivatives, can't remember if I copy/pasted that into KF or if I just did the linear version....

#### hobold

• Fractal Fluff
• Posts: 397
##### Re: iterating corners to find period
« Reply #5 on: October 09, 2020, 03:23:43 PM »
he makes some weird statement about an odd number of vertices crossing one half of either axis

This is really a "point in polygon" test, namely the ray casting method: https://en.wikipedia.org/wiki/Point_in_polygon, with a half-axis as the ray.

#### marcm200

• 3c
• Posts: 925
##### Re: iterating corners to find period
« Reply #6 on: October 09, 2020, 03:32:41 PM »
A nice implementation of this odd-even algorithm has been done by Glauner et. al. https://github.com/pglauner/point_in_polygon with some special considerations in case one or more vertices lie on the axis-to-follow (U, S-form).

"A Simple and Correct Even-Odd Algorithm for the Point-in-Polygon Problem for Complex Polygons", Galetzka M, Glauner P. 2017.

#### quaz0r

• Fractal Feline
• Posts: 152
##### Re: iterating corners to find period
« Reply #7 on: October 09, 2020, 05:35:14 PM »
if we say 3 of 4 vertices can cross one half of an axis and still surround the origin, are we referring to something like my image here?  i dont see what else it could be.  i guess that confused me since we are speaking in terms of vertices, this doesnt look like a box any more.  how about when munafo says you can derive a relative position of the root?  do you just forget about the original orientation of the points and let it be a square?  if so, what's the orientation?  like in my image, would you just let those points make a box, and the root is located toward the lower left corner?  or maybe the orientation changed and its near the upper right corner, or something else?

#### marcm200

• 3c
• Posts: 925
##### Re: iterating corners to find period
« Reply #8 on: October 09, 2020, 06:40:10 PM »
Is your image an iteration as described in the link? Or is it a drawn image by yourself? As claude pointed out, when the initial shape is not crossing edges, then it should'nt during iteration (squaring) unless it once enclosed the origin (not crossing), then afterwards it can. So if one does directly stop at the first origin-in-polygone event, this problem doesn't occur. I think the odd-number axis-crossings might be a general idea for larger polygones than 4 vertices - maybe the latter doesn't have 3 axis-crossings without edges crossing themselves (one could do a brute force placing of points above and below the axis in all combinations considering order to (dis)prove that).

#### quaz0r

• Fractal Feline
• Posts: 152
##### Re: iterating corners to find period
« Reply #9 on: October 09, 2020, 06:50:55 PM »
ok i figured out a shape that doesnt cross itself.  i dont know whether one, having started out as a box, would actually encounter this though.

in general as far as this period finding method goes, can one extend this past the first period?  like say you surround the origin after 200 iterations, so you say 200 is your period.  then you keep iterating and surround the origin after 50 more iterations.  would that be the next closest mini you passed to get to your location, having a period of 50?

#### gerrit

• 3f
• Posts: 2333
##### Re: iterating corners to find period
« Reply #10 on: October 09, 2020, 08:18:11 PM »
1st order ball method:
Code: [Select]
function p  = findPeriodM3(c0,dx,dy,n,doCont,mpow)% in ball centered on c0 find period (up to n) of nucleus% use 1nd order Taylor ball% M-power mpow set% doCont = 0 normallyr0 = min(abs(dx),abs(dy));z = c0*0;r = (r0);p = [];maxR = 1e5;az = abs(z);for k=1:n    r = (az+r).^mpow - az.^mpow + r0;    z = z.^mpow + c0;    az = abs(z);    if(r>az)        p = [p k];        fprintf('findPeriodBallM3: N-period found: %d\n',k);        if(~doCont)            break;        end    end    if(az>maxR | r>maxR)        fprintf('Ball: escaping\n',k);        break;    endend2nd order ball method:
Code: [Select]
function [p c1] = findPeriodBall2(c0,dx,dy,n,doCont)% in ball centered on c0 find period (up to n) of nucleus% use 2nd order Taylor balls, always smaller than 1st orderr0 = min(abs(dx),abs(dy));r0 = r0;z = c0*0;c1 = c0;dz = 0*c0;r = r0;p = [];maxR = 1e5;az = abs(z);adz = abs(dz);minz = 1e16;minIter = -1;for k=1:n    r = r.^2+2*(az+r0.*adz)*r + r0.^2.*adz.^2;    dz = 2*z.*dz + 1;    z = z.^2 + c0;    az = abs(z);    adz = abs(dz);    if(double(az)<minz)        minz = double(az);        minIter = k;     end    if((r+r0*adz)>az)        p = [p k];        fprintf('Ball: N-period found: %d, atom: %d\n',k,minIter);        if(~doCont)            break;        end    end    if(az>maxR | r>maxR)        fprintf('Ball: escaping\n',k);        break;    endend

#### hobold

• Fractal Fluff
• Posts: 397
##### Re: iterating corners to find period
« Reply #11 on: October 09, 2020, 08:33:27 PM »
ok i figured out a shape that doesnt cross itself.  i dont know whether one, having started out as a box, would actually encounter this though.
I don't think you can ever get a non-convex quadrangle when you start from a square that is wholly contained in one atom domain and do mandelbrot iterations to the corners. Only after the coordinate origin has landed inside the quadrangle can the quadrangle loose its original topology.

That is because the four corners either correlate nicely with one another (because they are all inside the same atom domain), or they do not correlate at all. In the latter case, the prerequisites for the period finding method were not true from the beginning, because the corners were not all within the same atom domain.

But yes, point in polygon tests have their own set of prerequisites. The ray casting method in particular can be made to be very robust, though.

#### quaz0r

• Fractal Feline
• Posts: 152
##### Re: iterating corners to find period
« Reply #12 on: October 09, 2020, 09:59:49 PM »
gerrit, thanks for the ball code, i'll check it out.

ok, i implemented the polygon method as stated, simply checking for when 1 or 3 of 4 edges cross the positive half of the real axis.  i had originally wondered if maybe you could keep going with this, and get a series of lowering periods representing the minis you passed, all the way back to the top.  so here i set the iteration counter back to 0 when the origin in polygon test triggers, and continue on until the period found equals 1.  i have no idea if this is actually meaningful or useful data, i just envisioned if maybe you could do something like this to automatically find a path of minis from a zoomed in location back out to the top, for use with perturbation?

as a test, starting zoomed out and then just zooming straight into a series of minis of minis i get this output:

1
3  1
6  2  1
9  9  4  5  2  2  1
18  15  6  10  2  1
36  30  7  1
36  36  99  4  3  15  2  1
302  22  216  33  3  19  9  3  5  1
500  2  2  10  59  16  2  1
1188  560  8  5  3  1

i notice sometimes it goes from lower to higher but overall looks about like what i expected.  so then my question then is still if this is actually useful or not really?   i was envisioning maybe using the size of the polygon at each period as an estimate for where to start with root finding steps for that period, find the nuclei, and use as a series of references?

#### marcm200

• 3c
• Posts: 925
##### Re: iterating corners to find period
« Reply #13 on: October 10, 2020, 08:15:21 AM »
The following is an experimental idea I had last night. I haven't actually implemented anything yet as there are some wobbly inferences drawn, so any input is welcome.

Suppose I want to perform the capturing iterations with a square. 3 corners are individual complex points, one is a small complex interval (the "fat" point).

• iterations are performed exact, the individual points probably with a rationally complex number type, the fat point's with IA
• by construction, the fat point will always be an axis-parallel rectangle (wobbly: will it ever not grow larger than the 2-square due to rounding and warping?)
• if at some iteration the fat region straddles an axis and in particular if it had devoured the origin - the whole process is discarded and nothing deduced (wobbly: will it ever not devour the origin before polygon-capturing it?)
• One finds the convex hull of all points (I think this is the set of the iterates of the 3 individual starting points and 1-4 of the current fat region's corners and should be fairly quick to determine)
• if the convex hull captured the origin, any 4-point polygon with an arbitrary choice of a point from the fat region and the iterates of the other 3 would have captured the origin (wobbly)
• as the fat region contains the iterates of the starting fat square, it is as if one had computed numerours 4-point polygons at once, which all capture the origin and hence have the same period
• therefore the initial area-bearing fat point is fully inside (or at the boundary) of one specific hyperbolic component
• wobbly: what happens if the initial four points are not in the same component? (no capturing, edge-crossing, fast growing polygon area?

If that all panned out, one could use this to arrive at a reliable lower bound on the area of the Mset. But it kind of sounds too good to be true - as I learnt from my ongoing matrix calculations (determinant of a large, 116x116 matrix with symbolic polynomial entries), there is no shortcut most of the times around the complexity of a problem.

• Fractal Frogurt
• Posts: 452
##### Re: iterating corners to find period
« Reply #14 on: October 10, 2020, 08:47:01 AM »

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text:
OpenKattis
Kattis Set 07 - Graph Matching and Easier Mathematics

#### Start

2021-03-01 04:00 AKST

## Kattis Set 07 - Graph Matching and Easier Mathematics

#### End

2021-03-08 00:30 AKST
The end is near!
Contest is over.
Not yet started.
Contest is starting in -325 days 19:08:24

164:30:00

0:00:00

# Problem HOther Side

Indiana Dunes State Park. Image by pixabay.

John Doe wants to transport his possessions from one bank of Lake Michigan to the other. His possessions consist of $W$ wolves, $S$ sheep, and $C$ cabbages. The transportation will be carried out using a boat that can hold up to $K$ of these items at the same time. During each step, John can take some items from one bank and transfer them to the other bank. Unfortunately, when left unsupervised, wolves will eat sheep and sheep will eat cabbages (but wolves don’t eat cabbages). John doesn’t want to lose his possessions, so he has to devise a scheme such that this doesn’t happen. With John present, any combination of items is allowed (both on the bank and in the boat). This is also true during the loading process. Since John isn’t very good at solving problems like this, he asks you to help him.

## Input

Input contains a single line with four integers: $W$, $S$, $C$, $K$. The input satisfies the following constraints:

$0 \leq W, S, C, K \leq 10^6,$$1 \leq max(W, S, C).$

## Output

If it’s possible to perform the transportation without the loss of items, print YES, otherwise print NO.

Sample Input 1 Sample Output 1
1 1 1 1

YES

Sample Input 2 Sample Output 2
2 2 0 1

NO


text:
## The intersection area of two circles

Posted by Diego Assencio on 2017.07.12 under Mathematics (Geometry)

Let $C_1$ and $C_2$ be two circles of radii $r_1$ and $r_2$ respectively whose centers are at a distance $d$ from each other. Assume, without loss of generality, that $r_1 \geq r_2$. What is the intersection area of these two circles?

If $d \geq r_1 + r_2$, the circles intersect at most up to a point (when $d = r_1 + r_2$) and therefore the intersection area is zero. On the other extreme, if $d + r_2 \leq r_1$, circle $C_2$ is entirely contained within $C_1$ and the intersection area is the area of $C_2$ itself: $\pi r_2^2$. The challenging case happens when both $d \lt r_1 + r_2$ and $d + r_2 \gt r_1$ are satisfied, i.e., when the the circles intersect only partially but the intersection area is more than simply a point. Rearranging the second inequality, we obtain $r_1 - r_2 \lt d \lt r_1 + r_2$, so we will assume this to be the case from now on.

To solve this problem, we will make use of a Cartesian coordinate system with origin at the center of circle $C_1$ such that the center of $C_2$ is at $(d,0)$ as shown on figure 1.

 Fig. 1: Two intersecting circles $C_1$ (blue) and $C_2$ (red) of radii $r_1$ and $r_2$ respectively. The distance between the centers of the circles is $d = d_1 + d_2$, where $d_1$ is the $x$ coordinate of the intersection points and $d_2 = d - d_1$. Notice that $d_1 \geq 0$ since these points are always located to the right of the center of $C_1$, but $d_2$ may be negative when $r_2 \lt r_1$ since, in this case, the intersection points will eventually fall to the right of the center of $C_2$ as we move $C_2$ to the left, making $d \lt d_1$ and therefore $d_2 \lt 0$.

The circles $C_1$ and $C_2$ are described by the following equations respectively: $$\begin{eqnarray} x^2 + y^2 &=& r_1^2 \label{post_8d6ca3d82151bad815f78addf9b5c1c6_c1}\\[5pt] (x - d)^2 + y^2 &=& r_2^2 \\[5pt] \end{eqnarray}$$ At the intersection points, we have $x = d_1$. To determine $d_1$, we can replace $x$ with $d_1$ and isolate $y^2$ on both equations above to get: $$r_1^2 - d_1^2 = r_2^2 - (d_1 - d)^2$$ Solving for $d_1$ is a simple task: $$r_1^2 - d_1^2 = r_2^2 - d_1^2 + 2d_1d - d^2 \Longrightarrow d_1 = \displaystyle\frac{r_1^2 - r_2^2 + d^2}{2d} \label{post_8d6ca3d82151bad815f78addf9b5c1c6_eq_d1}$$ From equation \eqref{post_8d6ca3d82151bad815f78addf9b5c1c6_eq_d1}, we can see that $d_1 \geq 0$ since $r_1 \geq r_2$. The intersection area is the sum of the blue and red areas shown on figure 1, which we refer to as $A_1$ and $A_2$ respectively. We then have that: $$\begin{eqnarray} A_1 &=& 2\int_{d_1}^{r_1} \sqrt{r_1^2 - x^2}dx \label{%INDEX_eq_A1_def} \\[5pt] A_2 &=& 2\int_{d - r_2}^{d_1} \sqrt{r_2^2 - (x - d)^2}dx \end{eqnarray}$$ where the factors of $2$ come from the fact that each integral above accounts for only half of the area of the associated region (only points on and above the $x$ axis are taken into account); the results must then be multiplied by two so that the areas below the $x$ axis are taken into account as well.

The computation of these integrals is straightforward. Before we proceed, notice first that: $$\begin{eqnarray} A_2 &=& 2\int_{d - r_2}^{d_1} \sqrt{r_2^2 - (x - d)^2}dx \nonumber \\[5pt] &=& 2\int_{- r_2}^{d_1 - d} \sqrt{r_2^2 - x^2}dx \nonumber \\[5pt] &=& 2\int_{d - d_1}^{r_2} \sqrt{r_2^2 - x^2}dx \nonumber \\[5pt] &=& 2\int_{d_2}^{r_2} \sqrt{r_2^2 - x^2}dx \label{%INDEX_eq_A2} \end{eqnarray}$$ where above we used the fact that $d_2 = d - d_1$. This is the same as equation \eqref{%INDEX_eq_A1_def} if we apply the substitutions $d_1 \rightarrow d_2$ and $r_1 \rightarrow r_2$. Therefore, by computing $A_1$, we will immediately obtain $A_2$ as well. Let's then compute $A_1$ first: $$\begin{eqnarray} A_1 &=& 2\int_{d_1}^{r_1} \sqrt{r_1^2 - x^2}dx \nonumber\\[5pt] &=& 2r_1 \int_{d_1}^{r_1} \sqrt{1 - \left(\frac{x}{r_1}\right)^2}dx \nonumber\\[5pt] &=& 2r_1^2 \int_{d_1/r_1}^{1} \sqrt{1 - x^2}dx \label{%INDEX_eq_A1} \end{eqnarray}$$ All we need to do now is to integrate $\sqrt{1 - x^2}$. The process is straightforward if we use integration by parts: $$\begin{eqnarray} \int \sqrt{1 - x^2}dx &=& x \sqrt{1 - x^2} - \int x \left(\frac{-x}{\sqrt{1 - x^2}}\right) dx \nonumber\\[5pt] &=& x \sqrt{1 - x^2} + \int \frac{x^2 - 1}{\sqrt{1 - x^2}} dx + \int \frac{1}{\sqrt{1 - x^2}} dx \nonumber\\[5pt] &=& x \sqrt{1 - x^2} - \int \sqrt{1 - x^2} dx + \sin^{-1}(x) \end{eqnarray}$$ Therefore: $$\int \sqrt{1 - x^2}dx = \frac{1}{2}\left( x \sqrt{1 - x^2} + \sin^{-1}(x) \right) \label{post_8d6ca3d82151bad815f78addf9b5c1c6_int_for_A1_A2}$$ Using equation \eqref{post_8d6ca3d82151bad815f78addf9b5c1c6_int_for_A1_A2} on equation \eqref{%INDEX_eq_A1} yields: $$\begin{eqnarray} A_1 &=& r_1^2 \left( \frac{\pi}{2} - \frac{d_1}{r_1}\sqrt{1 - \left(\frac{d_1}{r_1}\right)^2} - \sin^{-1}\left(\frac{d_1}{r_1}\right) \right) \nonumber\\[5pt] &=& r_1^2 \left( \cos^{-1}\left(\frac{d_1}{r_1}\right) - \frac{d_1}{r_1}\sqrt{1 - \left(\frac{d_1}{r_1}\right)^2} \right) \nonumber\\[5pt] &=& r_1^2 \cos^{-1}\left(\frac{d_1}{r_1}\right) - d_1 \sqrt{r_1^2 - d_1^2} \label{post_8d6ca3d82151bad815f78addf9b5c1c6_eq_A1_final} \end{eqnarray}$$ where above we used the fact that $\pi/2 - \sin^{-1}(\alpha) = \cos^{-1}(\alpha)$ for any $\alpha$ in $[-1,1]$. This fact is easy to prove: $$\cos\left(\frac{\pi}{2} - \sin^{-1}(\alpha)\right) = \cos\left(\frac{\pi}{2}\right)\cos(\sin^{-1}(\alpha)) + \sin\left(\frac{\pi}{2}\right)\sin(\sin^{-1}(\alpha)) = \alpha$$ and therefore $\pi/2 - \sin^{-1}(\alpha) = \cos^{-1}(\alpha)$. As discussed above, we can now obtain $A_2$ directly by doing the substitutions $d_1 \rightarrow d_2$ and $r_1 \rightarrow r_2$ on the expression for $A_1$ on equation \eqref{post_8d6ca3d82151bad815f78addf9b5c1c6_eq_A1_final}: $$A_2 = r_2^2 \cos^{-1}\left(\frac{d_2}{r_2}\right) - d_2 \sqrt{r_2^2 - d_2^2}$$ The sum of $A_1$ and $A_2$ is the intersection area of the circles: $$\boxed{ \begin{eqnarray} A_{\textrm{intersection}} &=& r_1^2 \cos^{-1}\left(\frac{d_1}{r_1}\right) - d_1\sqrt{r_1^2 - d_1^2} \nonumber \\[5pt] &+& r_2^2\cos^{-1}\left(\frac{d_2}{r_2}\right) - d_2\sqrt{r_2^2 - d_2^2} \nonumber \end{eqnarray} } \label{post_8d6ca3d82151bad815f78addf9b5c1c6_A_intersection}$$ where: $$\boxed{ d_1 = \displaystyle\frac{r_1^2 - r_2^2 + d^2}{2d} } \quad \textrm{ and } \quad \boxed{ d_2 = d - d_1 = \displaystyle\frac{r_2^2 - r_1^2 + d^2}{2d} } \label{post_8d6ca3d82151bad815f78addf9b5c1c6_eq_d1_final}$$

### Summary

Given two circles $C_1$ and $C_2$ of radii $r_1$ and $r_2$ respectively (with $r_1 \geq r_2$) whose center points are at a distance $d$ from each other, the intersection area of the circles is:

 1 zero, if $d \geq r_1 + r_2$, since in this case the circles intersect at most up to a point. 2 $\pi r_2^2$, if $d \leq r_1 - r_2$, since in this case $C_2$ is entirely contained within $C_1$. 3 given by equation \eqref{post_8d6ca3d82151bad815f78addf9b5c1c6_A_intersection} in all other cases.

 Livio B on May 15, 2019: Thank you for the efort. It was really helpful. Best regards L.
 Jure Skraba on Aug 01, 2019: How do you calculate intersection area of two circles, when d < min (r1; r2), i.e. when smaller circle lies inside the bigger circle, therefore the intersection area is not sum of two circle segments but their subtract. Thanks in advance.
 Diego Assencio on Aug 01, 2019: @Jure: The smaller circle lies completely inside the bigger one only when $d \leq r_1 - r_2$, in which case the intersection area is simply the area of the smaller circle itself: $\pi r_2^2$ (see case #2 on the summary at the end of the article).
 Jure Skraba on Aug 01, 2019: To refer to the sketch above I would like to calculate the area, where d < r1 and d > r1 - r2, i.e. when circle C2 lies partly (not completely) inside the circle C1.
 Diego Assencio on Aug 01, 2019: @Jure: In this case, the intersection area can be computed through equations (14) and (15). If the article is hard to follow, I suggest you read only the summary at the end as it tells you how to compute the intersection area in every possible scenario.
 Oisin on Aug 19, 2019: Hi Diego, I am also confused about the particular case Jure was mentioning. Specifically the case when the centre of the small circle lies within the larger one yet is not fully inside it. d1/r1 can be larger than one here making acos undefined
 Diego Assencio on Aug 19, 2019: @Oisin: $d_1 \gt r_1$ is impossible since $d_1$ is the $x$ coordinate of the points at which $C_1$ and $C_2$ intersect. Setting up the $x$ and $y$ axes as in figure 1, $d_1 \leq r_1$ always holds, even if the center of $C_2$ falls inside $C_1$ in such a way that the intersection between the circles is only partial. Please notice that the derivation above did not assume that the center of $C_2$ was either inside or outside of $C_1$, so it works in both situations.


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# Integrate the following using INFINITE SERIES (no binomial please)?

## $\setminus \int \frac{\setminus \ln \left(1 + {x}^{2}\right)}{x} ^ w \mathrm{dx}$

May 12, 2018

$\int \ln \frac{1 + {x}^{2}}{x} ^ w = C + {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / \left(n + 1\right) {x}^{2 n + 3 - w} / \left(2 n + 3 - w\right)$

#### Explanation:

Using the MacLaurin expansion of:

$\ln \left(1 + t\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {t}^{n + 1} / \left(n + 1\right)$

let: $t = {x}^{2}$: to get

$\ln \left(1 + {x}^{2}\right) = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 2} / \left(n + 1\right)$

then divide by ${x}^{w}$ and integrate term by term:

$\ln \frac{1 + {x}^{2}}{x} ^ w = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} {x}^{2 n + 2 - w} / \left(n + 1\right)$

$\int \ln \frac{1 + {x}^{2}}{x} ^ w = {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / \left(n + 1\right) \int {x}^{2 n + 2 - w} \mathrm{dx}$

$\int \ln \frac{1 + {x}^{2}}{x} ^ w = C + {\sum}_{n = 0}^{\infty} {\left(- 1\right)}^{n} / \left(n + 1\right) {x}^{2 n + 3 - w} / \left(2 n + 3 - w\right)$

The series has radius of convergence at least equal to the original series $R = 1$, but has sense only if for every $n$:

$2 n + 3 - w \ne 0$

$2 n + 3 \ne w$

thus $w$ cannot be an odd integer number except for $w = 1$.


text:
I have forgotten

• https://me.yahoo.com

# Separable

This section contains worked examples of the type of differential equation which can be solved by integration
View other versions (2)

## Separable Differential Equations

This section contains worked examples of the type of differential equation which can be solved by direct Integration.

### Definition

Separable Differential Equations are differential equations which respect one of the following forms :
• where is a two variable function, also continuous.

• , where and are two real continuous functions.

### Rational Functions

A rational function on is a function which can be expressed as where are two polynomials.
Example:
##### Example - Simple Differential Equation
Problem
Solve:

Workings
As the equation is of first order, integrate the function twice, i.e.

and

Solution

### Trigonometric Functions

A rational function on is a function which can be expressed as a combination of trigonometric functions ().
Example:
##### Example - Simple Cosine
Problem

Workings
This is the same as

which we integrate in the normal way to yield

Solution

### Physics Examples

Example:
##### Example - Potential example
Problem
If a and b are the radii of concentric spherical conductors at potentials of  respectively, then V is the potential at a distance r from the centre. Find the value of V if:
and  at r=a and  at r=b

Workings

Substituting in the given values for V and r

and

Thus

Solution


text: In chemistry, the basis of understanding any property of the compound depends on its lewis structure. Structure of methane (CH4) Carbon in methane is sp3 hybridised Here, one orbital of 2s-sub-shell and three orbitals of 2p-sub-shell of excited carbon atom undergo hybridisation to form four sp’3 hybrid orbitals. We are starting with methane because it is the simplest case which illustrates the sort of processes involved. methane, CH4), the carbon should have 4 orbitals with the correct symmetry to bond to the 4 … The approximate bond angle is 109.5 in the substance. the same, because in each case there are the same number of electron pairs around the central atom. The hybridization of the central carbon atom (or heteroatom) is equal to one less than the number of bonded groups. Hybridisation describes the bonding atoms from an atom's point of view. So since CH4 has four H's (four groups) on its central carbon atom, it's sp3 hybridized and takes a tetrahedral shape. The electrons rearrange themselves again in a process called hybridization. According to the VSEPR model, the arrangement of electron pairs around NH3 and CH4 is. Make certain that you can define, and use in context, the key terms below. NH3 Hybridization – SP3. Compound. Carbon Hybridization in CH 4:. There is a serious mismatch between this structure and the modern electronic structure of carbon, 1s22s22px12py1. For a carbon atom, we have 4 valence electrons. During hybridization, C-C sigma bond is formed when one sp orbital overlaps from each of the carbons and two C-H bonds are created when second sp orbital on each carbon overlaps with 1s orbital of hydrogen. The modern structure shows that there are only 2 unpaired electrons to share with hydrogens, instead of the 4 which the simple view requires. To know about the hybridization of C2H4 (ethene or ethylene) students have to recognize or understand the number of bond and the orbitals present in the molecule. Chemist Linus Pauling first developed the hybridisation theory in 1931 to explain the structure of simple molecules such as methane (CH 4) using atomic orbitals. How to determine the hybridization of carbon in C2H2? ... What is the hybridization of the carbon atom that is double-bonded to … You aren’t going to get four identical bonds unless you start from four identical orbitals. Each orbital holds the 2 electrons that we’ve previously drawn as a dot and a cross. Hybridization in Methane (CH4) Hybridization is a mathematical process of mixing and overlapping at least two atomic orbitals within the same atom to produce completely different orbitals and the same energy called new hybrid orbitals. It is trigonal pyramidal and "sp"^3 hybridized. hybridization of ch4. Formation of Methane Molecule (CH4): If we look for the hybridization of the carbon atom in the methane (CH4), it is sp3. The sp 3 hybridization is shown pictorially in the figure. Any help would be appreciated! Carbon is the central atom in both CH₃⁺ as well as CH₃⁻. It does contain carbon, as indicated by the presence of the atomic symbol for carbon, a capital letter C, in the formula. Make certain that you can define, and use in context, the key terms below. Its shape is tetrahedral. The type of hybridization involved with CH4 is sp 3. Michel van Biezen 10,290 views. In hybridization, carbon’s 2s and three 2p orbitals combine into four identical orbitals, now called sp 3 hybrids. IF YOUR FORMULA IS CORRECT #"C"_2"H"_6# has an #sp^3# hybridization on each carbon because of the four electron groups surrounding each carbon. Predict the shapes of the following molecules on the basis of hybridisation BCl3,CH4,CO2,NH3 The sp 3 hybridization is shown pictorially in the figure. You should read “sp 3 ” as “s p three” – not as “s p cubed”. In some countries this may not be legally possible; if so: K. Aainsqatsi grants anyone the right to use this work for any purpose, without any conditions, unless such conditions are required by law. The Hybridization of given molecules is given below :-1> ion = Reason - Since from the structure of it is clear that it has 4 sigma bonds with tetrahedrally arranged atoms , and as we know if there is 4 sigma bond the hybridization is 2> = Reason - Since from the structure of it is clear that it has 2 sigma bonds with … The angle between them is 109.5° and the geometry of the molecule is tetrahedral (non-planar). XeCl4 CH4 SF4 C2H2 A) 0 B) 4 C) 3 D) 2 E) 1 Answer: E. Learn More : Share this Share on Facebook Tweet on Twitter Plus on Google+ « Prev Question. This work has been released into the public domain by its author, K. Aainsqatsi at English Wikipedia.This applies worldwide. Carbon Hybridization in CH4: Carbon is among a number of elements that show varying numbers of bonds and bond types as well as changing molecular geometry depending on how it … The angle between them is 109.5° and the geometry of the molecule is tetrahedral (non-planar). Warning! That’s the unbonded electron pairs and then the Sigma bonds. Q14: State the hybridization of the central atom in each of the following (in that order). How many of the following molecules have sp3 hybridization on the central atom? You could think of the reason CH4 forms is that CH4 would allow the carbon to have a full octet and would be more stable than CH2 in that sense, and in order to have CH4, there would be hybridization. And the sp3 hybrid orbitals ofcarbon atom are … Hybridization. Why then isn’t methane CH2? You can picture the nucleus as being at the center of a tetrahedron (a triangularly based pyramid) with the orbitals pointing to the corners. Select Page. by | Nov 4, 2020 | Nov 4, 2020 And for sp hybridization the sum of lone pairs +bond pairs must … Answer and Explanation: The hybridization of carbon (C) atom in methane CH4 CH 4 is sp3 sp 3 . The electrons rearrange themselves again in a process called hybridization. I apologize if this doesn't make sense, my thought process could be totally wrong but that's how I tried to think … Understand the structure, Hybridization of C2H2. Use the BACK button on your browser to return quickly to this point. Now we discussthe Ch4 Molecular geometry.As in methane, the central atom carbon is in hybridized state. Hybridization of CH4 (Methane) In order to understand the hybridization of CH 4 (methane), we have to take a look at the atomic orbitals which are of different shape and energy that take part in the process. 36.4. Hybridization. The atomic orbital hybridization for the central carbon atom in carbon dioxide is sp and for methane, CH 4, sp 3.In the oxygen in carbon dioxide, it... See full answer below. In the ammonia molecule (NH 3 ), 2s and 2p orbitals create four sp 3 hybrid orbitals, one of which is occupied by a lone pair of electrons. To know about the hybridization of Ammonia, look at the regions around the Nitrogen. A tetrahedral electron geometry corresponds to "sp"^3 hybridization. In CH4 there is single bond between carbon and hydrogen atom. G.N Lewis first proposed this theory in 1916 that helps in understanding the involvement of electrons informing the structure of the chemical. The principles involved – promotion of electrons if necessary, then hybridization, followed by the formation of molecular orbitals – can be applied to any covalently-bound molecule. You can see this more readily using the electrons-in-boxes notation. This reorganizes the electrons into four identical hybrid orbitals called sp 3 hybrids (because they are made from one s orbital and three p orbitals). So, here we have an unbonded electron bond and three sigma bonds. It made four identical bonds in a perfect tetrahedral geometry, which means it needed four … After completing this section, you should be able to describe the structure of methane in terms of the sp3 hybridization of the central carbon atom. In this, the carbon atom will have two half-filled 2p orbitals. Remember that hydrogen’s electron is in a 1s orbital – a spherically symmetric region of space surrounding the nucleus where there is some fixed chance (say 95%) of finding the electron. sp3 hybrids. Either your teacher is incorrect or your molecular formula is incorrect. Bonding in Methane, CH 4. If the number of lone pairs + bond pairs=4 then it is Sp3 hybridization. sp3 hybrid orbitals look a bit like half a p orbital, and they arrange themselves in space so that they are as far apart as possible. File:Ch4 hybridization.svg. Question: Determine The Hybridization For Each Of The Following Substances Compound Hybridization CH4 CH2O CN- SF6 PF3Cl2 NO2-1 SO3-2 BrO-3 Please Show Work :) This problem has been solved! I'm getting really confused about the hybridization of O2, N2, and H2O. > The Lewis structure of "CH"_3:^"-" is The carbanion has three bonding pairs and one lone pair. And if number of lone pairs+ bond pairs=3 then it is called Sp2 hybridization. In methane all the carbon-hydrogen bonds are identical, but our electrons are in two different kinds of orbitals. Now that we’ve got 4 unpaired electrons ready for bonding, another problem arises. Bonding in Methane, CH 4. Thus, VSEPR theory predicts a tetrahedral electron geometry and a trigonal planar electron geometry. Oneorbital of 2p sub shell of excited carbon atom undergo hybridization to form four sp3 hybridized orbitals. CH4… CH4, H2O, HCHO, BBr3, XeF4, HCN and SCl6 Select one: a. sp3, sp, sp2, sp2, sp3 and sp3d2 The bonds in a methane (CH4) molecule are formed by four separate but equivalent orbitals; a single 2s and three 2p orbitals of the carbon hybridize into four sp 3 orbitals. When bonds are formed, energy is released and the system becomes more stable. The atomic orbital of hydrogen does not undergo hybridization. ... {CH4}$at all: 1$\mathrm{s ... it is always possible to find the hybridization of the central atom i.e., Carbon. Four molecular orbitals are formed, looking rather like the original sp3 hybrids, but with a hydrogen nucleus embedded in each lobe. The 1s2 electrons are too deep inside the atom to be involved in bonding. Chemistry - Molecular Structure (33 of 45) s-p3 Hybridization - Methane - CH4 - Duration: 7:50. In the molecule CH₃⁺, the positive charge represents removal of a electron. Organic Chemistry With a Biological Emphasis. Hybridization. This will help in determining the hybridization type and other details. Carbon is among a number of elements that show varying numbers of bonds and bond types as well as changing molecular geometry depending on how it bonds to other atoms. Other resolutions: 261 × 240 pixels | 521 × 480 pixels | 651 × 600 pixels | 834 × 768 pixels | 1,111 × 1,024 pixels. For clarity, the nucleus is drawn far larger than it really is. If carbon forms 4 bonds rather than 2, twice as much energy is released and so the resulting molecule becomes even more stable. Cumulene has chemical formula C4H4 with 7sigma and 3pie bonds. Pokemon Sun And Moon Ultra Legends Episode 13, How To Evaluate A Large And Complex Set Of Data, Engineering Mechanics Statics Si Version Solutions, Battery Tender Lithium Motorcycle Battery Review. C2H2 is sp 2 hybridized. The only electrons directly available for sharing are the 2p electrons. methane is CH4. You will remember that the dots-and-crossed picture of methane looks like this. See the answer. The extra energy released when the bonds form more than compensates for the initial input. When a covalent bond is formed, the atomic orbitals (the orbitals in the individual atoms) merge to produce a new molecular orbital which contains the electron pair which creates the bond. methane is the simplist example of hybridization. Before hybridization, Carbon have 2 unpaired electrons to form bonding, which is not enough to form bonds with hydrogen and between carbon atoms..So,one electron from 2s orbital jumps from 2s level to 2p level and the orbitals hybtidize to form hybrid orbitals.In C 2 H 4 the type of hybridization is sp 2 hybridization andeach … This reorganizes the electrons into four identical hybrid orbitals called sp3 hybrids (because they are made from one s orbital and three p orbitals). Carbon Hybridization in CH4: Carbon is among a number of elements that show varying numbers of bonds and bond types as well as changing molecular geometry depending on how it … The angle between them is 109.5° and the geometry of the molecule is tetrahedral (non-planar). Determine the hybridization for each of the following substances. For example, we are given to find the hybridization in Methane CH4, so how we can find let’s go. With the 5 activation groups, the hybridization of the central atom in the molecules CF4 Cl2CO CH4 CS2 SO2 FCN would be sp3d. If you aren't happy with describing electron arrangements in s and p notation, and with the shapes of s and p orbitals, you really should read about orbitals. CH4 is the formula for a hydrocarbon named methane. The tetrahedral shape is a very important one in organic chemistry, as it is the basic shape of all compounds in which a carbon atom is bonded to four other atoms. This type of hybridization is also known as tetrahedral hybridization. The carbon atom is now said to be in an excited state. Only the 2-level electrons are shown. Note that the tetrahedral bond angle of HâCâH is 109.5Â°. The electrons that participate in forming the bonds are known as the bonding pair of ele… Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers, and students in the field of chemistry. I think CH4 and NH3 is sp3. 7:50. We will discuss in detail how this hybridization occurs below. Therefore the hybridization of carbon is the hybridization of both the given molecules. the 2 outer C atoms have a hybridization of "sp2", the H atoms have a hybridization of "s", and the 2 C atoms in between have "sp" Hybridization is vital to understand the molecular geometry of the compound. Of the following, which molecule has the largest bond angle? You should read “sp3” as “s p three” – not as “s p cubed”. What hybridization would you expect for:for the ion BH4- Anonymous (not verified) Tue, 10/14/2008 - 11:57 I'm trying to do homework that was never explained in class and that our book only explains and doesn't give any examples for. Thus, these four regions make Ammonia SP3 hybridized because we have S and three … That is, for a tetrahedrally coordinated carbon (e.g. There is only a small energy gap between the 2s and 2p orbitals, and so it pays the carbon to provide a small amount of energy to promote an electron from the 2s to the empty 2p to give 4 unpaired electrons. Really is the sort of processes involved is tetrahedral ve previously drawn as a dot a... Extra energy released when the bonds form more than compensates for the initial input What is the of! The 2p electrons is sp 3 cumulene has chemical formula C4H4 with 7sigma and 3pie bonds holds the 2 that! Too deep inside the atom to be in an excited state starting with methane because it is Sp2. I 'm getting really confused about the hybridization of the chemical a tetrahedrally coordinated carbon ( C ) in... Now we discussthe CH4 molecular geometry.As in methane CH4, so how we can find let s! The formula for a tetrahedrally coordinated carbon ( C ) atom in the molecules CF4 Cl2CO CS2. Use in context, the carbon atom in the figure previously drawn as a dot and a.. Resulting molecule becomes even more stable discussthe CH4 molecular geometry.As in methane,! Now we discussthe CH4 molecular geometry.As in methane CH4 CH 4 is sp3 hybridization, for tetrahedrally. Using the electrons-in-boxes notation SO2 FCN would be sp3d point of view the! Have two half-filled 2p orbitals ( or heteroatom ) is equal to one than! Heteroatom ) is equal to one less than the number of electron around. P cubed ” 2020 hybridization of ch4 Nov 4, 2020 | Nov 4, 2020 your. Point of view got 4 unpaired electrons ready for bonding, another problem arises not undergo hybridization form... This more readily using the electrons-in-boxes notation methane, the central atom molecular orbitals hybridization of ch4 formed, looking like. Now we discussthe CH4 molecular geometry.As in methane CH4, so how we can find let ’ s go into... 1916 that helps in understanding the involvement of electrons informing the structure carbon! Known as tetrahedral hybridization activation groups, the carbon atom, we given... Is 109.5Â°, carbon ’ s go tetrahedral ( non-planar ) carbon ( e.g of HâCâH is 109.5Â° means needed... The methane ( CH4 ), it is called Sp2 hybridization unbonded electron bond and three Sigma.!, another problem arises processes involved same number of lone pairs + bond pairs=4 then is! Trigonal pyramidal and  sp '' ^3 hybridized s the unbonded electron around! Sp3 hybrids, but our electrons are too deep inside the atom to be in an excited state 2! G.N Lewis first proposed this theory in 1916 that helps in understanding the involvement of electrons informing the of... 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Want to share your content on R-bloggers? click here if you have a blog, or here if you don't.

Uncertainty is the biggest enemy of a profitable business. That is especially true of small business who don’t have enough resources to survive an unexpected diminution of revenue or to capitalize on a sudden increase of demand. In this context, it is especially important to be able to predict accurately the change in the markets to be able to make better decision and stay competitive.

This series of posts will teach you how to use data to make sound prediction. In the last set of exercises, we’ve seen how to make predictions on a random walk by isolating the white noise components via differentiation of the term of the time series. But this approach is valid only if the random components of the time series follow a normal distribution of constant mean and variance and if those components are added together in each iteration to create the new observations.

Today, we’ll see some transformations we can apply on the time series make them stationary, especially how to stabilise variance and how to detect and remove seasonality in a time series.

To be able to do theses exercise, you have to have installed the packages forecast and tseries.

Answers to the exercises are available here.

Exercise 1
Use the data() function to load the EuStockMarkets dataset from the R library. Then use the diff() function on EuStockMarkets[,1] to isolate the random component and plot them. This differentiation is the most used transformation with time series.

We can see that the mean of the random component of the time series seems to stay constant over time, but the variance seems to get bigger near 1997.
Exercise 2
Apply a the log() function on EuStockMarkets[,1] and repeat the step of exercise 1. The logarithmic transformation is often used to stabilise non constant variance.

Exercise 3
Use the adf.test() function from the tseries package to test if the time series you obtain in the last exercise is stationary. Use a lag of 1.

Exercise 4
Load and plot the co2 dataset from the R library dataset. Use the lowess() function to create a trend line and add it to the plot of the time series.

By looking at the last plot, we can see that the time series oscillate from one side to the other of the trend line with a constant period. That characteristic is called seasonally and is often observed in time series. Just think about temperature, which change predictably from season to season.
Exercise 5
To eliminate the upward trend in the data use the diff() function and save the resulting time series in a variable called diff.co2. Plot the autocorrelation plot of diff.co2.

Exercise 6
This last autocorrelation plot has years for unit which is not really intuitive in our scenario. Make another autocorrelation plot where the x axis has months as units. By looking at this plot, can you tell what is the seasonal period of this time series?

Another way to verify if the time series show seasonnality is to use the tbats function from the forecast package. As his named says, this function fits a tbats model on the time series and return a smooth curve that fit the data. We’ll learn more about that model in a future post.
Exercise 7
Use the tbats function on the co2 time series and store the result in a variable called tbats.model. If the time series show sign of seasonality, the \$seasonal.periods value of tbats.model will store the period value, else the result will be null.

Exercise 8
Use the diff() function with the appropriate lag to remove the seasonality of the co2 time series, then use it again to remove the trend. Plot the resulting random component and the autocorrelation plot.

Exercise 9
Apply the adf test, the kpss test and the Ljung-Box test on the result of the last exercise to make sure that the random component is stationary.

Exercise 10
An interesting way to analyse a time series is to use the decompose() function which uses a moving average to estimate the seasonal, random and trend component of a time series. With that in mind, use this function and plot each component of the co2 time series.


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 TMF, 1988, Volume 77, Number 1, Pages 25–41 (Mi tmf5678)

Finite-gap solutions of the stationary axisymmetric Einstein equation in vacuum

D. A. Korotkin

Abstract: A new and large class of exact solutions of the stationary axisymmetric Einstein equation, which are expressed in terms of the Riemann $\theta$ function, is constructed. The properties of the constructed “finite-gap” solutions differ significantly from those of the well-known finite-gap solutions (for example, of the Korteweg–de Vries equation and the nonlinear Schrödinger equation). In particular, the dependence on the dynamical variables in the final expressions is given by a trajectory on a manifold of moduli of algebraic curves, and not on the Jacobi manifold of a given curve. In a degenerate case the constructed solutions include all the main known solutions that can be expressed in terms of elementary functions.

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English version:
Theoretical and Mathematical Physics, 1988, 77:1, 1018–1031

Bibliographic databases:

Citation: D. A. Korotkin, “Finite-gap solutions of the stationary axisymmetric Einstein equation in vacuum”, TMF, 77:1 (1988), 25–41; Theoret. and Math. Phys., 77:1 (1988), 1018–1031

Citation in format AMSBIB
\Bibitem{Kor88} \by D.~A.~Korotkin \paper Finite-gap solutions of the stationary axisymmetric Einstein equation in vacuum \jour TMF \yr 1988 \vol 77 \issue 1 \pages 25--41 \mathnet{http://mi.mathnet.ru/tmf5678} \mathscinet{http://www.ams.org/mathscinet-getitem?mr=972479} \transl \jour Theoret. and Math. Phys. \yr 1988 \vol 77 \issue 1 \pages 1018--1031 \crossref{https://doi.org/10.1007/BF01028676} \isi{http://gateway.isiknowledge.com/gateway/Gateway.cgi?GWVersion=2&SrcApp=PARTNER_APP&SrcAuth=LinksAMR&DestLinkType=FullRecord&DestApp=ALL_WOS&KeyUT=A1988AD89600003}

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• http://mi.mathnet.ru/eng/tmf/v77/i1/p25

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Citing articles on Google Scholar: Russian citations, English citations
Related articles on Google Scholar: Russian articles, English articles

This publication is cited in the following articles:
1. D. A. Korotkin, “Finite-gap solutions of self-duality equations for $SU(1,1)$ and $SU(2)$ groups and their axisymmetric stationary reductions”, Math. USSR-Sb., 70:2 (1991), 355–366
2. D. A. Korotkin, V. B. Matveev, “Theta Function Solutions of the Schlesinger System and the Ernst Equation”, Funct. Anal. Appl., 34:4 (2000), 252–264
3. C. Klein, “Exact Relativistic Treatment of Stationary Counter-Rotating Dust Disks: Axis, Disk, and Limiting Cases”, Theoret. and Math. Phys., 127:3 (2001), 767–778
4. C. Klein, “Isomonodromy Approach to Boundary Value Problems for the Ernst Equation”, Theoret. and Math. Phys., 134:1 (2003), 72–85
5. C. Klein, “The Kerr Solution on Partially Degenerate Hyperelliptic Riemann Surfaces”, Theoret. and Math. Phys., 137:2 (2003), 1520–1526
6. Klein, C, “On explicit solutions to the stationary axisymmetric Einstein-Maxwell equations describing dust disks”, Annalen der Physik, 12:10 (2003), 599
7. Matveev, VB, “30 years of finite-gap integration theory”, Philosophical Transactions of the Royal Society A-Mathematical Physical and Engineering Sciences, 366:1867 (2008), 837
8. Aristophanes Dimakis, Nils Kanning, Folkert Müller-Hoissen, “The Non-Autonomous Chiral Model and the Ernst Equation of General Relativity in the Bidifferential Calculus Framework”, SIGMA, 7 (2011), 118, 27 pp.
•  Number of views: This page: 394 Full text: 117 References: 38 First page: 1


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org.apache.sis.referencing.operation.projection

## Enum NormalizedProjection.ParameterRole

• ### Enum Constant Summary

Enum Constants
Enum Constant and Description
`CENTRAL_MERIDIAN`
Maps the central meridian parameter (symbol: λ₀).
`FALSE_EASTING`
Maps the false easting parameter (symbol: FE).
`FALSE_NORTHING`
Maps the false northing parameter (symbol: FN).
`FALSE_SOUTHING`
Maps the false southing parameter (symbol: FS).
`FALSE_WESTING`
Maps the false westing parameter (symbol: FW).
`LATITUDE_OF_CONFORMAL_SPHERE_RADIUS`
Maps the parameter for the latitude where to compute the radius of conformal sphere (symbol: Rc).
`SCALE_FACTOR`
Maps the scale factor parameter (symbol: k₀).
`SEMI_MAJOR`
Maps the semi-major axis length parameter (symbol: a).
`SEMI_MINOR`
Maps the semi-minor axis length parameter (symbol: b).
• ### Method Summary

All Methods
Modifier and Type Method and Description
`static NormalizedProjection.ParameterRole` `valueOf(String name)`
Returns the enum constant of this type with the specified name.
`static NormalizedProjection.ParameterRole[]` `values()`
Returns an array containing the constants of this enum type, in the order they are declared.
• ### Methods inherited from class Enum

`clone, compareTo, equals, finalize, getDeclaringClass, hashCode, name, ordinal, toString, valueOf`
• ### Methods inherited from class Object

`getClass, notify, notifyAll, wait, wait, wait`
• ### Enum Constant Detail

• #### SEMI_MAJOR

`public static final NormalizedProjection.ParameterRole SEMI_MAJOR`
Maps the semi-major axis length parameter (symbol: a). This value is used for computing `NormalizedProjection.eccentricity`, and is also a multiplication factor for the denormalization matrix.

Unless specified otherwise, this is always mapped to a parameter named `"semi_major"`. `NormalizedProjection` subclasses typically do not need to provide a value for this key.

• #### SEMI_MINOR

`public static final NormalizedProjection.ParameterRole SEMI_MINOR`
Maps the semi-minor axis length parameter (symbol: b). This value is used for computing `NormalizedProjection.eccentricity`.

Unless specified otherwise, this is always mapped to a parameter named `"semi_minor"`. `NormalizedProjection` subclasses typically do not need to provide a value for this key.

`public static final NormalizedProjection.ParameterRole LATITUDE_OF_CONFORMAL_SPHERE_RADIUS`
Maps the parameter for the latitude where to compute the radius of conformal sphere (symbol: Rc). If this parameter is provided, then the radius of the conformal sphere at latitude φ will be used instead than the semi-major axis length in the denormalisation matrix. In other words, if provided then a is replaced by Rc below:
$MathML capable browser required$

This enumeration shall be used only when the user requested explicitely spherical formulas, for example the "Mercator (Spherical)" projection (EPSG:1026), but the figure of the Earth may be an ellipsoid rather than a sphere. In the majority of cases, this enumeration should not be used.

• #### CENTRAL_MERIDIAN

`public static final NormalizedProjection.ParameterRole CENTRAL_MERIDIAN`
Maps the central meridian parameter (symbol: λ₀). This value is subtracted from the longitude values before the map projections.

Some common names for this parameter are:

• Longitude of origin
• Longitude of false origin
• Longitude of natural origin
• Spherical longitude of origin
• Longitude of projection centre
• #### SCALE_FACTOR

`public static final NormalizedProjection.ParameterRole SCALE_FACTOR`
Maps the scale factor parameter (symbol: k₀). This is a multiplication factor for the (x,y) values obtained after map projections.

Some common names for this parameter are:

• Scale factor at natural origin
• Scale factor on initial line
• Scale factor on pseudo standard parallel
• #### FALSE_EASTING

`public static final NormalizedProjection.ParameterRole FALSE_EASTING`
Maps the false easting parameter (symbol: FE). This is a translation term for the x values obtained after map projections.

Some common names for this parameter are:

• False easting
• Easting at false origin
• Easting at projection centre
• #### FALSE_WESTING

`public static final NormalizedProjection.ParameterRole FALSE_WESTING`
Maps the false westing parameter (symbol: FW). This is the same x translation than `FALSE_EASTING`, but of opposite sign.

Actually, there is usually no parameter named "false westing" in a map projection. But some projections like "Lambert Conic Conformal (West Orientated)" are defined in such a way that their "false easting" parameter is effectively a "false westing". This enumeration value can be used for informing `NormalizedProjection` about that fact.

• #### FALSE_NORTHING

`public static final NormalizedProjection.ParameterRole FALSE_NORTHING`
Maps the false northing parameter (symbol: FN). This is a translation term for the y values obtained after map projections.

Some common names for this parameter are:

• False northing
• Northing at false origin
• Northing at projection centre
• #### FALSE_SOUTHING

`public static final NormalizedProjection.ParameterRole FALSE_SOUTHING`
Maps the false southing parameter (symbol: FS). This is the same y translation than `FALSE_NORTHING`, but of opposite sign.

Actually, there is usually no parameter named "false southing" in a map projection. But some projections like "Transverse Mercator (South Orientated)" are defined in such a way that their "false northing" parameter is effectively a "false southing". This enumeration value can be used for informing `NormalizedProjection` about that fact.

• ### Method Detail

• #### values

`public static NormalizedProjection.ParameterRole[] values()`
Returns an array containing the constants of this enum type, in the order they are declared. This method may be used to iterate over the constants as follows:
```for (NormalizedProjection.ParameterRole c : NormalizedProjection.ParameterRole.values())
System.out.println(c);
```
Returns:
an array containing the constants of this enum type, in the order they are declared
• #### valueOf

`public static NormalizedProjection.ParameterRole valueOf(String name)`
Returns the enum constant of this type with the specified name. The string must match exactly an identifier used to declare an enum constant in this type. (Extraneous whitespace characters are not permitted.)
Parameters:
`name` - the name of the enum constant to be returned.
Returns:
the enum constant with the specified name
Throws:
`IllegalArgumentException` - if this enum type has no constant with the specified name
`NullPointerException` - if the argument is null


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### Home > A2C > Chapter 11 > Lesson 11.2.2 > Problem11-62

11-62.

For each equation below, state the amplitude, period, and locator point, and then sketch two cycles of the graph.

1. $y = \operatorname{tan}\left(x\right)$

Amplitude: N/A
Period: π
Locator point: $\left(0,0\right)$

1. $y = \operatorname{tan}\left(x − π\right)$

The graph is shifted π radians to the right. How does this affect all of the properties of the graph listed above?


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# Test for non random-walk

I have a set of sequential data around of around 12k points. Each is a 1 or -1. They are split something like 53%/47%. I want to test the hypothesis that this sequence comes from a random walk with a constant parameter. What sort of statistical test should I use?

(I'd also be grateful for any book recommendations in this general area)

Thanks very much

-
It sounds to me that what you have are the steps of a process that you want to test whether it is a time-homogeneous simple random walk (possibly asymmetric). So, can the problem be reasonably codified as follows? You observe $\xi_1,\xi_2,\ldots$ each in $\{+1,-1\}$ with $\mathbb P(\xi_i = +1) = p(i)$. You want to test the hypothesis that $p(\cdot)$ is actually a constant function. Is this right? If so, do you know (i.e., can you assume) anything further about the properties of $p(\cdot)$, e.g., a single change in level or that it is smooth, etc? –  cardinal Jan 5 '12 at 12:28
Yes, spot on. At this stage I'm trying simply to arrive at a confidence value with which I can reject the hypothesis of p(.) being a constant process. I don't know a priori the value of the asymmetry, although I'm happy that it's roughly 53/47. –  Omnium Jan 5 '12 at 14:04
Note that the variance of the sample mean is maximized when $p(\cdot)$ is constant. In other words, suppose $Y_1, Y_2,\ldots$ are $\mathrm{Ber}(p_i)$ in $\{0,1\}$. Then $\mathbb E \overline{Y} = n^{-1} \sum_i p_i = \bar{p}$ and $\mathrm{Var}(\overline{Y}) = n^{-2} \sum_i p_i (1-p_i) \leq n^{-1} \bar{p}(1-\bar{p})$ by Jensen's inequality. The right-hand side is the variance of $\overline{Y}$ if $p(\cdot)$ is constant (always equal to $\bar{p}$). Hence, heuristically we'd expect the walk to be "too close" to the "trend" $\sum_{i=1}^n p_i$ on average if $p(\cdot)$ were not constant. –  cardinal Jan 5 '12 at 15:49
You've rather lost me there I'm afraid. –  Omnium Jan 5 '12 at 17:16

You could try Runs Test. Let $n_1$ bet the number of +1 runs and $n_2$ bet the number of -1 runs. You could use the test as in the wiki page.

-
Thanks for amazingly fast response! I don't think it's quite what I'm looking for. It seems to be looking more at non-independence of neighbouring points. I know that my data points are entirely independent. The cumulative graph wanders away from the 53% trend. I don't believe that the 1/-1 split early on (data is time ordered) is generated in the same mix as later. I've produced simulated graphs and they don't wander away from the trend like the data - I'm convinced that it isn't a constant valued random walk, but I can't think how to assess the degree of certainty of that conclusion. –  Omnium Jan 5 '12 at 11:19
Are you interested to test the data come from 50%/50%.What did you mean by constant parameter? –  vinux Jan 5 '12 at 11:24
A random walk consisting of a +1 move 53% of the time, and a -1 move 47% of the time. –  Omnium Jan 5 '12 at 11:32


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