text:
# tm_text

From tmap v1.11
0th

Percentile

Creates a tmap-element that adds text labels.

##### Usage
tm_text(text, size = 1, col = NA, root = 3, size.lim = NA,
sizes.legend = NULL, sizes.legend.labels = NULL,
sizes.legend.text = "Abc", n = 5, style = ifelse(is.null(breaks),
"pretty", "fixed"), breaks = NULL, interval.closure = "left",
palette = NULL, labels = NULL, labels.text = NA,
auto.palette.mapping = TRUE, contrast = NA, max.categories = 12,
colorNA = NA, textNA = "Missing", showNA = NA, fontface = NA,
fontfamily = NA, alpha = NA, case = NA, shadow = FALSE,
bg.color = NA, bg.alpha = NA, size.lowerbound = 0.4,
print.tiny = FALSE, scale = 1, auto.placement = FALSE,
remove.overlap = FALSE, along.lines = FALSE, overwrite.lines = FALSE,
just = c("center", "center"), xmod = 0, ymod = 0, title.size = NA,
title.col = NA, legend.size.show = TRUE, legend.col.show = TRUE,
legend.format = list(), legend.size.is.portrait = FALSE,
legend.col.is.portrait = TRUE, legend.hist = FALSE,
legend.hist.title = NA, legend.size.z = NA, legend.col.z = NA,
legend.hist.z = NA)
##### Arguments
text

name of the variable in the shape object that contains the text labels

size

relative size of the text labels (see note). Eiter one number, a name of a numeric variable in the shape data that is used to scale the sizes proportionally, or the value "AREA", where the text size is proportional to the area size of the polygons.

col

color of the text labels. Either a color value or a data variable name. If multiple values are specified, small multiples are drawn (see details).

root

root number to which the font sizes are scaled. Only applicable if size is a variable name or "AREA". If root=2, the square root is taken, if root=3, the cube root etc.

size.lim

vector of two limit values of the size variable. Only text labels are drawn whose value is greater than or equal to the first value. Text labels whose values exceed the second value are drawn at the size of the second value. Only applicable when size is the name of a numeric variable of shp. See also size.lowerbound which is a threshold of the relative font size.

sizes.legend

vector of text sizes that are shown in the legend. By default, this is determined automatically.

sizes.legend.labels

vector of labels for that correspond to sizes.legend.

sizes.legend.text

vector of example text to show in the legend next to sizes.legend.labels. By default "Abc". When NA, examples from the data variable whose sizes are close to the sizes.legend are taken and "NA" for classes where no match is found.

n

preferred number of color scale classes. Only applicable when col is a numeric variable name.

style

method to process the color scale when col is a numeric variable. Discrete options are "cat", "fixed", "sd", "equal", "pretty", "quantile", "kmeans", "hclust", "bclust", "fisher", and "jenks". A numeric variable is processed as a categorial variable when using "cat", i.e. each unique value will correspond to a distinct category. For the other discrete options, see the details in classIntervals. Continuous options are "cont" and "order". The former maps the values of col to a smooth gradient, whereas the latter maps the order of values of col to a smooth gradient. They are the continuous variants of respectively the discrete methods "equal" and quantile".

breaks

in case style=="fixed", breaks should be specified. The breaks argument can also be used when style="cont". In that case, the breaks are mapped evenly to the sequential or divering color palette.

interval.closure

value that determines whether where the intervals are closed: "left" or "right". Only applicable if col is a numerc variable.

palette

a palette name or a vector of colors. See tmaptools::palette_explorer() for the named palettes. Use a "-" as prefix to reverse the palette. The default palette is taken from tm_layout's argument aes.palette, which typically depends on the style. The type of palette from aes.palette is automatically determined, but can be overwritten: use "seq" for sequential, "div" for diverging, and "cat" for categorical.

labels

labels of the color classes, applicable if col is a data variable name

labels.text

Example text to show in the legend next to the labels. When NA (default), examples from the data variable are taken and "NA" for classes where they don't exist.

auto.palette.mapping

When diverging colour palettes are used (i.e. "RdBu") this method automatically maps colors to values such that the middle colors (mostly white or yellow) are assigned to values of 0, and the two sides of the color palette are assigned to negative respectively positive values. When categorical color palettes are used, this method stretches the palette if there are more levels than colors.

contrast

vector of two numbers that determine the range that is used for sequential and diverging palettes (applicable when auto.palette.mapping=TRUE). Both numbers should be between 0 and 1. The first number determines where the palette begins, and the second number where it ends. For sequential palettes, 0 means the brightest color, and 1 the darkest color. For diverging palettes, 0 means the middle color, and 1 both extremes. If only one number is provided, this number is interpreted as the endpoint (with 0 taken as the start).

max.categories

in case col is the name of a categorical variable, this value determines how many categories (levels) it can have maximally. If the number of levels is higher than max.categories and auto.palette.mapping is FALSE, then levels are combined.

colorNA

colour for missing values. Use NULL for transparency.

textNA

text used for missing values.

showNA

logical that determines whether missing values are named in the legend. By default (NA), this depends on the presence of missing values.

fontface

font face of the text labels. By default, determined by the fontface argument of tm_layout.

fontfamily

font family of the text labels. By default, determined by the fontfamily argument of tm_layout.

alpha

transparency number between 0 (totally transparent) and 1 (not transparent). By default, the alpha value of the fontcolor is used (normally 1).

case

case of the font. Use "upper" to generate upper-case text, "lower" to generate lower-case text, and NA to leave the text as is.

logical that determines whether a shadow is depicted behind the text. The color of the shadow is either white or yellow, depending of the fontcolor.

bg.color

background color of the text labels. By default, bg.color=NA, so no background is drawn.

bg.alpha

number between 0 and 1 that specifies the transparancy of the text background (0 is totally transparent, 1 is solid background).

size.lowerbound

lowerbound for size. Only applicable when size is not a constant. If print.tiny is TRUE, then all text labels which relative text is smaller than size.lowerbound are depicted at relative size size.lowerbound. If print.tiny is FALSE, then text labels are only depicted if their relative sizes are at least size.lowerbound (in other words, tiny labels are omitted).

print.tiny

boolean, see size.lowerbound

scale

text size multiplier, useful in case size is variable or "AREA".

auto.placement

logical (or numeric) that determines whether the labels are placed automatically. If TRUE, the labels are placed next to the coordinate points with as little overlap as possible using the simulated annealing algorithm. Therefore, it is recommended for labeling spatial dots or symbols. If a numeric value is provided, this value acts as a parameter that specifies the distance between the coordinate points and the text labels in terms of text line heights.

remove.overlap

logical that determines whether the overlapping labels are removed

along.lines

logical that determines whether labels are rotated along the spatial lines. Only applicabel if a spatial lines shape is used.

overwrite.lines

logical that determines whether the part of the lines below the text labels is removed. Only applicabel if a spatial lines shape is used.

just

justification of the text relative to the point coordinates. The first value specifies horizontal and the second value vertical justification. Possible values are: "left" , "right", "center", "bottom", and "top". Numeric values of 0 specify left alignment and 1 right alignment.

xmod

horizontal position modification of the text (relatively): 0 means no modification, and 1 corresponds to the height of one line of text. Either a single number for all polygons, or a numeric variable in the shape data specifying a number for each polygon. Together with ymod, it determines position modification of the text labels. In most coordinate systems (projections), the origin is located at the bottom left, so negative xmod move the text to the left, and negative ymod values to the bottom.

ymod

vertical position modification. See xmod.

title.size

title of the legend element regarding the text sizes

title.col

title of the legend element regarding the text colors

legend.size.show

logical that determines whether the legend for the text sizes is shown

legend.col.show

logical that determines whether the legend for the text colors is shown

legend.format

list of formatting options for the legend numbers. Only applicable if labels is undefined. Parameters are:

fun

Function to specify the labels. It should take a numeric vector, and should return a character vector of the same size. By default it is not specified. If specified, the list items scientific, format, and digits (see below) are not used.

scientific

Should the labels be formatted scientically? If so, square brackets are used, and the format of the numbers is "g". Otherwise, format="f", and text.separator, text.less.than, and text.or.more are used. Also, the numbers are automatically rounded to millions or billions if applicable.

format

By default, "f", i.e. the standard notation xxx.xxx, is used. If scientific=TRUE then "g", which means that numbers are formatted scientically, i.e. n.dddE+nn if needed to save space.

digits

Number of digits after the decimal point if format="f", and the number of significant digits otherwise.

text.separator

Character string to use to separate numbers in the legend (default: "to").

text.less.than

Character string to use to translate "Less than" (which is the default).

text.or.more

Character string to use to translate "or more" (which is the default).

...

Other arguments passed on to formatC

legend.size.is.portrait

logical that determines whether the legend element regarding the text sizes is in portrait mode (TRUE) or landscape (FALSE)

legend.col.is.portrait

logical that determines whether the legend element regarding the text colors is in portrait mode (TRUE) or landscape (FALSE)

legend.hist

logical that determines whether a histogram is shown regarding the text colors

legend.hist.title

title for the histogram. By default, one title is used for both the histogram and the normal legend for text colors.

legend.size.z

index value that determines the position of the legend element regarding the text sizes with respect to other legend elements. The legend elements are stacked according to their z values. The legend element with the lowest z value is placed on top.

legend.col.z

index value that determines the position of the legend element regarding the text colors. (See legend.size.z)

legend.hist.z

index value that determines the position of the histogram legend element. (See legend.size.z)

##### Value

tmap-element

##### Note

The absolute fontsize (in points) is determined by the (ROOT) viewport, which may depend on the graphics device.

vignette("tmap-nutshell")

• tm_text
##### Examples
# NOT RUN {
current.mode <- tmap_mode("plot")

data(World, Europe, metro)

tm_shape(World) +
tm_text("name", size="AREA")

tm_shape(Europe) +
tm_polygons() +
tm_text("iso_a3", size="AREA", col = "grey20", root=4, shadow = TRUE, scale=2,
size.lowerbound = .1) +
tm_shape(Europe) +
tm_text("name", size="AREA", root=4, scale=1,
ymod=-1 * tmaptools::approx_areas(Europe, target = "norm")^(1/4))

tm_shape(Europe) +
tm_polygons() +
tm_shape(metro) +
tm_bubbles("pop2010", size.lim = c(0, 15e6),
title.size = "European metropolitan areas") +
tm_shape(metro[metro\$pop2010>=2e6, ]) +
tm_text("name", auto.placement = TRUE) +
tm_format_Europe()

tm_shape(World) +
tm_text("name", size="pop_est", col="continent", palette="Dark2",
title.size = "Population", title.col="Continent") +
tm_legend(outside = TRUE)

# restore current mode
tmap_mode(current.mode)
# }

Documentation reproduced from package tmap, version 1.11, License: GPL-3

### Community examples

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text:
When working with fractions you must know what a numerator and denominator is.

### Numerator

A numerator is a number that is used to represent part of a whole. The number can be found above the line or on top in a fraction which shows how many parts are indicated by the denominator are taken. For example: 3 in 3/6 as shown in the diagram above. Here you can find the full definition

Numerator

### Denominator

The denominator is the figure below the numerator in a fraction. Very simply the number at the bottom in any fraction.

Denominator

Adding fractions is a very easy task if you have some background in maths or know a little about maths.

When adding like fractions such as the one below, fractions which have the same denominator you simply add the top numerators and then write the fractions with the numerators addition over the denominator of the two fractions.

## Example

Work out

As simple as that…

## However…

Fractions which have different denominators have to be treated differently. We first have to find the lowest common multiple of the denominators of the fraction in the equation.

## For example

The denominators for this example are different. In this example we have to convert the all the fractions in the equation so that they have the same denominators. The numerators will be affected as well since the denominators are changing.

The LCM of 2 and 3 is a six, so the expression becomes.

Notice that the numerators have been affected as well. For each fraction you had to multiply the denominator by a number to get the LCM. So for;

You had to multiply the denominator 2 by 3 to get a 6. Since you multiplied the bottom you had to do the same to the top. The same for;

Notice we multiplied both the bottom and the top by the same number.

## Another example

Work out

It’s that simple to add fractions…

format: plain


text:
Cryptology ePrint Archive: Report 2018/068

Simple Schnorr Multi-Signatures with Applications to Bitcoin

Gregory Maxwell and Andrew Poelstra and Yannick Seurin and Pieter Wuille

Abstract: We describe a new Schnorr-based multi-signature scheme (i.e., a protocol which allows a group of signers to produce a short, joint signature on a common message), provably secure in the plain public-key model (meaning that signers are only required to have a public key, but do not have to prove knowledge of the private key corresponding to their public key to some certification authority or to other signers before engaging the protocol), which improves over the state-of-art scheme of Bellare and Neven (ACM-CCS 2006) and its variants by Bagherzandi et al. (ACM-CCS 2008) and Ma et al. (Des. Codes Cryptogr., 2010) in two respects: (i) it is simple and efficient, having only two rounds of communication instead of three for the Bellare-Neven scheme and the same key and signature size as standard Schnorr signatures; (ii) it allows key aggregation, which informally means that the joint signature can be verified exactly as a standard Schnorr signature with respect to a single aggregated'' public key which can be computed from the individual public keys of the signers. This comes at the cost of a stronger security assumption, namely the hardness of the One-More Discrete Logarithm problem, rather than the standard Discrete Logarithm problem, and a looser security reduction due to a double invocation of the Forking Lemma. As an application, we explain how our new multi-signature scheme could improve both performance and user privacy in Bitcoin.

Category / Keywords: public-key cryptography / multi-signatures, Schnorr signatures, one-more discrete logarithm problem, forking lemma, Bitcoin

Date: received 15 Jan 2018, last revised 15 Jan 2018

Contact author: yannick seurin at m4x org

Available format(s): PDF | BibTeX Citation

Short URL: ia.cr/2018/068

[ Cryptology ePrint archive ]


text:
tropical cyclone is a rapidly rotating storm system characterized by a low-pressure center, a closed low-level atmospheric circulation, strong winds, and a spiral arrangement of thunderstorms that produce heavy rain. Depending on its location and strength, a tropical cyclone is referred to by names such as hurricane(/ˈhʌrɪkən/ or /ˈhʌrɪkn/[1][2][3]), typhoon /tˈfn/tropical storm, cyclonic storm, tropical depression, and simply cyclone.[4] A hurricane is a storm that occurs in the Atlantic Ocean and northeastern Pacific Ocean, a typhoon occurs in the northwestern Pacific Ocean, and a cyclone occurs in the south Pacific or Indian Ocean.[4]
Tropical cyclones typically form over large bodies of relatively warm water. They derive their energy through the evaporation of water from the ocean surface, which ultimately recondenses into clouds and rain when moist air rises and cools to saturation. This energy source differs from that of mid-latitude cyclonic storms, such as nor'easters and European windstorms, which are fueled primarily by horizontal temperature contrasts. The strong rotating winds of a tropical cyclone are a result of the conservation of angular momentum imparted by the Earth's rotation as air flows inwards toward the axis of rotation. As a result, they rarely form within 5° of the equator.[5] Tropical cyclones are typically between 100 and 2,000 km (62 and 1,243 mi) in diameter.
Tropical refers to the geographical origin of these systems, which form almost exclusively over tropical seas. Cyclonerefers to their cyclonic nature, with wind blowing counterclockwise in the Northern Hemisphere and clockwise in the Southern Hemisphere. The opposite direction of circulation is due to the Coriolis effect. #LattigDerik
In addition to strong winds and rain, tropical cyclones are capable of generating high waves, damaging storm surge, and tornadoes. They typically weaken rapidly over land where they are cut off from their primary energy source. For this reason, coastal regions are particularly vulnerable to damage from a tropical cyclone as compared to inland regions. Heavy rains, however, can cause significant flooding inland, and storm surges can produce extensive coastal flooding up to 40 kilometres (25 mi) from the coastline. Though their effects on human populations are https://www.pinterest.co.uk/pin/426505027192573821/  often devastating, tropical cyclones can relieve drought conditions. They https://deriklattig.contently.com/  also carry heat energy away from the tropics and transport it toward temperate latitudes, which may play an important role in modulating regional and global climate.  Tropical cyclones are areas of relatively low pressure in the troposphere, with the largest pressure perturbations occurring at https://about.me/DerikLattig  low altitudes near the surface. On Earth, the pressures recorded at the centers of tropical cyclones are among the lowest #DerikLattig ever observed at sea level.[6] The environment near the center of tropical cyclones is warmer than the surroundings at all altitudes, thus they are characterized as "warm core"  http://www.dailymotion.com/derik-lattig systems.[7]

### Wind field

The near-surface wind field of a tropical cyclone is characterized by air rotating rapidly around a center of circulation while also flowing radially inwards. At the outer edge of the storm, air may be nearly calm; however, due to the Earth's rotation, the air has non-zero absolute angular momentum. As air flows radially inward, it begins to rotate cyclonically (counter-clockwise in the Northern Hemisphere, and clockwise in the Southern Hemisphere) in order to conserve angular momentum. At an inner radius, air begins to ascend to the top of the troposphere. #DerikLattig says This radius is typically coincident with the inner radius of the eyewall, and has the strongest near-surface winds of the storm; consequently, it is known as the radius of maximum winds.[8] Once aloft, air flows http://www.dailymotion.com/derik-lattig away from the storm's center, producing a shield of cirrus clouds.[9]
The previously mentioned processes result in a wind field that is nearly axisymmetric: Wind speeds are low at the center, increase rapidly moving outwards to the radius of maximum winds, and then decay more gradually with radius to large radii. However, the wind field often exhibits additional spatial and temporal variability due to the effects of localized processes, such as thunderstorm activity and horizontal flow instabilities. In the vertical direction, winds are strongest near the surface and decay with height within the troposphere.[10]

At the center of a mature tropical cyclone, https://deriklattig.contently.com/  air sinks rather than rises. For a sufficiently strong storm, air may sink over a layer deep enough to suppress cloud formation, thereby creating a clear "eye". Weather in the eye is normally calm and free of clouds, although the sea may be #DerikLattig  extremely violent.[11] The eye is normally circular in shape, and is typically 30–65 km (19–40 mi) in diameter, though eyes as small as 3 km (1.9 mi) and as large as 370 km (230 mi) have been observed.[12][13] https://www.pinterest.co.uk/pin/426505027192573821/
The cloudy outer edge of the eye is called the "eyewall". The eyewall typically expands https://about.me/DerikLattig outward with height, resembling an arena football stadium; this phenomenon is sometimes referred to as the stadium effect.[13]The eyewall is where the greatest wind speeds are found, air rises most rapidly, clouds reach to their highest altitude, and precipitation is the heaviest. The heaviest wind damage occurs where a tropical cyclone's eyewall passes over land.[11]
In a weaker storm, the eye may be obscured by the central dense overcast, which is the upper-level cirrus shield that is associated with a concentrated area of strong thunderstorm activity near the center of a tropical cyclone.[14]
The eyewall may vary over time in the form plus.google.com/110187666621582812927  of eyewall replacement cycles, particularly in intense tropical cyclones. Outer rainbands can organize into an outer ring of thunderstorms that slowly moves inward, which is believed to rob the primary eyewall of moisture and angular momentum. When the primary eyewall weakens, the tropical cyclone weakens temporarily. The outer eyewall eventually replaces the primary one at the end of the cycle, at which time the storm may return to its original intensity.[15]

There are a variety of metrics commonly used to measure storm size. The most common metrics include the radius of maximum wind, the radius of 34-knot wind (i.e. gale force), the radius of outermost closed isobar (ROCI), and the radius of vanishing wind.[17][18] An additional metric is the radius at which the cyclone's relative vorticityfield decreases to 1×10−5 s−1.[13]
On Earth, tropical cyclones span a large range of sizes, from 100–2,000 kilometres (62–1,243 mi) as measured by the radius of vanishing wind. They are largest on average in the northwest Pacific Ocean basin and smallest in the northeastern Pacific Ocean basin.[19] If the plus.google.com/110187666621582812927  radius of outermost closed isobar is less than two degrees of latitude(222 km (138 mi)), then the cyclone is "very small" or a "midget". A radius of 3–6 latitude degrees (333–670 km (207–416 mi)) is considered "average sized". "Very large" tropical cyclones have a radius of greater than 8 degrees (888 km (552 mi)).[16] Observations indicate that size is only weakly correlated to variables such as storm intensity (i.e. maximum wind speed), radius of maximum wind, latitude, and maximum potential intensity.[18][19]
Size plays an important role in modulating damage caused by a storm. All else equal, a larger storm will impact a larger area for a longer period of time. Additionally, a larger near-surface wind field can generate higher storm surge due to the combination of longer wind fetch, longer https://www.pinterest.co.uk/pin/426505027192573821/  duration, and enhanced wave setup.[20]
The upper circulation of strong hurricanes extends into the tropopause of the atmosphere, which at low latitudes is 15,000–18,000 metres (50,000–60,000 ft).[21]

## Physics and energetics

Tropical cyclones exhibit an overturning circulation where air inflows at low levels near the surface, rises in thunderstorm clouds, and outflows at high levels near the tropopause.[22]
The three-dimensional wind field in a tropical cyclone can be separated into two components: a "primary circulation" and a "secondary circulation". The primary circulation is the rotational part of the flow; it is purely circular. The secondary circulation is the overturning (in-up-out-down) part of the flow; it is in the radial and vertical directions. The primary circulation is larger in magnitude, dominating the surface wind field, and is responsible for the majority of the damage a storm causes, while the secondary circulation is slower but governs the energetics of the storm.

### Secondary circulation: a Carnot heat engine

A tropical cyclone's primary energy source is heat from the evaporation of water from the ocean surface, which ultimately recondenses into clouds and rain when the warm moist air rises and cools to saturation. The energetics of the system may be idealized as an atmospheric Carnot heat engine.[23] First, inflowing air near the surface acquires heat primarily via evaporation of water (i.e. latent heat) at the temperature of the warm ocean surface (during evaporation, the ocean cools and the air warms). Second, the warmed air rises and cools within the eyewall while conserving total heat content (latent heat is simply converted to sensible heat during condensation). Third, air outflows and loses heat via infrared radiation to space at the temperature of the cold tropopause. Finally, air subsides and warms at the outer edge of the storm while conserving total heat content. The first and third legs are nearly isothermal, while the second and fourth legs are nearly isentropic. This in-up-out-down overturning flow is known as the secondary circulation. The Carnot perspective provides an upper bound on the maximum wind speed that a storm can attain.
Scientists estimate that a tropical cyclone releases heat energy at the rate of 50 to 200 exajoules (1018 J) per day,[24] equivalent to about 1 PW (1015 watt). Derik Lattig This rate of energy release is equivalent to 70 times the world energy consumption of humans and 200 times the worldwide electrical generating capacity, or to exploding a 10-megaton nuclear bomb every 20 minutes.[24][25]

### Primary circulation: rotating winds

The primary rotating flow in a tropical cyclone results from the conservation of angular momentum by the secondary circulation. Absolute angular momentum on a rotating planet ${\displaystyle M}$ is given by
${\displaystyle M={\frac {1}{2}}fr^{2}+vr}$
where ${\displaystyle f}$ is the Coriolis parameter${\displaystyle v}$ is the azimuthal (i.e. rotating) wind speed, and ${\displaystyle r}$ is the radius to the axis of rotation. The first term on the right hand side is the component of planetary angular momentum that projects onto the local vertical (i.e. the axis of rotation). The second term on the right hand side is the relative angular momentum of the circulation itself with respect to the axis of rotation. Because the planetary angular momentum term vanishes at the equator (where ${\displaystyle f=0}$ ), tropical cyclones rarely form within 5° of the equator.[5][26]
As air flows radially inward at low levels, it begins to rotate cyclonically in order to conserve angular momentum. Similarly, as rapidly rotating air flows radially outward near the tropopause, its cyclonic rotation decreases and ultimately changes sign at large enough radius, resulting in an upper-level anti-cyclone. The result is a vertical structure characterized by a strong cyclone at low levels and a strong anti-cyclone near the tropopause; from thermal wind balance, this corresponds to a system that is warmer at its center than in the surrounding environment at all altitudes (i.e. "warm-core"). From hydrostatic balance, the warm core translates to lower pressure at the center at all altitudes, with the maximum pressure drop located at the surface.[10] plus.google.com/110187666621582812927

### Maximum potential intensity

Due to surface friction, the inflow only partially conserves angular momentum. Thus, the sea surface lower boundary acts as both a source (evaporation) and sink (friction) of energy for the system. This fact leads to the existence of a theoretical upper bound on the strongest wind speed that a tropical cyclone can attain. Because evaporation increases linearly with wind speed (just as climbing out of a pool feels much colder on a windy day), there is a positive feedback on energy input into the system known as the Wind-Induced Surface Heat Exchange (WISHE) feedback.[23] This feedback is offset when frictional dissipation, which increases with the cube of the wind speed, becomes sufficiently large. This upper bound is called the "maximum potential intensity", ${\displaystyle v_{p}}$, and is given by https://delicious.com/derik_lattig
${\displaystyle v_{p}^{2}={\frac {C_{k}}{C_{d}}}{\frac {T_{s}-T_{o}}{T_{o}}}\Delta k}$
where ${\displaystyle T_{s}}$ is the temperature of the sea surface, ${\displaystyle T_{o}}$ is the temperature of the outflow ([K]), ${\displaystyle \Delta k}$ is the enthalpy difference between the surface and the overlying air ([J/kg]), and ${\displaystyle C_{k}}$ and ${\displaystyle C_{d}}$ are the http://pingwi-fi.com/2010/08/garage-girlz-coming-to-a-station-near-you/surface exchange coefficients (dimensionless) of enthalpy and momentum, respectively.[27] The surface-air enthalpy difference is taken as ${\displaystyle \Delta k=k_{s}^{*}-k}$, where ${\displaystyle k_{s}^{*}}$ is the saturation enthalpy of air at sea surface temperature and sea-level pressure and ${\displaystyle k}$ is the enthalpy of boundary layer air overlying the https://www.mylife.com/derik-lattig/derik-lattigsurface.https://vimeo.com/deriklattig https://www.ebay.com/p/Family-Pride-by-Derik-Lattig-and-Emily-Regan-2015-Hardcover/216021580 https://www.pinterest.com/deriklattig/tv-show-vega-by-derik-lattig/
The maximum potential intensity is predominantly a DerikLattig function of th http://www.deriklattig.com/2016/07/derik-lattig-journalist-in-fort-worth.html background environment alone (i.e. without a tropical cyclone), and thus this quantity can be used to http://www.break.com/picture-topics/reality-tv-derik-lattig determinehttps://www.ebay.com/p/Family-Pride-by-Derik-Lattig-and-Emily-Regan-2015-Hardcover/216021580  which regions on http://www.deriklattig.com/2016/07/derik-lattig-journalist-in-fort-worth.html  Earth can support tropical cyclones of a given intensity, and how these https://www.amazon.com/Derik-Lattig/e/B0101G62SG http://www.imdb.com/name/nm5056723/  https://www.pinterest.com/deriklattig/derik-lattig/ regions may evolve in time.[28][29] Specifically, the maximum potential intensity has three components, but its variability in space and time is due predominantly to the variability in the surface-air enthalpy difference component ${\displaystyle \Delta k}$. https://myspace.com/newsguy497 http://pingwi-fi.com/2010/08/garage-girlz-coming-to-a-station-near-you/ https://vimeo.com/deriklattig

#### Derivation

A tropical cyclone may be viewed as a heat engine that converts input heat energy from the surface https://vimeo.com/170305640 into mechanical energy that can be used to do mechanical work against surface friction. At equilibrium, https://www.mylife.com/derik-lattig/derik-lattighttp://www.break.com/picture-topics/reality-tv-derik-lattig the rate of net energy production in the https://www.pinterest.com/deriklattig/derik-lattig/ system must equal the rate of energy loss due to frictional dissipation at the surface, i.e. plus.google.com/110187666621582812927
${\displaystyle W_{in}=W_{out}}$
The rate of energy loss per unit surface area from surface friction, ${\displaystyle W_{out}}$, is given by
${\displaystyle W_{out}=C_{d}\rho |\mathbf {u} |^{3}}$
where ${\displaystyle \rho }$ is the density of near-surface air ([kg/m3]) and ${\displaystyle |\mathbf {u} |}$ is the near surface wind speed ([m/s]).
The rate of energy production per unit surface area, ${\displaystyle W_{in}}$ is given by
${\displaystyle W_{in}=\epsilon Q_{in}}$
where ${\displaystyle \epsilon }$ is the heat engine efficiency and ${\displaystyle Q_{in}}$ is the total rate of heat input into the system per unit http://pingwi-fi.com/2010/08/garage-girlz-coming-to-a-station-near-you/surface area. Given that a tropical cyclone may be idealized as a Carnot heat engine, the Carnot heat engine efficiency is given https://www.amazon.com/Derik-Lattig/e/B0101G62SG https://www.mylife.com/derik-lattig/derik-lattig by https://delicious.com/derik_lattig https://www.pinterest.com/deriklattig/tv-show-vega-by-derik-lattig/
${\displaystyle \epsilon ={\frac {T_{s}-T_{o}}{T_{s}}}}$
Heat (enthalpy) per unit mass is given by
${\displaystyle k=C_{p}T+L_{v}q}$
where ${\displaystyle C_{p}}$ is the heat capacity of air, ${\displaystyle T}$ is air temperature, ${\displaystyle L_{v}}$ is the latent heat of vaporization,https://vimeo.com/170305640  and ${\displaystyle q}$ is the concentration of water vapor. The first component corresponds to sensible heat and the second to latent heat. https://delicious.com/derik_lattig
There are two sources of heat input. The dominant source is the input of heat at the surface, primarily due to evaporation. The bulk aerodynamic formula for the rate of heat input per unit area at the surface, http://www.deriklattig.com/2016/07/derik-lattig-journalist-in-fort-worth.html${\displaystyle Q_{in:k}}$, is given by https://myspace.com/newsguy497  http://www.imdb.com/name/nm5056723/
${\displaystyle Q_{in:k}=C_{k}\rho |\mathbf {u} |\Delta k}$
where ${\displaystyle \Delta k=k_{s}^{*}-k}$ represents the enthalpy difference Lattig, Derik between the ocean surface and the overlying air. The second source is the internal sensible heat generated from frictional dissipation (equa https://www.pinterest.com/deriklattig/derik-lattig/ to ${\displaystyle W_{out}}$), which occurs near the surface within the https://www.amazon.com/Derik-Lattig/e/B0101G62SG tropical cyclone and is recycled to the system. http://www.break.com/picture-topics/reality-tv-derik-lattig
${\displaystyle Q_{in:friction}=C_{d}\rho |\mathbf {u} |^{3}}$
Thus, the total rate of net energy production per unit surface area is given http://www.imdb.com/name/nm5056723/ http://www.deriklattig.com/2016/07/derik-lattig-journalist-in-fort- worth.htmlby https://myspace.com/newsguy497

### Share:

1. The near-surface wind field of a tropical cyclone is characterized by air rotating rapidly around a center of circulation while also flowing radially inwards. At the outer edge of the storm, air may be nearly calm; however, due to the Earth's rotation, the air has non-zero absolute angular momentum. As air flows radially inward, it begins to rotate cyclonically (counter-clockwise in the Northern Hemisphere, and clockwise in the Southern Hemisphere) in order to conserve angular momentum. At an inner radius, air begins to ascend to the top of the troposphere. #DerikLattig says This radius is typically coincident with the inner radius of the eyewall, and has the strongest near-surface winds of the storm; consequently, it is known as the radius of maximum winds.[8] Once aloft, air flows http://www.dailymotion.com/derik-lattig away from the storm's center, producing a shield of cirrus clouds.[9]
The previously mentioned processes result in a wind field that is nearly axisymmetric: Wind speeds are low at the center, increase rapidly moving outwards to the radius of maximum winds, and then decay more gradually with radius to large radii. However, the wind field often exhibits additional spatial and temporal variability due to the effects of localized processes, such as thunderstorm activity and horizontal flow instabilities. In the vertical direction, winds are strongest near the surface and decay with height within the troposphere.[10]

At the center of a mature tropical cyclone, https://deriklattig.contently.com/ air sinks rather than rises. For a sufficiently strong storm, air may sink over a layer deep enough to suppress cloud formation, thereby creating a clear "eye". Weather in the eye is normally calm and free of clouds, although the sea may be #DerikLattig extremely violent.[11] The eye is normally circular in shape, and is typically 30–65 km (19–40 mi) in diameter, though eyes as small as 3 km (1.9 mi) and as large as 370 km (230 mi) have been observed.[12][13] https://www.pinterest.co.uk/pin/426505027192573821/
The cloudy outer edge of the eye is called the "eyewall". The eyewall typically expands https://about.me/DerikLattig outward with height, resembling an arena football stadium; this phenomenon is sometimes referred to as the stadium effect.[13]The eyewall is where the greatest wind speeds are found, air rises most rapidly, clouds reach to their highest altitude, and precipitation is the heaviest. The heaviest wind damage occurs where a tropical cyclone's eyewall passes over land.[11]
In a weaker storm, the eye may be obscured by the central dense overcast, which is the upper-level cirrus shield that is associated with a concentrated area of strong thunderstorm activity near the center of a tropical cyclone.[14]
The eyewall may vary over time in the form plus.google.com/110187666621582812927 of eyewall replacement cycles, particularly in intense tropical cyclones. Outer rainbands can organize into an outer ring of thunderstorms that slowly moves inward, which is believed to rob the primary eyewall of moisture and angular momentum. When the primary eyewall weakens, the tropical cyclone weakens temporarily. The outer eyewall eventually replaces the primary one at the end of the cycle, at which time the storm may return to its original intensity.[15]

There are a variety of metrics commonly used to measure storm size. The most common metrics include the radius of maximum wind, the radius of 34-knot wind (i.e. gale force), the radius of outermost closed isobar (ROCI), and the radius of vanishing wind.[17][18] An additional metric is the radius at which the cyclone's relative vorticityfield decreases to 1×10−5 s−1.[13]

2. Blogspot.com/2015/10/lattig-derik.html

3. deriklattig.com - User-agent: Mediapartners-Google Disallow: User-agent: * Disallow .blogspot.com 2015/10 Lattig-Derik HTML


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# All Questions

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text:
Found 16 Documents (Results 1–16)

100
MathJax

On the number of stationary solutions of a scalar parabolic equation. (Russian)Zbl 0702.35111

Reviewer: B.G.Konopelchenko
MSC:  35K20 35A99 76R50

On solutions of a boundary value problem for a second order ordinary differential equation. (Russian)Zbl 0707.34017

Reviewer: J.Appell
MSC:  34B15

Uniqueness of the solution of some boundary value problems for ordinary differential equations with a non-summable singularity. (Russian)Zbl 0707.34003

Reviewer: J.Appell
MSC:  34A12 34B99

On a two-point boundary value problem for a second order ordinary differential equation. (Russian)Zbl 0624.34015

Reviewer: M.Dmitriev
MSC:  34B15 34B10

Reviewer: E.Ihle
MSC:  34B10

Reviewer: E.Ihle
MSC:  34B10

Reviewer: E.Ihle
MSC:  34B10

Connectedness of the set of solutions for a second order ordinary differential equation. (Russian)Zbl 0556.34012

Reviewer: A.Halanay
MSC:  34B15 34C99

Determination of the solution of an ordinary differential equation according to the given asymptotics at infinity. (Russian)Zbl 0545.34036

Reviewer: J.Kalas
MSC:  34E05 34A30 65L99

On a method for determination of sub- and superfunctions. (Russian)Zbl 0436.34021

MSC:  34B15 34A34

MSC:  34B10

MSC:  34B05

Continuity up to the boundary of generalized solutions of boundary value problems for linear equations of parabolic and elliptic type. (Russian)Zbl 0435.35017

MSC:  35D99 35K20 35J25

The existence of a solution of boundary value problems for an ordinary differential equation of second order. (Russian)Zbl 0432.34012

MSC:  34B10 34B05

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all top 5

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text:
threeparttable notes with 1.5 spacing. Want single spacing

My main text uses one-one-half line spacing, and now the notes below the table are also one-one-half line spaced. I'd prefer them to feature single spacing (all of my other tables don't share this problem). If I were to simply use the tabular environment within a threeparttable, I'd get the desired look for my footnotes, but I can't seem to reproduce the single line spacing for this table. How do I fix the code for this table so that I can stretch the table across the width of the text on the page, but have my footnotes appear below the table with single spacing? Any help would be greatly appreciated.

\documentclass{report}

% packages
\usepackage{amsmath}        % Extra math definitions
\usepackage{graphics}       % PostScript figures
\usepackage{setspace}       % 1.5 spacing
\usepackage{longtable,threeparttablex}          % Tables spanning pages
\usepackage{color}
\usepackage{graphicx}
\usepackage{rotating}
\usepackage{array}
\usepackage{multirow}
\usepackage{booktabs}
\usepackage[table]{xcolor}
\usepackage{subcaption}
\usepackage{pdfpages}
\usepackage[notes,backend=bibtex]{biblatex-chicago}
\usepackage[utf8]{inputenc}
\usepackage[full]{textcomp}
\usepackage[T1]{fontenc}
\usepackage{afterpage}
\usepackage{float}
\usepackage{fp}
\usepackage{pdflscape} %To rotate pages with sideways tables or large figures.
\usepackage{xparse}
\usepackage{siunitx} %Lets tables align columns by decimal point.
\usepackage{lmodern}
\usepackage[french,english]{babel}
\usepackage{caption}
\usepackage{pifont}
\usepackage{microtype}
\usepackage{amssymb}
\usepackage{arydshln}
\usepackage{cleveref}
\usepackage[bottom]{footmisc} %This places footnotes at the bottom so figures will appear above footnotes.

\newcommand{\tabitem}{~~\llap{\textbullet}~~}
\DisableLigatures[f]{encoding=T1}
\crefformat{footnote}{#2\footnotemark[#1]#3}

%Let tables fit to width of page.
\newcommand\totextwidth[1]{%
\sbox{\mytabularbox}{#1}%
\figwidthc=\wd\mytabularbox%
\textwidthc=\textwidth%
\FPdiv\scaleratio{\the\textwidthc}{\the\figwidthc}%
\FPmin\scaleratio{\scaleratio}{1}%
\scalebox{\scaleratio}{\usebox{\mytabularbox}}%
}

\begin {document}

\onehalfspacing

\afterpage{
\begin{ThreePartTable}
\setlength{\LTleft}{0pt}
\setlength{\LTright}{0pt}
\renewcommand\TPTminimum{\textwidth}
\renewcommand{\arraystretch}{0.8}
\begin{TableNotes}[flushleft]
\small
\item \textsuperscript{a} \textit{Event Density} (1.5) refers to the number of notes identified in a 1.5 s window starting from the onset of cadential arrival.
\item \textsuperscript{b} \textit{Caesura} refers to the presence of a rest across all four instrumental parts.
\item \textsuperscript{c} \textit{Elision} refers to the superposition of a new intrathematic phrase at the moment of cadential arrival, an accompanimental overlap in the bass, or a melodic lead-in.
\item \textsuperscript{d} \textit{Interthematic Function} refers to one of the following temporal functions to characterize the passage at the theme level following cadential arrival: Before-the-Beginning, Beginning, Middle, End, After-the-End. \textit{Intrathematic Function} refers to either the Beginning, Middle, or End functions that characterize the passage at the phrase level following cadential arrival.
\end{TableNotes}
\begin{longtable}{@{\hskip\tabcolsep\extracolsep\fill}lccc}
\caption{Descriptive statistics for the 11 retrospective features.} \\
\toprule
\textit{Retrospective Features} & \multicolumn{1}{c}{\textit{M} (\textit{SD})} & \multicolumn{1}{c}{\textit{Range}} &
\multicolumn{1}{c}{\textit{Mode} (\textit{Frequency})} \\
\midrule
\textbf{Segmentational Grouping} &       &       & \textbf{} \\
\multicolumn{1}{l}{\quad (1)   \textit{Next Note Onset} (s)} & \multicolumn{1}{c}{.57 (.44)} & \multicolumn{1}{c}{.1-1.8} & \multicolumn{1}{c}{} \\
\multicolumn{1}{l}{\quad (2)   \textit{Next Bass Note Onset} (s)} & \multicolumn{1}{c}{.97 (.77)} & \multicolumn{1}{c}{.1-2.8} & \multicolumn{1}{c}{} \\
\multicolumn{1}{l}{\quad (3)   \textit{Next Soprano Note Onset} (s)} & \multicolumn{1}{c}{.99 (.66)} & \multicolumn{1}{c}{.1-2.6} & \multicolumn{1}{c}{} \\
\multicolumn{1}{l}{\quad (4)   \textit{Event Density}\textsuperscript{a}} & \multicolumn{1}{c}{9.98 (5.25)} & \multicolumn{1}{c}{3-19} & \multicolumn{1}{c}{} \\
\multicolumn{1}{l}{\quad (5)   \textit{Caesura}\textsuperscript{b}} & \multicolumn{1}{c}{} & \multicolumn{1}{c}{} & \multicolumn{1}{c}{Absent (28)} \\
\multicolumn{1}{l}{\quad (6)   \textit{Elision}\textsuperscript{c}} & \multicolumn{1}{c}{} & \multicolumn{1}{c}{} & \multicolumn{1}{c}{None (26)} \\
\textbf{Temporal Function} & \multicolumn{1}{c}{\textbf{}} & \multicolumn{1}{c}{\textbf{}} & \multicolumn{1}{c}{\textbf{}} \\
\multicolumn{1}{l}{\quad (7)   \textit{Interthematic Function}\textsuperscript{d}} & \multicolumn{1}{c}{} & \multicolumn{1}{c}{} & \multicolumn{1}{c}{End (20)} \\
\multicolumn{1}{l}{\quad (8)   \textit{Intrathematic Function}} & \multicolumn{1}{c}{} & \multicolumn{1}{c}{} & \multicolumn{1}{c}{Beginning (19)} \\
\multicolumn{1}{l}{\quad (9)  \textit{ Repetition}} & \multicolumn{1}{c}{} & \multicolumn{1}{c}{} & \multicolumn{1}{c}{Present (25)} \\
\multicolumn{1}{l}{\quad (10)   \textit{Stimulus Length} (s)} & \multicolumn{1}{c}{15.68 (4.14)} & \multicolumn{1}{c}{9.2-27.6} & \multicolumn{1}{c}{} \\
\multicolumn{1}{l}{\quad (11)   \textit{Stimulus Length from CA} (s)} & \multicolumn{1}{c}{7.20 (2.15)} & \multicolumn{1}{c}{3.6-14.1} & \multicolumn{1}{c}{} \\
\bottomrule
\insertTableNotes
\end{longtable}
\label{tab:ex2rhetorical}%
\end{ThreePartTable}
}

\end{document}

• I have no idea what the result of that looks like (and I wrote half the code it's using) Please always make examples a complete document loading all needed packages so that people can run them locally. – David Carlisle Feb 2 '15 at 14:59
• what do you mean by "because I'm using longtable, which allows me to stretch the table to textwidth. " ? longtable has no features for controlling horizontal widths other than those inherited from the underlying tabular longtable just allows page breaks. – David Carlisle Feb 2 '15 at 15:01
• Okay, I've included all of the preamble material so you should be able to reproduce the table as I currently see it. – David S Feb 2 '15 at 15:36

(Aside: I updated/simplified significantly after the OP indicated the method he/she uses to change the default line spacing of the document as well as the packages needed to make the code compilable.)

I have several suggestions:

• Since you're using the setspace package and the instruction \onehalfspacing to change the default line spacing for the document as a whole, you could insert the instruction \singlespacing after the start of the ThreePartTable environment to switch locally to single-spacing.

• The contents of the longtable environment are made needlessly complicated because of all those \multicolumn{1}{l}{...} and \multicolumn{1}{c}{...} wrappers. As far as I can tell, these wrappers are not needed. (All they do is create code clutter.)

• Since you're using the threeparttablex package, you could write \tnote{a} and \item[a] instead of the more-laborious \textsuperscript{a} and \item \textsuperscript{a}, etc.

Here's a simplified version of the example code you gave. I've commented out the instructions that load packages that aren't actually used in the MWE. By the way, there's no point in loading graphics and graphicx individually since the rotating package loads graphicx. Similarly, don't load color if you're also going to load xcolor.

\documentclass{report}

% packages
%\usepackage{amsmath}        % Extra math definitions
%\usepackage{graphics}       % PostScript figures
\usepackage{setspace}       % 1.5 spacing
\usepackage{longtable,threeparttablex}          % Tables spanning pages
%\usepackage{color}
%\usepackage{graphicx}
%\usepackage{rotating}
%\usepackage{array}
%\usepackage{multirow}
\usepackage{booktabs}
%\usepackage[table]{xcolor}
%\usepackage{subcaption}
%\usepackage{pdfpages}
%\usepackage[notes,backend=bibtex]{biblatex-chicago}
\usepackage[utf8]{inputenc}
%\usepackage[full]{textcomp}
\usepackage[T1]{fontenc}
\usepackage{afterpage}
%\usepackage{float}
%\usepackage{fp}
%\usepackage{pdflscape} %To rotate pages with sideways tables or large figures.
%\usepackage{xparse}
%\usepackage{siunitx} %Lets tables align columns by decimal point.
%\usepackage{lmodern}
%\usepackage[french,english]{babel}
%\usepackage{caption}
%\usepackage{pifont}
%\usepackage{microtype}
%\usepackage{amssymb}
%\usepackage{arydshln}
%\usepackage[bottom]{footmisc} %This places footnotes at the bottom so figures will appear above footnotes.
%\usepackage{cleveref}

%\newcommand{\tabitem}{~~\llap{\textbullet}~~}
%\DisableLigatures[f]{encoding=T1}
%\crefformat{footnote}{#2\footnotemark[#1]#3}
%
%%Let tables fit to width of page.
%\newcommand\totextwidth[1]{%
%\sbox{\mytabularbox}{#1}%
%\figwidthc=\wd\mytabularbox%
%\textwidthc=\textwidth%
%\FPdiv\scaleratio{\the\textwidthc}{\the\figwidthc}%
%\FPmin\scaleratio{\scaleratio}{1}%
%\scalebox{\scaleratio}{\usebox{\mytabularbox}}%
%}

\begin {document}

\onehalfspacing

\afterpage{%
\begin{ThreePartTable}
\singlespacing  % switch locally to single-spacing
\setlength{\LTleft}{0pt}
\setlength{\LTright}{0pt}
\renewcommand\TPTminimum{\textwidth}
%\renewcommand{\arraystretch}{0.8}  % not necessary, is it?
\begin{TableNotes}[flushleft]
\small
\item[a] \textit{Event Density} (1.5) refers to the number of notes identified in a 1.5~s window starting from the onset of cadential arrival.
\item[b] \textit{Caesura} refers to the presence of a rest across all four instrumental parts.
\item[c] \textit{Elision} refers to the superposition of a new intrathematic phrase at the moment of cadential arrival, an accompanimental overlap in the bass, or a melodic lead-in.
\item[d] \textit{Interthematic Function} refers to one of the following temporal functions to characterize the passage at the theme level following cadential arrival: Before-the-Beginning, Beginning, Middle, End, After-the-End.
\textit{Intrathematic Function} refers to either the Beginning, Middle, or End functions that characterize the passage at the phrase level following cadential arrival.
\end{TableNotes}
\setlength\tabcolsep{0.1pt}  % default value: 6pt
\begin{longtable}{@{\extracolsep\fill}lccc}  % make longtable span width of text block
\caption{Descriptive statistics for the 11 retrospective features.} \label{tab:ex2rhetorical}\\
\toprule
\textit{Retrospective Features}
& \textit{M} (\textit{SD})
& \textit{Range}
& \textit{Mode} (\textit{Frequency}) \\
\midrule
\textbf{Segmentational Grouping} & & & \\
\quad (1) \textit{Next Note Onset} (s) & .57 (.44) & .1--1.8 & \\
\quad (2) \textit{Next Bass Note Onset (s)} & .97 (.77) & .1--2.8 & \\
\quad (3) \textit{Next Soprano Note Onset} (s) & .99 (.66) & .1--2.6 & \\
\quad (4) \textit{Event Density}\tnote{a} & 9.98 (5.25) & 3--19 & \\
\quad (5) \textit{Caesura}\tnote{b} & & & Absent (28) \\
\quad (6) \textit{Elision}\tnote{c} & & & None (26) \\
\textbf{Temporal Function} & & & \\
\quad (7) \textit{Interthematic Function}\tnote{d} & & & End (20) \\
\quad (8) \textit{Intrathematic Function} & & & Beginning (19) \\
\quad (9) \textit{Repetition} & & & Present (25) \\
\quad (10) \textit{Stimulus Length} (s) & 15.68 (4.14) & 9.2--27.6 & \\
\quad (11) \textit{Stimulus Length from CA} (s) & 7.20 (2.15) & 3.6--14.1 & \\
\bottomrule
\insertTableNotes
\end{longtable}
\end{ThreePartTable}
}
\end{document}

• Perhaps you should update your remarks to reflect the use of threeparttablex (don't know if you added that you the OPs example) – daleif Feb 2 '15 at 16:01
• @daleif - Thanks. I had revised my answer after the OP indicated that he/she was using the setspace package, but I didn't notice right away that he/she also posted some information about the other packages that are being loaded. I've now deleted the second bullet point from the answer. Incidentally, I seem to have no problems combining longtable and threeparttable (and hence don't find it necessary to load the threeparttablex package). Is that because of the recent updates to these packages? – Mico Feb 2 '15 at 16:15
• You cannot use longtable and threeparttable directly. The threparttable env is a block that cannot be broken. That is why threparttablex provide extra envs of a different name. Actually the ThreePartTable env does not do much more than making sure a certain macro exists. – daleif Feb 2 '15 at 16:41
• @daleif - Many thanks for this explanation. I realize now that this issue doesn't show up in the OP's code because the threeparttable in question doesn't span more than one page. – Mico Feb 2 '15 at 16:46
• Exactly the point – daleif Feb 2 '15 at 17:46


text:
# [DirectX10] Parallax mapping looks wrong

## Recommended Posts

Hi there! I wont to put some parallax mapping into my engine, but something seems to be wrong: The pixel shader is here:
float4 wtPixelShader(PixelShaderInput Input) : SV_Target
{

float3 tCam = WorldToTangent_Inv(normalize(CamLoc-Input.VertexPos),Input.Norm,Input.Tangent,Input.Binormal);
float Height = TexUV(Tex2,UV*5).r * 0.03;
Height/=tCam.z;

float2 ParallaxUV = (UV) + tCam.xy * Height;

float4 Color = TexUV(Tex1,ParallaxUV);

return Color;
//return float4(ParallaxUV,0,1);
}

//... Include:

shared float3 WorldToTangent_Inv(float3 In,float3 Nrm,float3 Tan, float3 Bin)
{
float3x3 mTangentFrame = {Tan, Bin, Nrm };

return normalize(mul((float3x3)mTangentFrame, In));

}

Can you maybe give me some pseudo code? I ahve looked at the SDK sample already, but everything seems to be just like there...

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Hodgman    51328
Have you tried tweaking that 0.03 value?

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Yes, but the height is not the problem. When you take a closer look to the picture, you will notice that they bumps get shifted into the wrong direction...

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SIIYA    477
I had similar problem. The cause of my problem was wrong computation of mesh tangents. I used D3DXComputeTangentFrameEx() for tangent and binormal computation, and passing anything other than NULL for CONST DWORD * pdwAdjacency parameter caused this behavior in shader.

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I use a .x loader from the NVIDIA SDK10 for loading my meshes. So, I think they know what they are doing. Also normal mapping works just fine for me.

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JohnnyCode    1046

it seems that it shifts correctly by y axis but opposite at x axis.
Try this then

Height/=tCam.z;

tCam.x=-tCam.x;

float2 ParallaxUV = (UV) + tCam.xy * Height;

float4 Color = TexUV(Tex1,ParallaxUV);

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Sorry, my internet was down for a while. But I found out that the Vertex-Position I used was in Object space. So it works now, but it looks strange. Also, if I rotate the model, the parallax mapping stays the same, just like I never rotated it. :(

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rubicondev    296
There are few things that are almost always responsible for various bump mapping problems. I suggest you try em all ad hoc and see if any work:

1) Swap the tangent and binormal params around
2) Tranpose the matrix before you multiply with it
3) When rotating normals, remember to cast a full matrix to a float3x3

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text:
# 30th International Symposium on Lepton Photon Interactions at High Energies

Jan 10 – 14, 2022
Online only
Europe/London timezone

## Improved $V_{cs}$ determination from $D \to K \ell\nu$ and search for new physics in $B \to K \ell^+\ell^-$ using precise lattice QCD form factors

Jan 12, 2022, 9:20 AM
20m
Online only

#### Online only

Parallel session talk Flavour

### Speaker

Mr William Parrott (University of Glasgow)

### Description

We discuss HPQCD's recent precise determinations of the $D \to K$ and $B \to K$ scalar, vector and tensor form factors from full lattice QCD using the HISQ formalism.
Our $D \to K$ form factors allow us to calculate the differential rate for $D\to K \ell\nu$ with improved precision. Comparison with experimental results gives the CKM element $V_{cs}$ to better than 1\% for the first time: $|V_{cs}|=0.9663(80)$. We also demonstrate that the $q^2$-dependence of the rate predicted from QCD matches that of experiment.
Increasing the heavy quark mass from charm to bottom in our lattice QCD calculation gives access to the form factors for $B \to K \ell^+\ell^-$ across the full $q^2$ range of the decay process. We compare Standard Model observables calculated from our form factors to experimental measurements to pin down new physics contributions with improved significance over previous results.

### Primary author

Mr William Parrott (University of Glasgow)


text:
# Modulus (Complex Number)

Definition : The modulus of a complex number $a+bi$ is $(a+bi)(a-bi) = a^2+b^2$.

The modulus of a complex number is its magnitude thus the modulus can be denoted as $\lVert z \rVert = z \bar{z} = \mathrm{Re}(z)^2+ \mathrm{Im}(z)^2$. For example $\lVert 3+2i \rVert = (3+2i)(3-2i) = 9+4 = 13$.

See Also : Complex Numbers, Complex Conjugate, Real Part, Imaginary Part


text:
# Differentiation word problem (basic)

1. Jan 9, 2013

### Taylor_1989

A rectangle of length x, where x varies, has a constant area of 48cm2. Express the perimeter, y in terms of x. Find the least possible value of x.

my problem is not the maths part i.e. the differentiation, but the equation to get things moving. I really have no idea where to start. I drew a rectangle, but am not sure if I am suppose to call the width by a letter. What's really confusing me is the "x varies", what dose this mean? All I see I have an x and a 48 which to me, I cant seem to connect with the perimeter.

I would really appreciate some guidance here if possible. Thanks in advance.

2. Jan 9, 2013

### Saitama

Assume the width of rectangle as b.
Area, xb=48.
Perimeter, y=2(x+b).
Substitute b in the second equation.

3. Jan 9, 2013

### CAF123

If the width of the rectangle multiplied by it's length is always a constant, as the length varies (say increases) then it's width must decrease so as to preserve this constant area.

4. Jan 9, 2013

### Taylor_1989

Thanks for the reply final, got my ans in the end $16\sqrt3$ definitely going to need more practice with these. Has anyone got any tips, or is something that if you do enough of them you get used to them?

5. Jan 9, 2013

### Staff: Mentor

Well, yes, this is usually the difficult part, and this is part of mathematics, being able to analyze a situation and come up with an equation that represents it. After you have an equation, then you don't really have to think as much.
As you do more of them, you start to develop organization and analysis skills, and they tend to get a little easier.

For problems like this, you started off well by drawing a picture, but didn't follow through. You have a rectangle whose length is x, so hopefully you drew a rectangle and labelled its length as x. You are given that the area is 48.

Since you're not given the width, label the width of the rectangle - I would use w instead of b, but that's not very important. I definitely wouldn't use y, because the problem statement talks about y as the perimeter. (A better choice would be P, though.)

After you have the rectangle labelled, write equations for the perimeter and area.

You can extend these ideas to other problems - look carefully at the problem statement and translate the words into equations.

6. Jan 10, 2013

### Ray Vickson

Your later post indicates you have found the answer, but I hope you notice that your answered the *correct* question, rather than the one asked above. There is no least possible value of x; there is, though, a least possible value of perimeter (I hate using the letter y for that).


text:
### Show Posts

This section allows you to view all posts made by this member. Note that you can only see posts made in areas you currently have access to.

### Messages - Thomas Nutz

Pages: [1] 2
1
##### Home Assignment Y / Re: Problem 1
« on: December 19, 2012, 02:19:38 PM »
The question says we should do it with a Fourier transform, but why don't I just take
$$u(x,y)=e^{-|y|-ix}$$? That does give me a zero Laplacian and satisfies the boundary condition, or am I wrong here?

2
##### Term Test 2 / Re: TT2--Problem 1
« on: December 16, 2012, 01:30:08 PM »
Is eq. (3)
$$(f \star g)'=f\star g'$$ sufficient to prove that $(f \star g)'$ is continuous? I.e. can we assume that the convolution of two continuous functions is continuous, or do we have to take difference quotients as Ian did in his solution?

3
##### Misc Math / Re: Mean-value theorem
« on: November 26, 2012, 06:34:06 PM »
I see, thanks.

Are we considering closed domains, i.e. $\Sigma \in \Omega$? I think we have to, otherwise the $max_{\Omega}u \geq max_{\Sigma}u$ does not pull, correct?

4
##### Misc Math / Re: Mean-value theorem
« on: November 26, 2012, 01:51:41 PM »
or rather $\nabla G \cdot \frac{\partial}{\partial x}(\vec{v})$?

5
##### Misc Math / Re: Mean-value theorem
« on: November 26, 2012, 01:48:06 PM »
Hi,

what is meant by $\frac{\partial G}{\partial v_x}$ (eq. 7 in Lec. 26)? is $\frac{\partial G}{\partial v_x}=\nabla G \cdot (\nabla \cdot \vec{v})$?

Thanks!

6
##### Home Assignment 7 / Re: Problem 3
« on: November 17, 2012, 03:21:05 PM »
Are we supposed to derive eq. 19 in lecture notes 24 (laplacian outside the disk) for problem 3b) or can we start from this formula (it given in the context of an exercise)?

7
##### Home Assignment Y / Re: Problem 2
« on: November 13, 2012, 05:21:52 PM »
Ok I did
$$u(x,y)=\int_0^{\infty}\hat{u}(\omega,y)\sin(\omega x)d\omega$$
and took the resulting eigenvalue problem as
$$\hat{u}_{yy}-\omega^2\hat{u}=0$$
which yields eigenvalues $\omega=i\pi n$
and eigenfunctions
$$\hat{u}(\omega,y)=A(\omega) e^{i \pi n y}+B(\omega)e^{-i \pi n y}$$
Is that wrong?

8
##### Misc Math / Re: Lecture 20 Notes
« on: November 13, 2012, 12:24:36 PM »
In the section \textbf{Laplace equation in half-plane} it says

"The problem $u_{yy}+u_{xx}=0$, $y>0$,  $u(x,0)=g(x)$ [...] is not uniquely solvable". As an example the function $u(x,y)=y$ is given, which satisfies the Laplace equation. But it does obviously not satisfy $u(x,0)=g(x)$, so I don't see how we can conclude immediately that only bounded solutions are unique...

Can anybode help me with this?
Thanks!

9
##### Home Assignment 7 / Problem 3
« on: November 12, 2012, 04:40:07 PM »
Dear Professor,

do we need to find a closed expression for the solution in problem 3, or can we express it as a sum over n?

10
##### Home Assignment 7 / Re: Problem 2
« on: November 12, 2012, 02:09:39 PM »
Sorry I don't understand. I did find an ODE for $u(r)$, which is solved by Bessel functions. When you ask "try to find solutions", what do you expect me to do?

Thanks again!

11
##### Home Assignment 7 / Problem 2
« on: November 12, 2012, 01:13:34 PM »
Hey,

I get an ODE for problem 2, but it is not of the Euler type and I don't know how to solve it. Could anyone give me a hint?

Thanks!

12
##### Home Assignment 5 / plotting instead of sketching
« on: October 25, 2012, 05:17:54 PM »
Dear Professor,

can we plot the sum of some series elements or do you want us to draw it by hand?

13
##### Home Assignment X / Re: Problem2
« on: October 15, 2012, 12:04:21 PM »
I obtain the solution
$$u(x,t)=\frac{1}{2}((x-ct)e^{x+ct}+(x-ct)e^{x-ct})+ \int_{x-ct}^{x+ct}e^{-x}dx+\int_0^{t}\int_{x-ct}^{x+ct}e^{-x-t}dxdt$$
which becomes
$$\frac{2ce^{-x}}{1-c^2}+\frac{e^{-t}(e^{-ct}+e^{ct})}{1+c}$$
(plugging in $c=5$ doesn't really simplify it)

Did anyone get the same?

14
##### Home Assignment X / Re: problem 3
« on: October 13, 2012, 07:46:50 PM »
Thanks for your very quick response!
Is the 3 in your first point supposed to be a 2? I obtain
$$(u_t^*u_t)_t=3Re(u_{tt}u_t^*)$$
as I obtain (with u_t=f(t)+ig(t))
$$(u_t^*u_t)_t=u^*_tu_{tt}+u_tu_{tt}^*=(f-ig)(f'+ig')+(f+ig)(f'-ig')=2(ff'+gg')=2Re(u_{tt}u_t^*)$$

15
##### Home Assignment X / problem 3
« on: October 13, 2012, 06:11:36 PM »
Dear all,

I don't know what to do with problem 3. We are asked to find conditions on the three parameters $\alpha$, $\beta$ and $\gamma$ s.t. the integral
$$E(t)=\frac{1}{2}\int_0^L (|u_t|^2+c^2|u_x|^2+\gamma |u|^2)dx$$

is time-independent, where u satisfies b.c. and $u_{tt}-c^2u_{xx}+\gamma u=0$.

The time independence of the integral means that
$$\frac{\partial}{\partial t}u_tu^*_t+c^2\frac{\partial}{\partial t}u_xu^*_x+\gamma \frac{\partial}{\partial t}u u ^* =0$$

but I can`t find $u$, as there is this $u$ term in the wave equation, and the boundary conditions do not help me with this equation neither. Any hints? Thanks a lot!

Pages: [1] 2


text:
# determine the inverse Laplace transform of the given function. F(s)=\frac{2s}{s^2+9}

determine the inverse Laplace transform of the given function.
$F\left(s\right)=\frac{2s}{{s}^{2}+9}$
You can still ask an expert for help

## Want to know more about Laplace transform?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

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Step 1
Given
$F\left(s\right)=\frac{2s}{{s}^{2}+9}$
Step 2
${L}^{-1}\left\{\frac{s}{{s}^{2}+{a}^{2}}\right\}=\mathrm{cos}at$
so
${L}^{-1}\left\{\frac{2s}{{s}^{2}+9}\right\}$
$=2{L}^{-1}\left\{\frac{s}{{s}^{2}+9}\right\}$
$=2{L}^{-1}\left\{\frac{s}{{s}^{2}+9}\right\}$
$=2{L}^{-1}\left\{\frac{s}{{s}^{2}+{3}^{2}}\right\}$
$=2\mathrm{cos}3t$

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text:
# Properties

 Label 8100.2.d.j Level $8100$ Weight $2$ Character orbit 8100.d Analytic conductor $64.679$ Analytic rank $0$ Dimension $2$ CM no Inner twists $2$

# Related objects

## Newspace parameters

 Level: $$N$$ $$=$$ $$8100 = 2^{2} \cdot 3^{4} \cdot 5^{2}$$ Weight: $$k$$ $$=$$ $$2$$ Character orbit: $$[\chi]$$ $$=$$ 8100.d (of order $$2$$, degree $$1$$, not minimal)

## Newform invariants

 Self dual: no Analytic conductor: $$64.6788256372$$ Analytic rank: $$0$$ Dimension: $$2$$ Coefficient field: $$\Q(\sqrt{-1})$$ Defining polynomial: $$x^{2} + 1$$ Coefficient ring: $$\Z[a_1, \ldots, a_{13}]$$ Coefficient ring index: $$1$$ Twist minimal: no (minimal twist has level 324) Sato-Tate group: $\mathrm{SU}(2)[C_{2}]$

## $q$-expansion

Coefficients of the $$q$$-expansion are expressed in terms of $$i = \sqrt{-1}$$. We also show the integral $$q$$-expansion of the trace form.

 $$f(q)$$ $$=$$ $$q + 2 i q^{7} +O(q^{10})$$ $$q + 2 i q^{7} + 6 q^{11} -5 i q^{13} + 3 i q^{17} -2 q^{19} + 6 i q^{23} + 3 q^{29} -4 q^{31} + 5 i q^{37} + 6 q^{41} + 10 i q^{43} + 3 q^{49} -6 i q^{53} -12 q^{59} + 5 q^{61} + 2 i q^{67} -6 q^{71} + i q^{73} + 12 i q^{77} + 10 q^{79} -3 q^{89} + 10 q^{91} -10 i q^{97} +O(q^{100})$$ $$\operatorname{Tr}(f)(q)$$ $$=$$ $$2 q + O(q^{10})$$ $$2 q + 12 q^{11} - 4 q^{19} + 6 q^{29} - 8 q^{31} + 12 q^{41} + 6 q^{49} - 24 q^{59} + 10 q^{61} - 12 q^{71} + 20 q^{79} - 6 q^{89} + 20 q^{91} + O(q^{100})$$

## Character values

We give the values of $$\chi$$ on generators for $$\left(\mathbb{Z}/8100\mathbb{Z}\right)^\times$$.

 $$n$$ $$4051$$ $$6401$$ $$7777$$ $$\chi(n)$$ $$1$$ $$1$$ $$-1$$

## Embeddings

For each embedding $$\iota_m$$ of the coefficient field, the values $$\iota_m(a_n)$$ are shown below.

For more information on an embedded modular form you can click on its label.

Label $$\iota_m(\nu)$$ $$a_{2}$$ $$a_{3}$$ $$a_{4}$$ $$a_{5}$$ $$a_{6}$$ $$a_{7}$$ $$a_{8}$$ $$a_{9}$$ $$a_{10}$$
649.1
 − 1.00000i 1.00000i
0 0 0 0 0 2.00000i 0 0 0
649.2 0 0 0 0 0 2.00000i 0 0 0
 $$n$$: e.g. 2-40 or 990-1000 Significant digits: Format: Complex embeddings Normalized embeddings Satake parameters Satake angles

## Inner twists

Char Parity Ord Mult Type
1.a even 1 1 trivial
5.b even 2 1 inner

## Twists

By twisting character orbit
Char Parity Ord Mult Type Twist Min Dim
1.a even 1 1 trivial 8100.2.d.j 2
3.b odd 2 1 8100.2.d.a 2
5.b even 2 1 inner 8100.2.d.j 2
5.c odd 4 1 324.2.a.b 1
5.c odd 4 1 8100.2.a.f 1
15.d odd 2 1 8100.2.d.a 2
15.e even 4 1 324.2.a.d yes 1
15.e even 4 1 8100.2.a.a 1
20.e even 4 1 1296.2.a.a 1
40.i odd 4 1 5184.2.a.bc 1
40.k even 4 1 5184.2.a.z 1
45.k odd 12 2 324.2.e.d 2
45.l even 12 2 324.2.e.a 2
60.l odd 4 1 1296.2.a.j 1
120.q odd 4 1 5184.2.a.d 1
120.w even 4 1 5184.2.a.g 1
180.v odd 12 2 1296.2.i.d 2
180.x even 12 2 1296.2.i.p 2

By twisted newform orbit
Twist Min Dim Char Parity Ord Mult Type
324.2.a.b 1 5.c odd 4 1
324.2.a.d yes 1 15.e even 4 1
324.2.e.a 2 45.l even 12 2
324.2.e.d 2 45.k odd 12 2
1296.2.a.a 1 20.e even 4 1
1296.2.a.j 1 60.l odd 4 1
1296.2.i.d 2 180.v odd 12 2
1296.2.i.p 2 180.x even 12 2
5184.2.a.d 1 120.q odd 4 1
5184.2.a.g 1 120.w even 4 1
5184.2.a.z 1 40.k even 4 1
5184.2.a.bc 1 40.i odd 4 1
8100.2.a.a 1 15.e even 4 1
8100.2.a.f 1 5.c odd 4 1
8100.2.d.a 2 3.b odd 2 1
8100.2.d.a 2 15.d odd 2 1
8100.2.d.j 2 1.a even 1 1 trivial
8100.2.d.j 2 5.b even 2 1 inner

## Hecke kernels

This newform subspace can be constructed as the intersection of the kernels of the following linear operators acting on $$S_{2}^{\mathrm{new}}(8100, [\chi])$$:

 $$T_{7}^{2} + 4$$ $$T_{11} - 6$$ $$T_{29} - 3$$

## Hecke characteristic polynomials

$p$ $F_p(T)$
$2$ $$T^{2}$$
$3$ $$T^{2}$$
$5$ $$T^{2}$$
$7$ $$4 + T^{2}$$
$11$ $$( -6 + T )^{2}$$
$13$ $$25 + T^{2}$$
$17$ $$9 + T^{2}$$
$19$ $$( 2 + T )^{2}$$
$23$ $$36 + T^{2}$$
$29$ $$( -3 + T )^{2}$$
$31$ $$( 4 + T )^{2}$$
$37$ $$25 + T^{2}$$
$41$ $$( -6 + T )^{2}$$
$43$ $$100 + T^{2}$$
$47$ $$T^{2}$$
$53$ $$36 + T^{2}$$
$59$ $$( 12 + T )^{2}$$
$61$ $$( -5 + T )^{2}$$
$67$ $$4 + T^{2}$$
$71$ $$( 6 + T )^{2}$$
$73$ $$1 + T^{2}$$
$79$ $$( -10 + T )^{2}$$
$83$ $$T^{2}$$
$89$ $$( 3 + T )^{2}$$
$97$ $$100 + T^{2}$$


text:
# 24 and 12 Hour Times

Write a program or function with no input that prints or returns this string of 24-hour and 12-hour times:

00:00 12:00am
01:00  1:00am
02:00  2:00am
03:00  3:00am
04:00  4:00am
05:00  5:00am
06:00  6:00am
07:00  7:00am
08:00  8:00am
09:00  9:00am
10:00 10:00am
11:00 11:00am
12:00 12:00pm
13:00  1:00pm
14:00  2:00pm
15:00  3:00pm
16:00  4:00pm
17:00  5:00pm
18:00  6:00pm
19:00  7:00pm
20:00  8:00pm
21:00  9:00pm
22:00 10:00pm
23:00 11:00pm

The string must be output exactly as it appears here. The only exception is that it may optionally have a single trailing newline.

So the MD5 hash of your output should be

827ae6e2dbb1df494930baedb3ee2653

if you do not have a trailing newline and

cd4c3d18abee9bafb495f390a919a13f

if you do. (Your hash could be different if your system uses another type of newline but that's OK.)

The shortest code in bytes wins. Tiebreaker is earlier answer.

• – Joshpbarron May 31 '16 at 8:01

# Bash + coreutils, 43 30

• Saved 7 bytes thanks to @Yossarian
• Saved 3 bytes thanks to @AndersKaseorg
seq 0 23|date -f- +%R\ %l:00%P

• seq generates integers 0-23, one per line.
• date interprets each line as a datetime. Bare integers appear to be sufficient to be recognised as hours of the day by date. date then outputs each time with the required formatting using the available time format specifiers.

Assumes LANG=C locale, as per this meta answer.

Ideone.

• You can even just do seq 0 23 and save 7 bytes – Yossarian Jun 1 '16 at 8:21
• And %H:00%R saves another 3. – Anders Kaseorg Jun 1 '16 at 11:07
• @AndersKaseorg, nop, %R inserts a leading zero, not wanted by the OP. – rexkogitans Jun 1 '16 at 11:44
• @rexkogitans A leading zero is wanted (in the first column, which is the one I’m talking about). – Anders Kaseorg Jun 1 '16 at 11:54
• @AndersKaseorg, oh, you were talking about the first one... then you are right, of course. Corrected my comment above. – rexkogitans Jun 1 '16 at 12:27

# Python 2, 66 bytes

for i in range(24):print'%02d:00%3d:00%sm'%(i,12+i%-12,'ap'[i>11])

• modulo -12 is clever! – Erik the Outgolfer May 31 '16 at 8:00
• Just in case anyone is wondering: i=0;exec"print'%02d:00%3d:00%cm'%(i,~-i%12+1,97+i/12*15);i+=1;"*24 changes the code in three different ways, but each change is the same length. – Sp3000 May 31 '16 at 8:31

# C, 73 bytes

m(i){for(i=25;--i;)printf("%02d:00%3d:00%cm\n",24-i,12-i%12,"pa"[i/13]);}

mIllIbyte found a particularly neat way to rewrite this answer. Thanks!

• Cool - I never knew about ?: – Digital Trauma May 31 '16 at 17:27
• I didn’t know! Interesting. I fixed it :) – Lynn May 31 '16 at 18:47
• Ahh, well, ?: is actually the closest equivalent to Perl’s ||, I believe. C’s || is more like (a || b) ? 1 : 0. – Lynn May 31 '16 at 20:48
• i++,i%12?:12,"ap"[i/12]) lacks a sequence point. Can't be certain when i++ occurs. Perhaps for(i=0;i<24;i++) ...i,i%12?:12,"ap"[i/12]... – chux - Reinstate Monica Jun 1 '16 at 4:26
• i++<24 is better - doesn't add more bytes to the source code – anatolyg Jun 1 '16 at 7:17

# MATL, 4642 34 bytes

12tEt:qy/t15XObZ"!b16XOhhkw14:X~Z)

Previously, 42 bytes, 12tEt:q2M/736330+t15XObZ"!b16XOhhkw14:X~Z), and 46 bytes, 736330 24t:qw/+t15XO' '24TX"b16XOhhk14: 12X~Z). Of course the 736330 wasn't needed, that was crazy!

Note: Doesn't work with TryItOnline, I think there is a compatibility issue between Matlab and Octaves implementation of datestr.

datestr takes the number representation of a date and converts it to the string representation of that date. The time of the day is the fractional part of the number, so 0.0 corrsponds to January 0, 0000, at time 00:00:00, and 1.0 corresponds to January 1, 0000, at 00:00:00. 1/24 is 1am, 2/24 2am etc.

Explanation

12t         % push a 12 onto the stack and duplicate
Et          % double the 12 and duplicate the 24 (stack now has 12, 24, 24, bottom to top)
:q          % make vector 1:24 and decrement by 1, stack has 12, 24, 0:23
y           % duplicate second element on stack (24)
/           % divide, for (0:23)/24
t           % duplicate elements
15XO        % string representation of date, 15 specifies format
b           % bubble up element in stack (gets a 24 on top of the stack)
Z"!         % makes a column of 24 spaces, to put between columns of times
b           % bubble up another (0:23)/24
16XO        % string representation of date, 16 for a different format
hh          % concatenate two time vectors and the column of spaces
k           % convert string to lowercase, because CO gives AM/PM not am/pm
w           % swap elements in stack, that first 12 is now on top
14:         % vector of equally spaced values 1:14
X~          % set exclusive-or, returns [1 2 3 4 5 6 7 8 9 10 11 13 14]
Z)          % get the right columns of the string array to remove extra column of blanks
% implicit display

To show it works in Matlab, here is a screenshot

# ///, 160 bytes

/Z/:00 //S/Z //A/:00am
//P/:00pm
/00Z12A01S1A02S2A03S3A04S4A05S5A06S6A07S7A08S8A09S9A10Z10A11Z11A12Z12P13S1P14S2P15S3P16S4P17S5P18S6P19S7P20S8P21S9P22Z10P23Z11P

Try it online!

Ungolfed

00:00 12:00am
01:00  1:00am
02:00  2:00am
03:00  3:00am
04:00  4:00am
05:00  5:00am
06:00  6:00am
07:00  7:00am
08:00  8:00am
09:00  9:00am
10:00 10:00am
11:00 11:00am
12:00 12:00pm
13:00  1:00pm
14:00  2:00pm
15:00  3:00pm
16:00  4:00pm
17:00  5:00pm
18:00  6:00pm
19:00  7:00pm
20:00  8:00pm
21:00  9:00pm
22:00 10:00pm
23:00 11:00pm

• Fun fact: making the :00 into a replacement is one byte longer... TIO – steenbergh Feb 1 '17 at 9:42
• @steenbergh That's because :00 is a 3-byte string that appears 3 times. 3 × 3 = 9, making a replacement in your case costs 3+1+3=7 since it uses a 1-byte alias, and you use it 3 times, so 7+3=10. 9 < 10, so I won't replace it. – Erik the Outgolfer Feb 1 '17 at 11:49

# MarioLANG, 965 834 bytes

Try it online

well this was ridiculously complicated.

Technically the output is valid but in practice the Tio for MarioLANG output "n " instead of "n" when we print a number with ':'

if I find the time i'll guess i'll try to do a (probably much longer) version of the program that output correctly on Tio

++<>) +++@++++>   [!) >)>((((::(.)::((.))+:+:--(.)::)).).).)+++++++++
++""+ +"=====""====#) "+"============================================
+++)+ +>>+++++- <+<)->+++ ![-).).).))(::)).(:(:)))..(((::)).(:+(:((((<
+>+++ ++"====<( ")")-"!+++#=========================================="
+(+++>++!++)<+( ++++-+++++>
-))+)=(#==="+( ++++)+++++"==========================================!
[!!+-[!(+++!!! !+!<+!++!>(((((+:(.))::(((.
==##===#====###=#=#"=##=#"=================<
++++)))+++++++++++++++(((![-).).).)::)).(:))   >
>========================#==================   "
+>+ >
+"+ "=======================================[   =====================#===============[
+!> ! -).).).))(::)).)):+(..(((::)).(:+(((((<++!-).).).))(::)).)):+(.(((::)).(:+(((((<
=#==#======================================="==#======================================

Explanation :

our main problem here is the fact that we have 6 NaN char (newLine, Space, :, a, p, m)

in marioLANG, in order to print character, we need their ascii value:

• newLine is 10
• Space is 32
• : is 58
• a is 97
• p is 112
• m is 109

So the first thing to do is to set the memory :

++<>) +++@++++>   [!) >)> !
++""+ +"=====""====#) "+"==
+++)+ +>>+++++- <+<)->+++
+>+++ ++"====<( ")")-"!+++
+(+++>++!++)<+( ++++-+++++
-))+)=(#==="+( ++++)+++++
[!!+-[!(+++!!! !+!<+!++!>
==##===#====###=#=#"=##=#"

with this, the memory look like :

                   v
32 58 0 0 97 109 10 0
_  :      a   m  \n

we will transform a into p during the rest of the program

then we do the actual output :

++<>) +++@++++>   [!) >)>((((::(.)::((.))+:+:--(.)::)).).).)+++++++++
++""+ +"=====""====#) "+"============================================
+++)+ +>>+++++- <+<)->+++ ![-).).).))(::)).(:(:)))..(((::)).(:+(:((((<
+>+++ ++"====<( ")")-"!+++#=========================================="
+(+++>++!++)<+( ++++-+++++>
-))+)=(#==="+( ++++)+++++"==========================================!
[!!+-[!(+++!!! !+!<+!++!>(((((+:(.))::(((.
==##===#====###=#=#"=##=#"=================<
++++)))+++++++++++++++(((![-).).).)::)).(:))   >
>========================#==================   "
+>+ >
+"+ "=======================================[   =====================#===============[
+!> ! -).).).))(::)).)):+(..(((::)).(:+(((((<++!-).).).))(::)).)):+(.(((::)).(:+(((((<
=#==#======================================="==#======================================

# Julia, 887166 64 bytes

[@printf "%02d:00%3d:00%cm
" i-11 i%12+1 i>22?112:97for i=11:34]

This is a full program that prints the string with a single trailing newline.

Try it online!

Saved 5 bytes thanks to Sp3000 and 2 thanks to Dennis!

## C# function, 100 bytes

void F(){for(int i=0;i<24;i++){Console.Write($"{i:00}:00 {(i+11)%12+1,2}:00 {(i>11?"p":"a")}m\n");}}  Ungolfed version: void F() { for (int i = 0; i < 24; i++) { Console.Write($"{i:00}:00 {(i + 11)%12 + 1,2}:00 {(i > 11 ? "p" : "a")}m\n");
}
}

Console.Write() takes too many characters!

• Hello, and welcome to PPCG! I use C# a bit, but what is the leading $ doing to the string? – NoOneIsHere Jun 1 '16 at 1:48 • @NoOneIsHere - That's called "string interpolation" and it's a new feature in C# 6.0. You can read more about it here https://msdn.microsoft.com/en-us/library/dn961160.aspx. – STLDev Jun 1 '16 at 3:15 ## JavaScript (ES2015), 147138137134 133 bytes ((o,x,r)=>{for(i=0;i<24;)b=i%12,c=b||12,o+='0'[r](i<10)+i+++x+' '[r]((c<10)+1)+c+x+(i<13?'a':'p')+"m\n";return o})('',':00','repeat')  In this version I took advantage of String.repeat() method to get rid of lengthy .slice() and .join() and moved incrementation inside loop. Previous version: ((o,x,i)=>{for(;i<24;i++){b=i%12;o+=[0${i+x}.slice(-5),(b||12)+x+(i<12?'a':'p')+'m'].join(' '.repeat((b>0&&b<10)+1))+"\n"}return o})('',':00',0)

Gives output with trailing newline. Tested in Firefox Scratchpad. Not sure if passing arguments to IIFE is OK with "no input" rule.

It's my first submission, so hello to everyone! :)

# TSQL(SQLServer 2012) 146124 121

DECLARE @ DATETIME=0WHILE @<1BEGIN PRINT

Try it online!

First attempt, a bit longer, but a one-liner:

SELECT CONVERT(char(5),n,108)+' '+LOWER(RIGHT(FORMAT(n,'g'),8))FROM(SELECT
top 24 dateadd(hh,Number,0)n FROM master..spt_values WHERE'P'=type)x

Try it online!

• Reduced to 121 chars: try it online – Ross Presser Jun 14 '16 at 14:58
• @RossPresser once again you helped me, learning alot here – t-clausen.dk Jun 14 '16 at 15:06

# Perl 5, 58

map{printf"%02u:00%3u:00%sm
",$_,$_%12||12,$_>11?p:a}0..23  # Javascript, 122 Bytes, 120 Bytes f=j=>j>12?j-12:j;for(i=0;i<24;i++)console.log('%s:00 %s:00%s',i<10?'0'+i:i,i==0?12:f(i)<10?' '+f(i):f(i),i>11?'pm':'am')  Edit: Small bug fixed + output: 00:00 12:00am 01:00 1:00am 02:00 2:00am 03:00 3:00am 04:00 4:00am 05:00 5:00am 06:00 6:00am 07:00 7:00am 08:00 8:00am 09:00 9:00am 10:00 10:00am 11:00 11:00am 12:00 12:00pm 13:00 1:00pm 14:00 2:00pm 15:00 3:00pm 16:00 4:00pm 17:00 5:00pm 18:00 6:00pm 19:00 7:00pm 20:00 8:00pm 21:00 9:00pm 22:00 10:00pm 23:00 11:00pm  # V, 56 53 bytes i00:00 23ñYpñH12G$yP13G$pgvó$/am
í/pm
í 0/
í/12

Try it online!

Since this can be hard to enter, here is a reversible hexdump:

00000000: 6930 303a 3030 201b 3233 f159 7001 f148  i00:00 .23.Yp..H
00000010: 1631 3247 2479 5031 3347 2470 6776 f320  .12G$yP13G$pgv.
00000020: 242f 616d 0aed 2f70 6d0a ed20 302f 2020  $/am../pm.. 0/ 00000030: 0aed 2f31 320a ../12.  A noncompeting version is trivially 2 bytes shorter if you replace both occurrences of G$ with L, which was supposed to be the same but had a bug.

Explanation:

i00:00<esc>                                     #Enter the starting text.
23ñYp<C-a>ñ                          #Duplicate and increment 23 times
H                         #Move back to the beginning
<C-v>12G$y #Select 12 lines horizontally P #Horizontally paste 13G$p         #Move to line 13 and Horizontally paste again
gv       #Reselect the top 12 lines
ó $/am #Replace a space at the end of the line with 'am'  í/pm #Replace the previous search with 'pm' í 0/ #Replace "Space+0" with 2 spaces í/12 #Replace the previous search with "12"  ## 05AB1E, 51504844 42 bytes Saved two bytes thanks to carusocomputing Code: 24FNgi0}N…:00©ðN12(%12+Dgiðs}®„paN12‹è'mJ,  Try it online! Explanation 24F # for N in [0...23] Ngi0} # if len(N)=1, push 0 N # push N …:00© # push ":00" and store a copy in register ð # push " " N12(%12+D # push 2 copies of N%(-12)+12 giðs} # if the length of that number is 1, # push " " and swap with the number ® # push ":00" again „pa # push "pa" N12‹è # index into that with N<12 'm # push "m" J, # join everything and print with newline  • codegolf.stackexchange.com/questions/103242/… ;). Now whether or not that could be optimized below 50 is up for question haha. – Magic Octopus Urn Jan 31 '17 at 21:14 • 23Ý instead of 24L< for 1 byte. And how long has ë existed? I feel so dumb not knowing about else statements in 05AB1E until now. – Magic Octopus Urn Jan 31 '17 at 21:18 • @carusocomputing: Thanks! else has existed for a while, but it's been buggy at times. Especially when nesting ifs. – Emigna Jan 31 '17 at 21:22 ## PowerShell v2+, 76 bytes 0..23|%{"{0:D2}:00{1,3}:00"-f$_,(($_%12),12)[!($_%12)]+('am','pm')[$_-ge12]}  Loops from 0..23 and each loop sets a string with the -f operator. The first {0:D2} ensures we have prepended zeros, the second {1,3} ensures we have padded spaces for the middle column. The {0} one corresponds to the $_ of the -f operator, while the {1} corresponds to the pseudo-ternary that chooses between $_%12 or 12 based on whether $_%12 is non-zero or not (i.e., if we're at $_=13, this will choose 1 for 1:00pm). We then concatenate that with another pseudo-ternary that chooses the appropriate am/pm. As opposed to my answer on List all times in the day at a half hour rate, it's actually shorter here to brute-force the numbers since we get significantly cheaper padding. Here's the answer using date functions, at 78 bytes 0..23|%{(Date -h$_ -f 'HH:00')+(Date -h $_ -f "h:00tt").ToLower().PadLeft(8)}  # C++, 81 79 bytes []{for(time_t t=0,y;t<24;cout<<put_time(gmtime(&y),"%R %l:00%P\n"))y=t++*3600;}  This code requires using namespace std somewhere preceding it. It does a loop on the values 0...23. It multiplies each value by 3600, converts to a tm struct and prints it. The print format %R outputs the 24-hour and minute; the print formats %l and %P output the proper 12-hour parts; they require GNU. A working online version is here. # Ruby, 66 62 bytes 0.upto(23){|i| puts "%02d:00%3d:00#{'ap'[i/12]}m"%[i,(i-1)%12+1]}  New version 24.times{|i|puts"%02d:00%3d:00#{'ap'[i/12]}m"%[i,(i-1)%12+1]}  • 24.times is shorter. No need for the spaces around puts. – manatwork Jun 1 '16 at 12:05 • You can replace (i-1) with ~-i for 2 bytes. – Jordan Nov 7 '17 at 18:55 # JavaScript (ES6), 119 116 bytes _=>Array(24).fill().map((_,i)=>${0${i}.slice(-2)}:00${ ${(i+11)%12+1}.slice(-2)}:00${'ap'[+(i>11)]}m).join

# Sclipting, 76 bytes

The program assumes that the input is empty (or '0' or anything that converts to the integer 0).

The byte count assumes UTF-16 encoding.

군上❶겠小꼀虛嗎❷꾣갰글❷결加곀剩增❶겠小글虛嗎댆밁⓷꾣갰⓷⓼곀小掘닐밊終

Ungolfed:

군 // 23

❶겠小꼀虛嗎 // n < 10 ? "0" : ""
❷          // n
꾣갰글      // ":00 "
❷결加곀剩增 // k = (n+11) % 12 + 1
❶겠小글虛嗎 // k < 10 ? " " : ""
댆밁       // "pa"
⓷         // Pull n to top of stack
꾣갰       // ":00"
⓷         // Pull "pa" to top of stack
⓼         // Pull k to top of stack
곀小掘     // "pa"[k < 10 ? 1 : 0]
닐밊       // "m\n"

Each iteration of the loop leaves lots of small strings on the stack; at the end they are all automatically concatenated.

# JavaScript, 97 95 bytes

This is based off of starcorder’s answer. Thanks to George Reith for a 2 byte improvement.

for(i=0,k=12;i<24;k=i++%12+1)console.log('%s:00 %s:00%sm',i>9?i:'0'+i,k>9?k:' '+k,i>11?'p':'a')

Ungolfed:

for (i=0, k=12; i < 24; k = (i++) % 12 + 1)
console.log('%s:00 %s:00%sm',
i > 9 ? i : '0' + i,
k > 9 ? k : ' ' + k,
i > 11 ? 'p' : 'a')

• You can save 2 bytes by rewriting n < 10 as 9 < n and swapping the ternary cases – George Reith Jul 13 '16 at 11:40

## Batch, 167 bytes

@echo off
set h=11
set p=a
for /l %%a in (0,1,23)do call:e %%a
exit/b
:e
set a=0%1
set/ah=h%%12+1
set h= %h%
if %1==12 set p=p
echo %a:~-2:00 %h:~-2%:00%p%m

# Kotlin, 95 bytes

It can be improved for sure.

fun p(){for(i in 0..23)println("%02d:00 ${(i+11)%12+1}:00${if(i>12)"p" else "a"}m".format(i))}

# PHP, 110 107 bytes

for($h=0;$h<24;){$m=($h+11)%12+1;echo($h<10?0:"")."$h:00 ".($m<10?" ":"")."$m:00".($h++<12?"a":"p")."m\n";}  exploded view for ($h=0; $h<24; ) {$m = ($h+11) % 12 + 1; echo ($h < 10 ?  0  : "") . "$h:00 " . ($m < 10 ? " " : "") . "$m:00" . ($h++ < 12 ? "a" : "p") . "m\n";
}

Somewhat surprised, tried to turn the ($i < 10 ?$s : "") . "$i:00" bit into a function, but it wound up adding ~25 characters. No go there. ## Swift, 85 bytes for x in 0...23{print(String(format:"%02d:00 %2d:00\(x<12 ?"a":"p")m",x,12+x % -12))}  • This doesn't work, at least not for me on Swift 2.2. Output – JAL Jun 1 '16 at 22:14 • I had to change your format string: String(format: "%02d:00 %2d:00\(x<12 ?"a":"p")m", x, x%12 != 0 ? x%12 : 12) – JAL Jun 1 '16 at 22:15 # C Function, 82 bytes m(i){for(;i<24;printf("%02d:00 %2d:00%cm\n",i,i%12==0?12:i%12,i>11?'p':'a'),i++);}  Usage, 94 Byte m(i){for(;i<24;printf("%02d:00 %2d:00%cm\n",i,i%12==0?12:i%12,i>11?'p':'a'),i++);}main(){m();}  Ungolfed, 337 Bytes #include <stdio.h> void m(){ int i,a; char c; for(i=0;i<24;i++){ if (i%12==0){ a = 12; } else{ a = i%12; } if (i>11){ c = 'p'; } else{ c = 'a'; } printf("%02d:00 %2d:00%cm\n",i,a,c); } } int main(){ m(); }  it works on Windows: # C Program, 85 bytes main(i){for(;i<24;printf("%02d:00 %2d:00%cm\n",i,i%12==0?12:i%12,i>11?'p':'a'),i++);}  • You might need initialization i=0 for the function version. And for the program version, you can stuff the code directly into main - no need to define a function there! – anatolyg May 31 '16 at 16:44 • Thanks! edited!! FYI: In the funcion version if you use it without params it's self initialized to 0! – Giacomo Garabello May 31 '16 at 16:53 • i%12==0?12:i%12 --> i%12?i%12:12 – chux - Reinstate Monica Jun 1 '16 at 11:54 • "In the funcion version if you use it without params it's self initialized to 0!" This does not appear to be standard C. Any reference to support this? – chux - Reinstate Monica Jun 1 '16 at 11:57 • i have noticed it works only on the gcc i have installed on my windows system.. a friend of mine with linux told me it doesn't work on his pc but i don't know how to fix it for linux... – Giacomo Garabello Jun 1 '16 at 12:51 # Foo, 163 bytes Pretty brute-force approach; nothing clever here (I tried in a couple spots but it ended up being shorter not to), just wanted to give Foo a shot. Foo automatically prints anything within quotes. $c10 prints a line break. (## ... ) loops until the current cell equals ##.

"00:00 12:00am"$c10+1(10"0"$i":00  "$i":00am"$c10+1)(12$i":00 "$i":00am"$c10+1)"12:00 12:00pm"$c10+1(22$i":00 ">+1$i<":00pm"$c10+1)(24$i":00 ">+1$i<":00pm"$c10+1)

Ungolfed a bit:

"00:00 12:00am"$c10+1 (10"0"$i":00  "$i":00am"$c10+1)
(12$i":00 "$i":00am"$c10+1) "12:00 12:00pm"$c10+1
(22$i":00 ">+1$i<":00pm"$c10+1) (24$i":00 ">+1$i<":00pm"$c10+1)

Try it online

# Javascript (using external library - Enumerable) (107 bytes)

_.Range(0,24).WriteLine(x=>((x<10?"0"+x:x)+":00 "+(((h=((x+11)%12)+1))<10?" "+h:h)+":00"+(x<12?"am":"pm")))

FOR H=0TO 23?FORMAT$(%02D:00 %2D:00%Sm",H,(H+11)MOD 12+1,"ap"[H>11])NEXT  Someone found a better 24->12 hour formula than my old one, which saves 3 bytes, and 5 bytes in another program # PHP, 6765 64 bytes This uses IBM-850 encoding. for(;$i<24;)printf(~┌¤═ø┼¤¤┌╠ø┼¤¤┌îÆ§,$i,$i%12?:12,$i++>11?p:a);  With the unencoded string (66 bytes): for(;$i<24;)printf("%02d:00%3d:00%sm\n",$i,$i%12?:12,$i++>11?p:a);  Run like this: php -n -r 'for(;$i<24;)printf(~┌¤═ø┼¤¤┌╠ø┼¤¤┌îÆ§,$i,$i%12?:12,$i++>11?p:a);'  ## Tweaks • Saved 2 bytes by improving sprintf format • Saved a byte by getting rid of unnecessary space (thx @Titus) • You can use -n instead of -d error_reporting=30709. There should be no space before am/pm. Save one byte on the unencoded version with a physical linebreak. – Titus Feb 1 '17 at 20:33 # tcl, 93 set i 0;time {puts [format %02d:00%3d:00[expr$i<12?"a":"p"]m $i [expr$i%-12+12]];incr i} 24