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What's Wrong With This USA Today Item,

from the July 22 issue?

Guest:
Also, the graph only has 11 months of data.
8.2.2005 3:47pm
Fern R (mail):
Other than that the image is HUGE and I can't look at it on one screen?
8.2.2005 3:48pm
Cam (mail) (www):
Also, the graph itself is horrendous. They left out December, showing only 11 month's worth of data.
8.2.2005 3:49pm
lucia (mail) (www):
There is a second problem. Focusing on the plot itself, the arrows should point to the 15th day of the month, which should fall in midmonth. The way the plot is organized makes it appear there are 9 hours and 26 minutes of Daylight December 31 and 9 hours 45 on January 1.

Why do people use bar charts when they should use scatter plots?
8.2.2005 3:51pm
Steve:
I think you folks are looking at the bars as data points when you should be looking at the lines. This may make it a poor chart from a graphical standpoint, but it does look like they included 12 ides' worth of data.
8.2.2005 3:57pm
Shelby (mail):
The statement about north/south is inaccurate. It may be true as a total amount for the whole year, but my brother in Anchorage is getting a lot more daylight than this chart shows for Topeka in July.
8.2.2005 4:06pm
Greg Morrow (mail) (www):
As you get closer to the equator, you do get more sunlight, not in hours, but in intensity. The energy received from the sun scales by the cosine of latitude.
8.2.2005 4:10pm
UberTod (mail):
Ultimately it has to do with the 23.45 degree tilt of the planet on its' axis. But you COULD argue the mountain range and valley part too.
8.2.2005 4:38pm
DelVerSiSogna:
I don't know if it's a problem specific to certain web browsers, but the picture is enormous on mine, and it makes the entire rest of the page very difficult to read.
8.2.2005 4:47pm
lucia (mail) (www):
The display size of the image is partly Eugene's fault. He should specify height and width in his tag; I find specifying pixels works best. (He specified width= 80%, but that often works poorly.)
8.2.2005 4:51pm
Jack (mail) (www):
The graph isn't missing December's data. The graph shows daylight on the 15th day of the month. As Steve said above, the lines are the data points. The space in between is not data but filler.

And yes, the image is 'way too big!
8.2.2005 5:17pm
Eugene Volokh (www):
Sorry about the display problem -- I had assumed that the percentage width would work best, given varying display sizes, but now it seems that an absolute size is better. Hope it works for everyone now.
8.2.2005 5:40pm
pct:
Presumably the context was a discussion of extending daylight savings time, and USA Today caption referred to the minimum amount of daylight per year, not the mean. Hence the two boxes with arrows at the sides of the graph. The corresponding values for Anchorage are 6h 24 m (Jan) and 5h 32m (Dec).
8.2.2005 7:02pm
Kristen Chopra (mail):
I thought for sure the answer was "It's about Topeka, Kansas." (And as a native Kansan, that's from personal experience too.)
8.2.2005 7:36pm
Syd Henderson (mail):
Actually, the average amount of sunlight at a given location is slightly more than 12 hours, because the sun's light is refracted by the atmosphere. That means you can see the sun a few minutes earlier at dawn and later at sunset than you could if there were no atmosphere (ignoring the fact you'd also be dead).
8.2.2005 7:48pm
Dick King:
Indeed, it's only true in the fall and winter that you get more hours of sunlight as you move south, although you indeed get more energy from sunlight over the year as you move towards the equater.

There's an interesting mathematical problem here.

On a planet where the axis of rotation is parallel to the axis of the orbit, the equater gets a certain amount of sunlight but the poles get essentially none [considering the sun to be a point and considering the planet to not have a refractive atmosphere]. Such a planet is said to have an inclination of zero degrees.

On a planet where the axis of rotation is in the plane of the orbit, the poles get a certain amount of sunlight, the same amount as the equater gets on the zero inclination planet, and the equater gets less, so the poles get more than the equater. Such a planet has an inclination of ninety degrees.

The amount of sunlight reacing the pole and reaching the equater is a continuous monotonic function of the inclination, so there must be a unique inclination where the poles and the equater get the same amount of sunlight. Can anyone figure out that inclination? Can anyone figure out whether, on such a planet, the intermediate latitudes get that same amount of sunlight, or more, or less?

-dk
8.2.2005 9:27pm
bago (mail):
If you want to nitpick I suppose you could also point out that a linear connection between the ides is shown, when infact the curve would be a part of the greated sinoind if one were to properly interpolate.
8.2.2005 10:40pm
Zev Sero (mail) (www):
The image doesn't show up at all in Firefox.
8.3.2005 2:11am
Jim Anderson (mail) (www):
Works just great for me, Zev. On Firefox, that is.
8.3.2005 4:03am
Public_Defender:

As you get closer to the equator, you do get more sunlight, not in hours, but in intensity. The energy received from the sun scales by the cosine of latitude.

A nitpick to a nitpick. I love it. Professor Volokh should have written, "All places have, on average, the same amount duration of daylight. . . ."
8.3.2005 7:37am
Greg:
When above the arctic circle in June, the sun itself is visible low-ish on the horizon for most, if not all, of 24 hours. As you take the Topeka graph northward, the sine wave starts/finishes closer to zero as the hump moves toward 24 hours. You actually do have more hours of SUN in the summer as you move north of the equator. You have more hours of SUN in the winter (Dec) as you move south of the equator. On the equator the sine wave is a flat line.
8.3.2005 8:44am
Antares79:

The amount of sunlight reacing the pole and reaching the equater is a continuous monotonic function of the inclination, so there must be a unique inclination where the poles and the equater get the same amount of sunlight. Can anyone figure out that inclination? Can anyone figure out whether, on such a planet, the intermediate latitudes get that same amount of sunlight, or more, or less?

Defining "equator" to be the place farthest from the axis of rotation on the planet and the planet's period of rotation to be << than the period of revolution (orbit)...

If by "amount" you mean "duration" of sunlight, then any non-zero inclination will yield the same average hours of sunlight per year for any location (A pole, equatorial position, otherwise), just like on Earth.

If by "amount" you mean "energy" of sunlight, then a 90-degree axis of rotation (axis of rotation in plane of orbit) yields the same average intensity (energy/unit area) throughout the year for each pole and any equatorial location. Any lesser inclinations increase the equatorial location's average energy at the expense of the polar locations'.

On a planet where the axis of rotation is in the plane of the orbit, the poles get a certain amount of sunlight, the same amount as the equater gets on the zero inclination planet, and the equater gets less, so the poles get more than the equater. Such a planet has an inclination of ninety degrees.

This is wrong if we are speaking of total sunlight energy per year at a particular location. Are you wanting the calculation for both poles' duration or energy combined?
8.3.2005 2:19pm
Antares79:
Just in case you were wanting the inclination at which the combined yearly sunlight energy from both poles was equal to any given equatorial position's, it's 26.565 degrees.
8.3.2005 2:25pm
Don M (mail):
Is it to minor to point out that the graph says south and north when it should say closer to the equator and further from the equator.

Of course the poles get one "day" a year of 6 months length.

above the arctic circle gets some winter days with no daylight, just a twilight, and some summer days with the sun continuously in the sky.
8.3.2005 2:52pm
Public_Defender:

Is it to minor to point out that the graph says south and north when it should say closer to the equator and further from the equator.

It says "States south of Topeka get more [intense] daylight. . . ." This is true because there are no states south the of equator. But I don't know why the paper mentioned Topeka.
8.3.2005 4:22pm
fuloydo (mail) (www):
Just a quibble, concerning the size of the image.

As a webmaster back in the days when I wrote my html using notepad and my traffic consisted of ppl who were 99% dialup at 28.8 or slower......

Why is the original image that large to begin with?

You're forcing ppl to download, in bytes, an image larger than what they are displaying on their screen.

Photoshop, or whatever imageing software you prefer, can shrink such an image before your users download it.

I know that broadband is rampant these days but bandwidth is still not infinite. Never post an image larger than one size smaller than your average users screen resolution. Back in my day that was 800x600, (which meant 640x480 for "large" images), and I never posted images that large. Just thumbnails with links if I thought the image warrented such.

Just a thought...
8.3.2005 9:08pm
Blackwing1 (mail):
I must beg to differ with the proposed answer.

Taking as a source "Thermal Environmental Engineering", Threlkeld, 2nd Ed., and using information on sun angles, Equation 13.3 states:

sin(B) = cos(l)cos(h)cos(d) + sin(l)sin(d)

Where:

B is the sun's altitude angle
l is the latitude angle
h is the sun's hour angle
d is the sun's angle of declination

At sunrise/sunset, B = zero, so:

h = arccos[-tan(l)tan(d)]

Thus, for a given angle of declination (which varies with the season) the hour of sunrise/sunset DOES vary with latitude "l". This means that in turn, the length of day-lit period also varies, on a given day, as one travels from south to north.
8.4.2005 2:10pm
Antares79:

Thus, for a given angle of declination (which varies with the season) the hour of sunrise/sunset DOES vary with latitude "l". This means that in turn, the length of day-lit period also varies, on a given day, as one travels from south to north.

This is correct, the length of day does vary by lattitude for any given non-equinox day on Earth. The answer, however, states that on average (taken over at least a year) every location gets the same amount (~12 hours) of sunlight, which is also true.
8.5.2005 12:14pm