Size of shadow cast in summer vs winter?

[4myHennyPenny]

Ok, for all of the math whizzes and astronomy experts out there...

We're trying to figure out the best coop placement on our property. We can't move too far north in our yard because of the raised garden beds, short fencing (too open to alley), etc. At the same time, we need to be sure that we are far enough away from our neighbor's lot to the south that his large evergreens won't leave the coop in a deep freeze of shade this winter.

So if I go outside at noon tomorrow and measure the length of the evergreens' shadows, what equation can I use to calculate their length at the winter solstice? (Obviously, I can repeat the measurements at the summer solstice in a couple weeks to improve accuracy.) BTW, our latitude is approx. 40 degrees.

Thanks, and a gold star for anyone with the correct answer!

 

[BeckyLa]

 

[4myHennyPenny]

That's me exactly, BeckyLa. I consider it a triumph simply that we thought of this problem BEFORE the coop wound up stuck in ice all winter. Not sure that I ever remember calculating this in grade school (how sad). Interesting, though!

 

[BeckyLa]

I suppose you could call your local AG Extension office and ask them, or maybe the science dept of a local university?

 

[AK-Bird-brain]

Wow... that's a good one. Usually we know what areas get the most sun in the winter and go by that... I'm sure there's someone out there that knows how to calculate it. You probably need to give your latitude/longitude coordinates, too.

 

[CoyoteMagic]

Well....................I should know this I have taught science. I believe the degree of change is 30 degrees. But that's the angle of the sun's rays are coming in to the earth. hmmmmmm

Thinking......................... un see the smoke!!

45 degrees to the north-northeast of the shadow cast by the trees at about noon.

I think

 

[suburbanhomesteader]

Hmmm.... I think this is less science and more geometry or algebra. But Coyote was right about the angle of the sun.

I am not 100% sure on these calculations, but this was what I am thinking:

In solar calculation circles, the "altitude" is the angle formed by the sun and the horizon at any given point in time.

From what I've found, 40 degrees latitude gives a maximum altitude of approx 23-30 degrees. over the course of the day of the winter solstice.

We know (or closely guess) the height of the tree

We need the distance from the tree to the coop

I remember Sister Turkey (oops...Theresa Marie) teaching us, SOH CAH TOA.

I don't think SOH or CAH are useful for this, but wouldn't TOA be useful? Tan=Opposite divided by Adjacent.

So, we know that the tangent of 23 degrees is about 0.4, and the tangent of 30 degrees is about 0.6

If I'm doing this correctly (and PLEASE check me!),

.4=ht of tree/distance to tree OR, distance=ht of tree/0.4

I think this may be correct, because as I recall, the 30' trees on my south property line in Indiana threw a shadow at noon that was almost the entire depth of the 100' property.

And don't forget that the sun is not only low in the horizon, being blocked by things to the south, but doesn't come up past midpoint, so objects in the southeast and southwest could also throw shade on the arrangement, in the morning and evening.

Anyway, thanks a heck of a lot

because I was sleepy until your question came up, and I would NOT have been able to sleep until I solved this silly "word problem"! Now, I'm wide awake! Yikes!

 

[Mac in Wisco]

Calculate the sun's maximum altitude (Summer Solstice): AMAX = (90 - Latitude +23.44)°

Shadow Length = Object Height / tan (AMAX)

At 40 degs latitude the shortest shadow will be object height divided by 3.363



Calculate the sun's minimum altitude (Winter Solstice): AMIN = (90 - Latitude - 23.44)°

Shadow Length = Object Height / tan (90-AMIN)

At 40 degs latitude the longest shadow will be object height divided by 2.000


I'm not that smart, I got the math from a page on how to design sundials...

This won't happen exactly at noon either. To calculate solar noon you have to take into account daylight savings time, your position within your time zone, and azimuthal changes of the sun throughout the season. For each degree you are west of the center of the time zone it adds 4 minutes. I did know that much...

 

[4myHennyPenny]

By golly, unless I'm mistaken, I do believe that some of you are enjoying this brainteaser!

Ok, I came across something on the web which might help. (Actually, they were handouts for an 8th grade science experiment

) They were based in Chicago, which is virtually the same latitude (I realize this won't help me with the alzimuth but figure I'm close enough for the chicken coop).

For approx. 41 degrees latitude:
At noon on the summer solstice, sun is at approx. 72 degrees. So the shadow of an object 600' tall (in this case, the NBC Tower) is 200' long. At noon on the winter solstice, the sun has dropped to 25 degrees and the shadow is approx. 1300' long. So, by my math, approx. 6.5 x longer than the shadow cast at summer solstice.

Based on the figure of 6.5, I measured the noon-time shadow of the evergreens yesterday and calculated how far that shadow will stretch across my yard at winter solstice. This may not be as accurate as mac in abilene's calcs; does not take into account alzimuth or longitude. But hopefully gets us close enough! I'll have to do a little work and see how close my calculations come to those provided by mac and suburbanhomesteader.

Thanks again for all of the thought and time you put into this math stumper! And sorry to have kept you up, suburbanhomesteader; that's me, too--can't go to bed until I've figured it out.

 

[Mac in Wisco]

Yes, you have it correct. I don't know why I changed the math for the lowest altitude. That math was for a different type of sundial...

The sun's elevation at equinox is equal to 90 minus your latitude. At summer solstice it is 23.44 degs higher and at winter solstice it is 23.44 degs lower, for a total excursion of 46.88 degs.

Sun's max altitude: AMAX = (90 - Latitude +23.44)
Sun's min altitude: AMIN = (90 - Latitude - 23.44)

Max and Min Shadow Length = Object Height / tan (AMAX or AMIN)

At 41 degs latitude the highest the sun gets is 72.44 degs and the lowest is 25.56. A 600 ft object will throw shadows of 189 ft and 1254 ft respectively, for a ratio of 6.63.