how to calculate fence shadow
February 13, 2011 4:08 PM Subscribe
Is there an easy way to calculate the length and time of the shadow cast by my fence this spring and summer, for gardening purposes? Specs: 8 ft tall solid wood fence running almost exactly east to west, located in Washngton, DC.
The US Naval Observatory provides this sort of information for navigators.
This tool will give you height and direction information about the sun for a particular day. Once you know the height of sun it's simple trigonometry to find the length of the shadow.
Lots of other cool related tools here.
posted by Confess, Fletch at 4:57 PM on February 13, 2011
This tool will give you height and direction information about the sun for a particular day. Once you know the height of sun it's simple trigonometry to find the length of the shadow.
Lots of other cool related tools here.
posted by Confess, Fletch at 4:57 PM on February 13, 2011
Seconding sketchup! It's quite powerful, you just need to make sure your model is oriented the right way.
posted by defcom1 at 5:36 PM on February 13, 2011
posted by defcom1 at 5:36 PM on February 13, 2011
Best answer: Back in the day, when astronomy instruments were used for navigation, the Tropic of Cancer was the latitude where the sun would be exactly upwards at noon at the June solstice. It's about 23 1/2 degrees North.
Washington is about 38 1/2 degrees North. So the difference between them is 15 degrees, and that's how far away from directly vertical the sun would be in DC at the June solstice.
Now it's just about triangles. The tangent of 15 degrees is 0.268 so if the "adjacent" is 8 feet then the "opposite" is a bit over 2 feet, and that's as small as the shadow will ever get.
The Tropic of Capricorn is the equivalent latitude for the December solstice, about 23 1/2 degrees South. From DC, therefore, the sun would be 62 degrees off of vertical.
The tangent of 62 degrees is 1.88 so the shadow would be about eleven and a quarter feet.
posted by Chocolate Pickle at 5:42 PM on February 13, 2011 [1 favorite]
Washington is about 38 1/2 degrees North. So the difference between them is 15 degrees, and that's how far away from directly vertical the sun would be in DC at the June solstice.
Now it's just about triangles. The tangent of 15 degrees is 0.268 so if the "adjacent" is 8 feet then the "opposite" is a bit over 2 feet, and that's as small as the shadow will ever get.
The Tropic of Capricorn is the equivalent latitude for the December solstice, about 23 1/2 degrees South. From DC, therefore, the sun would be 62 degrees off of vertical.
The tangent of 62 degrees is 1.88 so the shadow would be about eleven and a quarter feet.
posted by Chocolate Pickle at 5:42 PM on February 13, 2011 [1 favorite]
Chocolate Pickle: I thought of that solution, and was about to post it, but that only works at noon. And away from noon, the sun is lower in the sky (lengthening the shadow) but also at an acute angle to the east-west fence (shortening the shadow). Which one wins?
posted by madcaptenor at 5:46 PM on February 13, 2011
posted by madcaptenor at 5:46 PM on February 13, 2011
Which one wins?
The sun wins. When the angle between the light and the object is 0°, the shadow is parallel to the ground and therefore "infinite" in length.
Its area will depend on your latitude and season: the sun is at most (Θ = 90 – your latitude + Earth's axial tilt = 90 – 38 + 23 = 75) degrees above the horizon during the summer and sets that many degrees south of due east.
Your fence will cast a shadow that is at most (height)/(tan Θ) feet long.
posted by Nomyte at 6:25 PM on February 13, 2011
The sun wins. When the angle between the light and the object is 0°, the shadow is parallel to the ground and therefore "infinite" in length.
Its area will depend on your latitude and season: the sun is at most (Θ = 90 – your latitude + Earth's axial tilt = 90 – 38 + 23 = 75) degrees above the horizon during the summer and sets that many degrees south of due east.
Your fence will cast a shadow that is at most (height)/(tan Θ) feet long.
posted by Nomyte at 6:25 PM on February 13, 2011
When the angle between the light and the object is 0°…
Should be "the light and the horizon," i.e. at the moment the sun sets.
posted by Nomyte at 6:26 PM on February 13, 2011
Should be "the light and the horizon," i.e. at the moment the sun sets.
posted by Nomyte at 6:26 PM on February 13, 2011
But say the sun sets due west, as it does on the equinoxes, therefore exactly lined up with the fence? Then it casts *no* shadow. (That being said, that's sort of a degenerate case...)
posted by madcaptenor at 8:24 PM on February 13, 2011
posted by madcaptenor at 8:24 PM on February 13, 2011
Did I just write that the sun sets in the east? Ah, I did.
The sun will not set due west anywhere north of the tropic of Cancer, by definition. It sets due west on the equinoxes at the equator. Then the width of the fence's shadow will be insignificant.
I concluded that the OP was concerned with where the shadow will affect garden plants most, which would be directly behind the fence. Basically, there will be a wedge-shaped area, smaller in the summer and bigger in winter, that is always covered with shade.
posted by Nomyte at 9:03 PM on February 13, 2011
The sun will not set due west anywhere north of the tropic of Cancer, by definition. It sets due west on the equinoxes at the equator. Then the width of the fence's shadow will be insignificant.
I concluded that the OP was concerned with where the shadow will affect garden plants most, which would be directly behind the fence. Basically, there will be a wedge-shaped area, smaller in the summer and bigger in winter, that is always covered with shade.
posted by Nomyte at 9:03 PM on February 13, 2011
Basically, there will be a wedge-shaped area, smaller in the summer and bigger in winter, that is always covered with shade.
No, you will have full sun on the garden in the morning and late afternoon assuming you have no trees blocking the east and west ends of the fence. The sun rises and sets north of the fence (on the garden side) between the spring and autumn equinoxes.
There is never perpetual shade cast by a straight fence no matter how tall or what direction it runs anywhere on the planet.
(We are all assuming that the garden is on the north side of the fence, else you probably wouldn't be asking the question.)
posted by JackFlash at 9:55 PM on February 13, 2011
No, you will have full sun on the garden in the morning and late afternoon assuming you have no trees blocking the east and west ends of the fence. The sun rises and sets north of the fence (on the garden side) between the spring and autumn equinoxes.
There is never perpetual shade cast by a straight fence no matter how tall or what direction it runs anywhere on the planet.
(We are all assuming that the garden is on the north side of the fence, else you probably wouldn't be asking the question.)
posted by JackFlash at 9:55 PM on February 13, 2011
Response by poster: Yes, the garden is on the north side of the fence!
Thanks for all the suggestions - I think I'll try SketchUp. But it still doesn't seem like the ideal solution - what would be really cool would be a calculator where I could just plug in the height of the fence, its orientation, my location, and the dates, and then get back a full account of how much shade will be cast over the course of the day for the full growing season. Barring that, I'm just going to try Sketchup for a few key dates and times over the season and go from there. But based on Chocolate Pickle's calculation of 11 ft of shade at noon on the June solstice, I think I already know that it will be a great place for lettuce (which needs to be shaded from the midday sun at midsummer) but a bad place for tomatoes!
posted by yarly at 7:26 AM on February 14, 2011
Thanks for all the suggestions - I think I'll try SketchUp. But it still doesn't seem like the ideal solution - what would be really cool would be a calculator where I could just plug in the height of the fence, its orientation, my location, and the dates, and then get back a full account of how much shade will be cast over the course of the day for the full growing season. Barring that, I'm just going to try Sketchup for a few key dates and times over the season and go from there. But based on Chocolate Pickle's calculation of 11 ft of shade at noon on the June solstice, I think I already know that it will be a great place for lettuce (which needs to be shaded from the midday sun at midsummer) but a bad place for tomatoes!
posted by yarly at 7:26 AM on February 14, 2011
You have 11 feet of shade at noon on the *December* solstice, not the *June* solstice.
posted by madcaptenor at 9:02 AM on February 14, 2011
posted by madcaptenor at 9:02 AM on February 14, 2011
Response by poster: Ugh, reading comprehension! So the max shadow is 2 ft in June?
posted by yarly at 9:35 AM on February 14, 2011
posted by yarly at 9:35 AM on February 14, 2011
There's an iPhone app called Mr. Sun that tells you the angle of the sun, sunrise and sunset for your location and date. It's pretty great.
posted by electroboy at 10:26 AM on February 14, 2011
posted by electroboy at 10:26 AM on February 14, 2011
The *minimum* shadow length at noon is two feet in June; the *maximum* at noon is eleven feet in December; the jury's still out on what happens at times which aren't noon, so I'd recommend using one of the software solutions other people have pointed out.
posted by madcaptenor at 10:35 AM on February 14, 2011
posted by madcaptenor at 10:35 AM on February 14, 2011
I can't believe this. I multiplied it wrong. 1.88 * 8 = about 15 feet for the December solstice.
posted by Chocolate Pickle at 3:04 PM on February 14, 2011
posted by Chocolate Pickle at 3:04 PM on February 14, 2011
This thread is closed to new comments.
posted by Paragon at 4:23 PM on February 13, 2011