I calculate the tree height to see if it is going to shade my solar panels.
I have not posted in a while because we moved to a new house.
The new house (actually 12 years old) has a great southern facing roof that may be perfect for a solar panel project. However there is a tree that might cause shadows in the winter time because of the lower sun angle.
So I needed to figure out the height of the tree first to see how big of a problem is might be.
I found a great site for finding the tree height and I decided to use the shadow method.
So I measured the tree’s shadow (456″) and my height (75″) and multiplied the two. Then I divided the result by my shadow (48″) and it came out to 59.5′ feet.
I learned from the grounds keeper that the tree is a Cottonwood so I looked it up online and found that the Cottonwoods can grow up to 80-100 feet tall. However there are different types of Cottonwoods and some don’t grow as tall, so I emailed the picture above to my town’s city hall to see if they knew what type it is. I also sent it to a local tree farm to see if they knew.
Once I find out the Cottonwood’s mature height I can then calculate if its shadow might cause on my solar panels to be shaded.
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Holy similar triangles Dave! I like the tree height estimation method. Next I’ll expect you to estimate the radius of the earth based on shadow observations at noon….
You need to get a hold of a solmetric or some other shadow analysis tool. David Fosdeck of REC Solar used one when he checked out my roof. It’s a cool product (basically integrates a PDA, a digital camera w/ fish-eye lens, a compass, and a level). I thought that the method for displaying the results (what part of roof is shaded and when (time of day and time of year)) was very cool (i.e. lots of info in one ‘image’)
Where is this tree relative to your house? (how far, what direction)
Eddie,
That is a very cool looking tool.
I am working on figuring out what the angle of the sun is at our latitude in our town, do you remember what that was from his readings - was it 26 degrees?
I will answer your questions on my next post.
I do not remember from his readings.
From geometry on my whiteboard and vague memories from reading books about sundials:
Neglecting the tilt of the earth, the noon time sun is (90 - latitude) degrees above the horizon. So for us that is almost exactly 50 degrees.
Factor in the tilt of the earth’s axis of rotation (23.5 degrees?), and the summer and winter extremes will be +/- 23.5 degrees of the (90 - latitude) number. That’s 26.5 to 73.5 degrees (winter to summer).
So I think that your 26 degree number is correct for the ‘winter low’ of the sun.
At least cottonwoods do not have leaves in the winter! (perhaps only partial shading….)
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