JUDGING TREE HEIGHT WITH GEOMETRY

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George Willer
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Postby George Willer » Wed Mar 14, 2007 10:32 pm

Jim Becker wrote: OK, next question, how to do it without the clinometer/protractor? Here is a hint, it will be easier next week.


On a sunny day? Measure the length of the tree's shadow and set up a ratio of that to the shadow of an upright stick of known length. :D
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Postby Jim Becker » Thu Mar 15, 2007 8:28 am

George Willer wrote:
Jim Becker wrote: OK, next question, how to do it without the clinometer/protractor? Here is a hint, it will be easier next week.


On a sunny day? Measure the length of the tree's shadow and set up a ratio of that to the shadow of an upright stick of known length. :D

That is the same as spiveyman's answer. I intended to also say without the stick. You need to know where you are.

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Postby spiveyman » Thu Mar 15, 2007 11:52 am

GW- Are you tryin to steal my answer?! :evil: lol....... I'm thinking hard, but I'm not sure where Jim is going... :(
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Postby George Willer » Thu Mar 15, 2007 12:23 pm

spiveyman wrote:GW- Are you tryin to steal my answer?! :evil: lol....... I'm thinking hard, but I'm not sure where Jim is going... :(


No, Sorry. I guess I just didn't read carefully enough.

I think Jim intends for you to know your latitude which will affect the length of the shadow. (the azimuth of the ecliptic is knowable if you know the latitude) I'll have to think about it a little to understand what effect the spring equinox will have.
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Postby Jim Becker » Sun Mar 18, 2007 12:24 pm

George, you are on the right track. I thought I would give everyone until the spring equinox (21st) to come up with an answer.

On the equinox, a bunch of complicating factors drop off. You can ignore all the complications described on this page:
http://www.jgiesen.de/analemma/


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