Estimating a tower's height using shadows

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In summary, the conversation discusses the use of similar triangles and ratios to estimate building height. The participants also mention the assumption of parallel sun rays and the concept of practical convenience. They also discuss the implications of the sun not being a point source of light. The conversation ends with a light-hearted comment about a mistyped message.
  • #1
jachyra
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Hi all!

I was reading the following article on howstuffworks.com:
http://science.howstuffworks.com/question379.htm

It says to apply the following procedure to estimate the building height:
1. Measure the length of the broomstick's shadow
2. Calculate the ratio of the broomstick's shadow length to the broomstick's height
3. Measure the tower's shadow
4. Apply the ratio to discover the tower's height.

Attached is a picture I drew showing why this simple calculation seems confusing to me. The black box represents the stick placed beside the building (white box). If the rays of the suns strike the ground at different angles, then how can a similar triangle approach using just ratios of the lengths be used to calculate the building height?

Is my drawing wrong? Have I assumed something wrong by drawing the sun as a point of light?
 

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  • #2
I think so. The sun's rays are parallel, remember, because the sun is so far away. Now try drawing the same diagram with the direction of sunlight same everywhere. Immediately you get similar triangles.
 
  • #3
Thats one thing that sometimes confuses me about physics, that you have to make assumptions or approximations sometimes to get better results.

This kind of thing happens in optics when you have lenses like in telescopes, you just assume that the stuff is so far away that the light coming from it is parallel.
 
  • #4
It's not so much of an assumption as a matter of practical convenience. Can you think of a simple experiment to show that the sun's rays are not parallel? (In reality, they are not exactly parallel.) For stars, they are parallel for almost all practical purposes.
 
  • #5
jachyra said:
Is my drawing wrong? Have I assumed something wrong by drawing the sun as a point of light?

Yes your drawing is wrong. The reason the ratio method works is because the two objects form similar triangles. Like Shooting Star mentioned, the hypotenuse of the triangles (sun's rays) are assumed to be parallel which gives the similar triangles.

This may help explain a little better...

http://mathforum.org/library/drmath/view/55238.html
 
  • #6
Shooting star said:
It's not so much of an assumption as a matter of practical convenience. Can you think of a simple experiment to show that the sun's rays are not parallel? (In reality, they are not exactly parallel.) For stars, they are parallel for almost all practical purposes.

You're right they're not. Two rays 20 feet apart diverge by 20 feet over 93 million miles or about 1 part in 24 billion, or 7 x 10^-13 degrees.

So, unless your estimation of the building height needs to be accurate to 13 decimal places, you'll be OK.




Actually, since we're getting nitpicky, we have to aco**** (that was supposed to be "account") for the fact that the sun is not a point source of light. Its rays come from a disc, which causes soft-edged shadows. This effect far overwhelms the divergence of the rays - actually there are more converging rays than there are diverging...
 
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  • #7
DaveC426913 said:
This effect far overwhelms the divergence of the rays - actually there are more converging rays than there are diverging...

Not the mention trying the measure the shadow! The instrumental error involved :P
 
  • #8
ohhhhhhhhhhhhhh!

It's all clear now. Thanks for all the information! I just joined this forum yesterday and I already love it! You guys rock.
 
  • #9
Does it mean something?

DaveC426913 said:
... we have to aco**** (that was supposed to be "account")...

Why? I mean why aco**** and then the explanation in the bracket?
 
  • #10
Shooting star said:
Why? I mean why aco**** and then the explanation in the bracket?
I typed my message, then hit preview, and this is what it did to my (obviously mistyped) message.

"Hah, what a silly bunt I am."
 
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  • #11
DaveC426913 said:
I typed my message, then hit preview, and thi sis what it did to my (obviously mistyped) message.

"Hah, what a silly bunt I am."

No need to go public on that. (The mistyping, I mean, of course :smile:)
 

Related to Estimating a tower's height using shadows

1. How do you estimate a tower's height using shadows?

To estimate a tower's height using shadows, you will need to measure the length of the shadow cast by the tower and the length of the shadow cast by an object of known height. By using the ratio of these two lengths, you can calculate the height of the tower.

2. What equipment do you need to estimate a tower's height using shadows?

To estimate a tower's height using shadows, you will need a measuring tape or ruler to measure the lengths of the shadows, a protractor to measure the angle of the sun's rays, and a calculator to calculate the height of the tower.

3. What is the best time of day to estimate a tower's height using shadows?

The best time of day to estimate a tower's height using shadows is when the sun is at its highest point in the sky, also known as solar noon. This is when the shadows are the shortest and the most accurate measurement can be obtained.

4. How accurate is estimating a tower's height using shadows?

The accuracy of estimating a tower's height using shadows depends on the precision of the measurements taken and the angle of the sun's rays. Generally, this method can provide a fairly accurate estimate within a few meters of the actual height.

5. Are there any limitations to using shadows to estimate a tower's height?

Yes, there are some limitations to using shadows to estimate a tower's height. The accuracy of the method is affected by factors such as the length of the shadow cast by the known object, the angle of the sun's rays, and the presence of obstacles that may cast shadows on the tower itself. Additionally, this method may not be suitable for estimating the height of very tall or irregularly shaped towers.

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