## Mathematical problem

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Mathematical problem

08-03-2009 8:52 AM

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If an object "appears" to be .5 of an inch (in size) when the known size is 48 inches,

what is the formula for working out how far away that object is? and what is the answer, please..

I suspect it is to do with geometry, and triangles, but it is so long since I did any stuff like that I have forgotten it all.... !

(and I am getting older every day)

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Re: Mathematical problem

08-03-2009 9:37 AM

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Re: Mathematical problem

08-03-2009 9:43 AM

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Re: Mathematical problem

08-03-2009 10:09 AM

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Re: Mathematical problem

08-03-2009 10:16 AM

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Re: Mathematical problem

08-03-2009 10:44 AM

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Split the problem into triangles. You know the size of the base of the big triangle and the size of the base of the small triangle. All other variables are unknowns. Unless you can provide at least one other variable (length or angle) then you don't have enough information to make any calculations about the proportional relationships.

If you vary the distance of the virtual measurement point from the eye then that 0.5" measurement changes so you can't make any headway without knowing how far it is from the eye OR how far it is from the object.

And if you really wanted to get accurate results then you would need to know the angle of the ground, the height of the eye from the ground, the angle you are holding the mesuring plane and any difference between that angle and the angle of the distant object etc.

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Re: Mathematical problem

08-03-2009 11:04 AM

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Thanks again...

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Re: Mathematical problem

08-03-2009 11:13 AM

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At arms length, my thumbnail is 28 inches from my eye, and is half an inch, the angle between the base of the nail, and the top of the nail is 5 degrees....

Now, is it possible to advance those two lines to an assumed 48 inches for the actual object size, then calculate the distance from the eye to the object?

Or am I going off on a tangent ? ? ? ?

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Re: Mathematical problem

08-03-2009 11:52 AM

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Quote from: The You need to know the distance from the eye to the virtual mesuring point too.

Quote from: shutter At arms length, my thumbnail is 28 inches from my eye,

That's the variable you failed to recognise before. You have all the measurements you need to calculate (roughly) the distance to your object.

We'll assume that the thumbnail and the object are on the same plane and we will just ignore ground level and elevation, just to keep things simple. So now we will consider the problem in the form of an issoceles triangle where the eye is at the angle opposite the base, and that base will be the height of the distant object.

We can now dissect the triangle with a line parallel to the base. This represents the thumbnail. The distance between your eye and the thumbnail will be represented by the side of the smaller triangle.

Lets call the bases B1 and B2

& we'll call the sides S1 and S2

B2/B1 = S2/S1

48/0.5 = S2/28

96 = S2/28

S2 = 28*96

S2 = 2688" or 224'

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Re: Mathematical problem

08-03-2009 12:08 PM

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. Now I can use the formula (and my thumbnail) when out and about, to give me the distance....

Thank you for the lesson..... much appreciated.

EDIT EDIT EDIT

Check out my blog, to see why I wanted to know how to do this ! ! !....

http://nemosphotography.blogspot.com/

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Re: Mathematical problem

08-03-2009 1:17 PM

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Astro nav was a real pain - but I never had to do it practise - even trying to help colleagues with their "homework" was bad enough

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Re: Mathematical problem

08-03-2009 2:39 PM

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But way back then, we had just finished Nav school, and it was all "fresh"....

Not sure if I could do it now 40 years on !

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Re: Mathematical problem

08-03-2009 3:03 PM

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I followed the course I was given ( honest ) and after an hour or so I spotted something ahead in the moonlite sea - and the noise level rose.

I got the navigotor up on deck, and showed him - he quickly told me to go about and head off at an angle of 90.

We had almost run onto the Casquets - a line of rocks just below or on the surface. It was the white foam as they water hit the rocks that I saw and heard.

I was vindicated - the nav had miscalculated the tidal flow !!!.

http://en.wikipedia.org/wiki/Casquets

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Re: Mathematical problem

08-03-2009 3:06 PM

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Re: Mathematical problem

09-03-2009 1:14 AM

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http://www.1728.com/angsize.htm

http://www.canon.com/bctv/calculator/calculator3.html

http://www.cfa.harvard.edu/webscope/activities/pdfs/measureSize.PDF

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