Math problem
May 4, 2018 06:58:49 #
May 4, 2018 07:29:17 #
Here's a link to a calculator: https://www.easycalculation.com/trigonometry/law-of-sines.php
- Use angle A as 1/2 of the FOV° (94°/2=47°).
- Use angle B as 180-90-(1/2 of FOV°) = 43°
- Use side A as 1/2 of the width of desired coverage. = 50'
- <calculate>
- Side B will be the height (46.62')
The basic formula is A/sine(a)=B/sine(b)
where A is the length of the side opposite angle a;
B is the length of the side opposite angle b.
B=(sin(b)*A)/sin(a)
(I think I got it all right. In spreadsheet formulas, just watch you are not using Radians instead of degrees.)
- Use angle A as 1/2 of the FOV° (94°/2=47°).
- Use angle B as 180-90-(1/2 of FOV°) = 43°
- Use side A as 1/2 of the width of desired coverage. = 50'
- <calculate>
- Side B will be the height (46.62')
The basic formula is A/sine(a)=B/sine(b)
where A is the length of the side opposite angle a;
B is the length of the side opposite angle b.
B=(sin(b)*A)/sin(a)
(I think I got it all right. In spreadsheet formulas, just watch you are not using Radians instead of degrees.)
May 4, 2018 07:42:17 #
May 4, 2018 07:49:39 #
Now, note! either 50*tan(47)=53.6 ft or 50*tan(42)=45 ft using either the 94 or 84 degree fov.
Boith assume the unlikely exact positioning at the 50 foot center mark, so if you assume say a 5 ft error in that positioning, use 55 as the target instead of 50. this yields either 55*tan(47) or 55*tan(42) again depending on the actual FOV. These numbers give 58.9 feet and 49.5 feet respectively
Boith assume the unlikely exact positioning at the 50 foot center mark, so if you assume say a 5 ft error in that positioning, use 55 as the target instead of 50. this yields either 55*tan(47) or 55*tan(42) again depending on the actual FOV. These numbers give 58.9 feet and 49.5 feet respectively
May 4, 2018 08:00:10 #
chem wrote:
The formula is 50Xtan(47)= height the 47 is in degrees (1/2 94)
Is that formula for an equilateral triangle? If so, it won't be the right height.
May 4, 2018 08:01:38 #
Orangebird wrote:
I will be shooting a aerial video of a 25 mile long by 100 feet wide patch of undeveloped forest land. I was wondering if anyone knew the formula to calculated the altitude I would need to fly at to only capture the 100 foot wide path? I'm using a 94 degree FOV camera with a 20 mm prime lens. I hope this is enough info to calculate. TIA.
If you want an entertaining way to consider what's involved you could watch the movie 'Hidden Figures' if you haven't already seen it.
Theoretical math needs to be mapped to reality.
Good luck.
May 4, 2018 08:15:05 #
omg, all that trig I forgot being used! But I am puzzled as to why the answers are so precise. Given the topography and who knows what other factor, I would think you would want to fly a bit higher than calculated then crop the picture to the required area in post.
May 4, 2018 08:20:49 #
Longshadow wrote:
Is that formula for an equilateral triangle? If so, it won't be the right height.
no. it is the right triangle formed by dropping a perpendicular from the camera point to the ground, touching the ground at a point 50 feet from either side of the 100 foot path. in the diagram "a" is the angle which is half of the 94 degree FOV or the 82 degree FOV of the lens.
May 4, 2018 08:23:14 #
vanbr wrote:
no. it is the right triangle formed by dropping a perpendicular from the camera point to the ground, touching the ground at a point 50 feet from either side of the 100 foot path. in the diagram "a" is the angle which is half of the 94 degree FOV or the 82 degree FOV of the lens.
So why does the rule of sines come up with a different value?
(I haven't used trig in a thousand years!)
May 4, 2018 08:33:59 #
ARRGGGH! of course the results are the same. I screwed up! x=50/tan(47) or x=50/tan(42) 46.6 and 55.5 ft and 55/tan(47) or 55/tan(42) yields 51.2 and 61 using the 5ft fudge factor.
May 4, 2018 08:36:55 #
vanbr wrote:
ARRGGGH! of course the results are the same. I screwed up! x=50/tan(47) or x=50/tan(42) 46.6 and 55.5 ft and 55/tan(47) or 55/tan(42) yields 51.2 and 61 using the 5ft fudge factor.
Wrong angle used?- not the 47 but 180-90-47 = 43.
Using 43 yields 46.62, the same as the sine rule. (Whew!)
May 4, 2018 08:41:27 #
ARRGGGH! of course the results are the same. I screwed up! x=50/tan(47) or x=50/tan(42) 46.6 and 55.5 ft and 55/tan(47) or 55/tan(42) yields 51.2 and 61 using the 5ft fudge factor.
May 4, 2018 08:47:23 #
FrankB
Loc: Sydney, Australia
Look at the diagram above. The angle at the top will be half of the quoted 94 degrees - i.e 47 degrees.
The tangent of an angle is the opposite side - 50 feet - divided by the adjacent side - x. i.e. tan 47 = 50 / x, so x = 50 feet / tan 47 = 43.48 feet.
Why has no-one else come up with this?
The tangent of an angle is the opposite side - 50 feet - divided by the adjacent side - x. i.e. tan 47 = 50 / x, so x = 50 feet / tan 47 = 43.48 feet.
Why has no-one else come up with this?
May 4, 2018 08:48:48 #
Mark Sturtevant wrote:
omg, all that trig I forgot being used! But I am puzzled as to why the answers are so precise. Given the topography and who knows what other factor, I would think you would want to fly a bit higher than calculated then crop the picture to the required area in post.
Um, the answers are so precise because of the math?
(I'm missing something here.)
It would be the distance above the land, following the topography.
May 4, 2018 08:54:15 #
FrankB wrote:
Look at the diagram above. The angle at the top will be half of the quoted 94 degrees - i.e 47 degrees.
The tangent of an angle is the opposite side - 50 feet - divided by the adjacent side - x. i.e. tan 47 = 50 / x, so x = 50 feet / tan 47 = 43.48 feet.
Why has no-one else come up with this?
The tangent of an angle is the opposite side - 50 feet - divided by the adjacent side - x. i.e. tan 47 = 50 / x, so x = 50 feet / tan 47 = 43.48 feet.
Why has no-one else come up with this?
Because if you use the tan or sine, the answer has to be the same.
The angle to use is the other (43), "a" may be in the wrong place in the diagram?
(Been so long since I used trig.)
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