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Math problem
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May 4, 2018 12:37:42   #
spaceylb Loc: Long Beach, N.Y.
 
PixHound wrote:
Trig Formula: Altitude = (Width/2) / (Tan(FOV/2))
So for a 94º FOV and 100' desired width: (100/2) x (Tan 47º) = 50/1.07) = 46.6' altitude
...and for 84º and 100': 50/0.9 = 55.5'
Cheers...



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May 4, 2018 13:30:09   #
RSpinney Loc: Maine
 
Greetings all. Sorry I have not read every single comment, but whoever used tan to solve the issue has the correct process. The vertical line is the distance above ground, not the hypotenuse - which is being used in the law of sines.

And by the way, if you're flying 47 feet above the ground and encounter a 60' tree, you're trip is going to be cut short.

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May 4, 2018 13:42:15   #
Longshadow Loc: Audubon, PA, United States
 
RSpinney wrote:
Greetings all. Sorry I have not read every single comment, but whoever used tan to solve the issue has the correct process. The vertical line is the distance above ground, not the hypotenuse - which is being used in the law of sines.

And by the way, if you're flying 47 feet above the ground and encounter a 60' tree, you're trip is going to be cut short.


That's interesting, as the rule of sines comes up with the same answer.....
I suppose it depends if you drop a vertical from the top (camera) point and use either half of the equilateral triangle you started with.

A/sine(a)=B/sine(b)=C/sine(c)

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May 4, 2018 14:16:22   #
boberic Loc: Quiet Corner, Connecticut. Ex long Islander
 
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.


I assume you will be using a drone, as flying a plane, or chopper at 50' is a good way to shorten your afternoon

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May 4, 2018 15:31:57   #
Peterff Loc: O'er The Hills and Far Away, in Themyscira.
 
A few things to consider: http://www.youtube.com/watch?v=EaXWy-HQefc

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May 4, 2018 16:01:56   #
OddJobber Loc: Portland, OR
 
RSpinney wrote:

And by the way, if you're flying 47 feet above the ground and encounter a 60' tree, you're trip is going to be cut short.

Never heard of obstacle avoidance?

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May 4, 2018 16:17:38   #
CatMarley Loc: North Carolina
 
OddJobber wrote:
Never heard of obstacle avoidance?

That will louse up your focus for sure!

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May 4, 2018 16:20:13   #
OddJobber Loc: Portland, OR
 
While we're at it, why is everyone concerned about 60 foot trees? In Oregon those are called "kindling". In "undeveloped forest land", you need to stay above 200 feet. :)

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May 4, 2018 16:36:02   #
SharpShooter Loc: NorCal
 
RSpinney wrote:

And by the way, if you're flying 47 feet above the ground and encounter a 60' tree, you're trip is going to be cut short.


Maybe the tree will be cut short?!?! LoL
Depends on whether the glass is half full or half empty!!!
SS

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May 4, 2018 17:04:31   #
donb17
 
CO wrote:
I just did the trigonometry on the data. The height would be 46.6 feet.


I agree, given the 94 degree FOV, not counting any tree height.

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May 4, 2018 18:14:36   #
DanielB Loc: San Diego, Ca
 
That would be above tree line I would think.
oregon don wrote:
with 60 ft trees

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May 4, 2018 18:16:05   #
R Kuykendall
 
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.


I am not a professional photographer, but I may have some help as an aviation professional. A number of years ago I wrote a program for airplanes with GPS to map rectangular areas using aircraft-mounted Hasselblad cameras; in that way, I learned some simple things about aerial photography. First of all is the fundamental equation:

(image size)/(focal length) = (subject size)/(distance from camera)

In the case of aerial photography, distance from camera equals altitude. For example, a 35-milimeter camera with a 70-milimeter lens could capture a 100-foot area from 200 feet. (These numbers were chosen to simplify the arithmatic -- for me --). Perhaps in your case, you could crop the image size to 10 milimeters; with a 20-milimeter focal length, the altitude would be 200 feet above ground level, a reasonable height for a drone. As has been pointed out, a 25-mile strip could be difficult to do legally with a drone; the difficulty lies in the legal restriction that the drone must stay within direct view of the operator. If I had a project of this sort, I would try to find a telephoto lens of an appropriate size and use a light airplane at some altitude between 500 and 1000 feet.
Good luck.

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May 4, 2018 18:17:43   #
DanielB Loc: San Diego, Ca
 
That's a Trigg'y question :-).
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.

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May 4, 2018 22:07:02   #
bmike101 Loc: Gainesville, Florida
 
Longshadow wrote:
In what direction?
bmike101 wrote:
I wonder....how far away from the plot of land do you need to be to capture all 100 feet??

In what direction?
Vertically, about 47 feet.Vertically, about 47 feet.


Well, if you are 1 fooi away from it (laterally) you aren't going to get it all.

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May 4, 2018 22:54:48   #
WB9DDF Loc: Knoxville, IL
 
I think higher will give a better perspective on the image of trees, hills, valleys etc. With that in mind do the problem in reverse. Drones are allowed to go up to 400 feet which isn't really high enough. Manned aircraft are supposed to maintain at least 500 ft over open terrain. I would think the ideal would be to fly a plane at 1000 feet and use a focal length a bit less than ten times the width of the image sensor. (For example 300mm for a 35mm sensor) This would cover a little more than 100 feet wide so you can trim it to exact width if that is important. Also you need to insure that the camera is mounted so it points straight down to not distort the image perspective. The camera should also be isolated as much as possible from the engine vibration.

Most aircraft will be moving well over a hundred feet per second so you will have to fire a few frames per second for the length of the flight. When the camera buffer is full the pilot will need to circle back and start another pass where the last one ended.

I once knew a pilot who did this kind of photography for land surveying back in the film days. His cameras were mounted on a gyro stabilized platform that shot through a hole in the belly of the aircraft. He was shooting from a few thousand feet taking in a mile wide strip and I think using IR film to see through the foliage and see things like buried pipelines.

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