Math problem (Page 3): I know you want the exact height but this would require the pilot to fly the exact center point or you will be off on one side or the other. I think you should...

Country Boy Loc: Beckley, WV

I know you want the exact height but this would require the pilot to fly the exact center point or you will be off on one side or the other. I think you should add a crop factor and increase the height to match. Something to consider anyway!

mikegreenwald Loc: Illinois

Personally, I use the distance above the tops of the trees, not the AGL altitude.
Van’s calculations and several of the others are correct. It always pays to carry extra altitude to cover errors in precision flying the drone.
My experience comes from flying real aircraft at considerably higher altitudes, with longer lenses - typically in the 50mm range for 35mm cameras, or 90mm for 120mm film cameras. More fun that way too.

FrankB Loc: Sydney, Australia

Longshadow wrote:

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.)

You can't use sine (Opposite side / Hypotenuse) as we don't know the length of the hypotenuse, but let's use the 43 degree angle. Once again tan = opposite side / adjacent side - tan 43 degrees = x / 50 feet, so x = tan 43 degrees * 50 = 43.48 feet.

Longshadow Loc: Audubon, PA, United States

FrankB wrote:

You can't use sine (Opposite side / Hypotenuse) as we don't know the length of the hypotenuse, but let's use the 43 degree angle. Once again tan = opposite side / adjacent side - tan 43 degrees = x / 50 feet, so x = tan 43 degrees * 50 = 43.48 feet.

What do you get if you use the rule of sines?
putting this formula into Excel yields 46.62
=50*TAN(RADIANS(43))

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)

Using:=(SIN(RADIANS(43))*50)/SIN(RADIANS(47))

Both yield 46.62

wrangler5 Loc: Missouri

Remember that a published lens angle of view can relate to image height (rarely, unless specified), width OR diagonal. On this chart https://www.scantips.com/lights/fieldofview2.html , for example, the angle of view of a 20mm lens on a full frame sensor (3:2 aspect ratio) is given as 84 degrees horizontal but 94 degrees diagonal. For purposes of the given math problem, I would assume the horizontal figure is the one to work from.

Mark Sturtevant Loc: Grand Blanc, MI

Longshadow wrote:

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.

If slight errors in the area captured in the photo are tolerable, then fine. I was just questioning whether one could do this task without some error.

Longshadow Loc: Audubon, PA, United States

Mark Sturtevant wrote:

If slight errors in the area captured in the photo are tolerable, then fine. I was just questioning whether one could do this task without some error.

Most likely not. I'd go some higher myself. I don't think the intent is to be exactly 100' wide on the capture.

spaceylb Loc: Long Beach, N.Y.

wrangler5 wrote:

Remember that a published lens angle of view can relate to image height (rarely, unless specified), width OR diagonal. On this chart https://www.scantips.com/lights/fieldofview2.html , for example, the angle of view of a 20mm lens on a full frame sensor (3:2 aspect ratio) is given as 84 degrees horizontal but 94 degrees diagonal. For purposes of the given math problem, I would assume the horizontal figure is the one to work from.

wrangler5 Loc: Missouri

Actually, the original question was how high to capture ONLY the 100 foot width. But the reality of controlling an airborne device is that it will be impossible to fly a precise enough path to capture only the desired width all the time. (I believe even the megazillion dollar terrain following systems in military attack aircraft only keep an altitude within tens of feet.) So the answer that's really needed is, what is the MINIMUM altitude needed to capture the desired view, from which the pilot can build in the extra tolerances for (inevitable) altitude and course deviations that will be needed to insure that the captured view always includes the subject of interest.

vanbr

the tan=opp/adj ratio has not changed. the incorrect calc was tan 47=x/50 that's adj/opp that was an error. the correct equation was tan 47=50/x thus x=50/tan 47 not 50*tan47.

Of course the answers are "precise" but equally "of course" they give a minimum height under ideal, perfectly flat conditions.

Longshadow Loc: Audubon, PA, United States

vanbr wrote:

the tan=opp/adj ratio has not changed. the incorrect calc was tan 47=x/50 that's adj/opp that was an error. the correct equation was tan 47=50/x thus x=50/tan 47 not 50*tan47.

Of course the answers are "precise" but equally "of course" they give a minimum height under ideal, perfectly flat conditions.

Gotcha. Thanks!
(Now I know I'm not going nuts!

)

CatMarley Loc: North Carolina

Longshadow wrote:

63 then.

Better be damned accurate about that tree height!

Longshadow Loc: Audubon, PA, United States

CatMarley wrote:

Better be damned accurate about that tree height!

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BebuLamar

First thing first the AOV or FOV of a 20mm lens on a 24x36mm format would cover 94 degrees diagonal but the OP would need the coverage for the longer side which is only 84 degrees. To cover 100ft wide path the camera must be 45ft above tree top.

PixHound Loc: Marietta, GA

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...