Height to Photo 1 Acre?

We use DroneDeploy for jobs like this. We recently did a large swath of land that included hotels, interstate, and woodlands. Scroll down this page to see the example I'm referencing. It took just a couple less than a 1,000 photos to make this none image.
 
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Does anyone have or know a formula to figure how high we wold have to fly in order to photograph 1 acre of land? What about 10 Acres or 50 acres? I live in a rural area and have been asked to photograph some farmland, all in one shot.

I was just wondering if there is a way to figure how high I would have to fly in order to photograph a certain size piece of land?

I realize of course that it would depend on the shape of the property. For the sake of making this easier, let's pretend the farm is square.

If I was asked to photograph a 50 acre farm, how high would I have to be?

Thanks in advance for your help.
Why not just Use Google Earth!!!
 
As long as you can see your Bird the 400 foot limit does not apply!

400' altitude is to be observed in the US unless you have a waiver that allows you an exception.
 
For a 4:3 ratio image from a camera with a diagonal field of view, Ø, the area, A, imaged from a height h is given by:

A = (24/25)..tan²(Ø/2)​

If h is in feet then A is in square feet. In acres:

Acres = A/43560​

Ø is approximately 90° for a Phantom.
According to my manual, FOV is 84 degrees.
 
Here is my take.

A = Altitude
W = Width of the shot pointing straight down
F =FOV Angle

Theoretically, if my math is correct:

A=.5W/tan(.5F)
and
W=2A*tan(.5F)

My Phantom 4 Pro has a FOV of 84 degrees. So for a shot straight down from 400ft, I would expect to capture 720 feet.

W=2*400*(tan(.5*84))
W=800*tan42
W=720 feet

Not sure how many acres 720ft by 720 ft is. Too lazy to do the math. But I am guessing it is much less that what you ultimately want to capture.

l tried my theory at a local high school football field. I wanted to see how high I needed to fly to capture the entire football field including both end zones (120 yards or 360 feet). Theoretically:

A=.5W/tan(.5F)
A=.5*360/tan(.5*84)
A=180/tan42
A=200 feet

When I went to the field the Altitude that I flew to in order to capture was a little more or less (can't remember which now) but it was pretty close. Not sure what caused the difference bit based on the response from Meta4, I think it might have been due to the Aspect Ratio setting. I have no clue what my aspect ratio was set at when I did my test. I will go back out and try my test again to see if this accounts for my error.

Long story made short, if you need to do 50 acres it would be best to get a mapping software that will fly autonomously taking many pictures. Those programs help you stitch the photos together.

EDIT NOTE: MY CALCULATIONS ABOVE ARE INCORRECT BECAUSE I DID NOT TAKE ASPECT RATIO SETTING INTO ACCOUND AND DID NOT TAKE INTO ACCOUNT THAT THE FOV OF 84 DEGREES IS THE DIAGONAL DISTANCE. SO FOLLOW UP POST FOR CORRECT EQUATIONS.
 
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Here is my take.

A = Altitude
W = Width of the shot pointing straight down
F =FOV Angle

Theoretically, if my math is correct:

A=.5W/tan(.5F)
and
W=2A*tan(.5F)

My Phantom 4 Pro has a FOV of 84 degrees. So for a shot straight down from 400ft, I would expect to capture 720 feet.

W=2*400*(tan(.5*84))
W=800*tan42
W=720 feet

Not sure how many acres 720ft by 720 ft is. Too lazy to do the math. But I am guessing it is much less that what you ultimately want to capture.

l tried my theory at a local high school football field. I wanted to see how high I needed to fly to capture the entire football field including both end zones (120 yards or 360 feet). Theoretically:

A=.5W/tan(.5F)
A=.5*360/tan(.5*84)
A=180/tan42
A=200 feet

When I went to the field the Altitude that I flew to in order to capture was a little more or less (can't remember which now) but it was pretty close. Not sure what caused the difference bit based on the response from Meta4, I think it might have been due to the Aspect Ratio setting. I have no clue what my aspect ratio was set at when I did my test. I will go back out and try my test again to see if this accounts for my error.

Long story made short, if you need to do 50 acres it would be best to get a mapping software that will fly autonomously taking many pictures. Those programs help you stitch the photos together.

Your equations are correct, but the FOV angle quoted is generally the diagonal of the sensor, so that is the distance given by your W, not the width or height.

I just noticed a factor of 2 error in my calculation above though, it should be:

A = (48/25)..tan²(Ø/2)​
 
Awesome, thanks for the clarification! I know very little about camera specs. So then does that mean that aspect ratio setting still need to be taken into account? Is that what your equation does?
 
Awesome, thanks for the clarification! I know very little about camera specs. So then does that mean that aspect ratio setting still need to be taken into account? Is that what your equation does?

The calculation I showed was for the particular case of a 4:3 aspect ratio (making a 3:4:5 triangle), where the diagonal is assumed to be the 84°.
 
There is an iPhone application called “PhotoPills” that can help you.
 
Does anyone have or know a formula to figure how high we wold have to fly in order to photograph 1 acre of land? What about 10 Acres or 50 acres? I live in a rural area and have been asked to photograph some farmland, all in one shot.

I was just wondering if there is a way to figure how high I would have to fly in order to photograph a certain size piece of land?

I realize of course that it would depend on the shape of the property. For the sake of making this easier, let's pretend the farm is square.

If I was asked to photograph a 50 acre farm, how high would I have to be?

Thanks in advance for your help.

Oblique or nadir?
 
Here are my revised formulas from my previous post for various Aspect Ratios:

F = Field of View Angle (84 degrees for the P4P)
W = Width Captured on the screen
A = Distance from the Object (Altitude)

Altitude (A) and Width (W) - Generic Formulas
4:3 Aspect Ratio - A=.625W/tan(.5F) and W = 1.600(A)tan(.5F)
3:2 Aspect Ratio - A=.600W/tan(.5F) and W = 1.667(A)tan(.5F)
16:9 Aspect Ratio - A=.574W/tan(.5F) and W = 1.742(A)tan(.5F)

So for a Phantom 4 Pro to capture a football field (including end zones):
4:3 Aspect Ratio - A=.625(360)/tan(.5(84)) = 250ft
3:2 Aspect Ratio - A=.600(360)/tan(.5(84)) = 240ft
16:9 Aspect Ratio - A=.574(360)/tan(.5(84)) = 229ft
 
Here are some numbers to give you a guide.
Assuming you flew a Phantom 4 pro and shot in 3:2 (other aspect ratios would capture a smaller area)
Altitude = 200 ft Area covered = 300 x 200 ft = 1.4 acres
Altitude = 400 ft Area covered = 600 x 400 ft = 5.5 acres
Altitude = 1200 ft Area covered = 1800 x 1200 ft = 49.6 acres

An acre is easy but you can see the problem with larger areas.
For big areas, the thing to do is to fly a grid, shooting a lot of precisely overlapping images and stitch them in a mapping program (a photo program won't make an accurate map necessarily).
The advantage of this is you get a huge composite image with tons of detail and you don't have to fly at crazy heights.
Here's an example of a 12 acre site I mapped from about 50 individual images.
i-KjJQpgb-X3.jpg
Thanks for this info.
 
Might be better to ask a pilot friend to do it for you..Buy him some gas and take a good high quality Digital camera along and shoot away
be sure it a clear skys day...and have him make several passes full length...same way each pass.

. Then also you could use the drone and take several pictures and piece the pictures together in photo shop....I've done that !

max altitude for us here in the States is 400 feet.
FAA regulations say drones shouldn't be flown above 400 feet. Higher than that, drones start to interfere with the national airspace. They can't be flown within a few miles of an airport, and they can be used only for fun — not for commercial purposes.
Had not heard they could not be used for commercial purposes. I saw one the other day that is used in surveying.
 
Is this a beauty shot? Like for real-estate? Or are you looking to create an orthophoto for forensics, architecture and/or construction? The differences between the two are profound. Without knowing what the photo is for, take all advice with a grain of salt.

If you made that clarification already, please repeat, as I don't have time to look through all of these comments. Let's cut to the chase.

D
 

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