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Author Topic: Real world dimensions of image obtained from drone flight  (Read 5946 times)

luisarvayo@gmail.com

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Real world dimensions of image obtained from drone flight
« on: May 31, 2018, 03:44:49 AM »
Hello,

How can I calculate the real world dimensiones, for instance METERS, for the width and height of an image obtained from a drone flight ?

Do I need to calculate the GSD or Ground Sample Distance for that purpose () ? If yes, I dont know what to do next once obtained the GSD.


Thanks


« Last Edit: May 31, 2018, 04:59:48 AM by luisarvayo@gmail.com »

SAV

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Re: Real world dimensions of image obtained from drone flight
« Reply #1 on: May 31, 2018, 07:19:12 AM »
Hi luisarvayo,

If you have 24MP images with 6,000 x 4000 pixels, then simply multiply the image width and height with the GSD.

For example, let's say your GSD is 5cm/pixel, then your image footprint for a nadir image on a flat surface is 30,000 x 20,000 cm = 300 x 200m.

If the camera is oriented at a certain angle, then you can calculate the footprint following this workflow:
https://photo.stackexchange.com/questions/56596/how-do-i-calculate-the-ground-footprint-of-an-aerial-camera?utm_medium=organic&utm_source=google_rich_qa&utm_campaign=google_rich_qa

All the best.

Regards,
SAV

luisarvayo@gmail.com

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Re: Real world dimensions of image obtained from drone flight
« Reply #2 on: May 31, 2018, 09:49:05 AM »
Hi SAV,

Thank you very much for the answer and the link.

I am calculating the GSD the following way:

GSDh = ( [Flight Height] * [Sensor Height] ) / ( [Focal Length] * [Image Height] )

GSDw = ( [Flight Height] * [Sensor Width] ) / ( [Focal Length] * [Image Width] )

GSD = Max Of ( GSDh, GSDw)

All units in centimeters,

For DJI Mavic PRO I am using :

Sensor Width = 4.7 mm
Sensor Height = 6.3 mm
Focal Length = 28 mm

Example Flight Height = 4790 cms

So I obtained :

GSD = 0.35925

That way, the real world dimensions are (assuming drone photo are 4000 x 3000 pixels):

Image width in cms = GSD * 4000  = 1437 cms
Image height in cms = GSD * 3000  = 1077.75 cms


But by simple viewing the photo you can see that the real world dimensions of the area is much bigger.

Do you think that the sensor and focal length parameters are wrong and is the reason of the wrong results ?


Also, additionally, after checking your link, I tried calculating with the formula provided but I have the doubt about the gimbal parameters.

The drone will give you the following angles:

GimbalRollDegree
GimbalPitchDegree
GimbalYawDegree

I am assuming

xgimbal = GimbalRollDegree
ygimbal = GimbalPitchDegree

But the results are completely wrong, example in c# language

                double xsensor = 4.7;
                double ysensor = 6.3;
                double focallen = 28;
                double altitude = 48;   // approx
                double xgimbal = 0;
                double ygimbal = -90;

                double xview = 2 * 57.296 * (Math.Atan(xsensor / (2 * focallen)));
                double yview = 2 * 57.296 * (Math.Atan(ysensor / (2 * focallen)));
                // from drone to bottom of picture
                double dbp = Math.Round(altitude * Math.Tan(0.01745 * (xgimbal - 0.5 * xview)), 2);
                // from drone to top of picture
                double dtp = Math.Round(altitude * Math.Tan(0.01745 * (xgimbal + 0.5 * xview)), 2);
                // from drone to left of picture
                double dlp = Math.Round(altitude * Math.Tan(0.01745 * (ygimbal - 0.5 * yview)), 2);
                // From drone to right of picture:
                double drp = Math.Round(altitude * Math.Tan(0.01745 * (ygimbal + 0.5 * yview)), 2);
                // height of photo footprint
                double heightOfPhotoFootprint = drp - dlp;
                // width of photo footprint
                double widthOfPhotoFootprint = dtp - dbp;

results are :
heightOfPhotoFootprint = -851.56;
widthOfPhotoFootprint = 8.04

0.01745 = conversion from degrees to radians
57.296 = conversion from radians to degrees

Thank you















« Last Edit: May 31, 2018, 09:56:12 AM by luisarvayo@gmail.com »