DJI H20T thermal radiometric jpeg file

[Drone Link]

Hey all,

I know im late to the party, but whats the latest on this issue? It seems to me the prevailing options are:
dji_h20t_rpeg_to_tif-main; by tejakattenborn
or traitement_temperature; by Remi

Is there anything more up to date? Would love to be updated on whats happened.
Thanks

[pierotofy]

I've started digging into this. To get the raw, uncorrected Celsius values one needs to apply the following formula to ThermalData:

Code Select Expand

celsius = ((raw_int16 >> 2) * 0.0625) - 273.15


Code Select Expand

from PIL import Image
import numpy as np

im = Image.open("/datasets/dji_thermal/images/DJI_20240520212354_0001_T.JPG")

# concatenate APP3 chunks
a = im.applist[3][1]
for i in range(4, 14):
    a += im.applist[i][1]

raw = np.array(Image.frombytes('I;16L', (640, 512), a), np.int16)
celsius = np.right_shift(raw, 2).astype(np.float32)
celsius *= 0.0625
celsius -= 273.15

Image.fromarray(celsius).save('/datasets/dji_thermal/celsius.tiff')

DJI applies some sort of correction based on temperature, distance, emissivity and humidity. I'm still looking into that.

[josemariagarcia]

Thanks for your contribution @pierotofy!!

I'm new to this world of processing thermographies and such low-level data and I was getting really frustrated.

Is there any other way to get this Celsius temperature directly from a R-JPEG? I thought that would be possible using the DJI SDK from Python, calling it with subprocess, but all the outputs I managed to produce, they couldn't either be opened by Image Viewer apps nor Python Image Handling modules.

Thanks in advance!

[Remo]

Very helpful work pierotofy, thank you very much. I applied your formula but still got some irregular differences compared to the values which I got from the DJI Thermal SDK. What exactly is the factor 0.0625? Is it stored in the r-jpg as well, or is it a kind of constant "DJI calibration factor"?