how to find height in an image using matlab

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KARTHIK B
KARTHIK B on 21 Apr 2014
Commented: Star Strider on 22 Apr 2014
like aircraft is moving, if it takes an image of ground. we have to calculate at what height the aircraft is please reply me

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Answers (2)

Walter Roberson
Walter Roberson on 21 Apr 2014
This is not generally possible. For example if you are flying over a large lake, unless you are very low, there are no distinguishing features that you could measure the size of in the image. Likewise if you are flying over rocks you will not generally be able to tell the scale. You will not always be able to tell the scale of human buildings or of trees in a forest.
Before GPS, altitude was usually measured by barometric pressure rather than visually.
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Star Strider
Star Strider on 22 Apr 2014
Still is. The pressure altimeter remains the standard, but variations in atmospheric pressure lead to interesting problems. Below 18000 feet, all aircraft use the local barometric pressure (‘altimeter setting’ or more simply ‘altimeter’), requiring VFR pilots to contact various Flight Service Stations en route for local updates. Above 18000 feet, all flight is IFR and all altimeters are set to 29.92 (inches of Hg).

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Image Analyst
Image Analyst on 22 Apr 2014
If you have two frames (a video) and can measure the angle from vertical to a distant landmark, and you know the velocity and the time to get to the landmark, you can do it. Let's say that the landmark is a distance d away and you know the angle from vertical to the landmark is theta. Then let's say you know your velocity and the time it takes until you're directly overhead.
|---------------------------------|
|.*...............................|
|.......*.........................|
|..............*..................| H
|....theta.............*..........|
|.............................*...|
|---------------------------------|
d = v * t
So tan(theta) = d/H or H = d / tan(theta) = (v*t)/tan(theta).
But with just a single image and no other assumptions, there's no way.
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Image Analyst
Image Analyst on 22 Apr 2014
Hint for #1: Start your triangle from the center of the earth.
Star Strider
Star Strider on 22 Apr 2014
I did, but maybe I did the initial calculations incorrectly. I calculated the radius as 4E6/(2*pi) meters, since If I remember correctly, the meter was originally defined as one-millionth the distance from the equator to the north pole, so assuming a sphere, my R calculation should be correct but obviously wasn’t.

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