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Asked: 2026-05-27T18:45:40+00:00

I am making a camera-based application for Android. My problem, is to find the

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I am making a camera-based application for Android. My problem, is to find the real angle subtended by an object in an image, on the camera, when I have :

1.The length of that object in the image, in pixels,

2.The vertical angle-of-view of the camera,

3.The length of the image,in pixels.

Being new to image processing, I have this (probably basic) doubt. Is there a linear relationship between the angle that an object subtends on the camera, and the number of pixels in the image of the object taken by the camera? Angle subtended: Join the topmost point of the object to camera, then join bottom most point of object to camera. The angle formed at the camera, is the angle subtended. For example: Consider an image that is vertically, 200 pixels in height. The vertical angle of view of the camera, is suppose 20 degrees. So, I can say, that this 20 degrees covers 200 pixels of the image. Now, there is a portion of the image (an object) which is 10 pixels in height. I want to find the angle subtended by this object on the camera, in physical world. Would I be right in telling that, since 200 pixels are covered by 20 degrees, 10 pixels would be covered by 1 degree? That is, this object subtends an angle of 1 degree on the camera?.

enter image description here

As shown in this diagram, I know the values of L (of image), P (of object in image), x (Vertical angle of view of camera), and I want to find y , the angle subtended by object on camera. My assumption is (y/P)=(x/L), that is directly proportional. Would this assumption be correct?

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1 Answer

  1. Editorial Team

    Draw horizontal line. It’s length D is distance to object.

    D * tg(X/2) = L/2

    and

    D * tg(Y/2) = P/2

    so

    Y = 2 * Arctg(tg(X/2) * P / L)

    (When angles are small, they are (rough) direct proportional)

    Addition: case of non-centered object.

    enter image description here

    P0, P1 – pixel coordinates of object ends relative to center point D

    distance Dx = |OD| = L * ctg (X/2)

    coordinate D=(Dx,0)

    Y = Angle(OP1, OP0) = ArcCos((OP1 x OP0)/(|OP1| * |OP0|)) , where x means scalar product

    Y = ArcCos((Dx*Dx+P1*P0)/Sqrt((Dx*Dx+P1*P1)(DxDx+P0*P0)))

    Example:
    L=L1-L0=768 pixels

    X = 42 degrees

    Dx = 1000 (rough)

    P1 position = 100 pixels

    P0 position = -50 pixels

    Y = ArcCos((1000*1000-100*50)/(Sqrt((1000*1000 +100*100)*(1000*1000+50*50)))) = 7 degrees (rough)

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