Could someone have a look at it and correct if something is wrong in calculation and formulas

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777rao Junior Member • Posts: 38
Could someone have a look at it and correct if something is wrong in calculation and formulas

Now i have 10m as well as 20m width with 100m length i am intersted in finding the field of view and focal length

My camera specifications

Pixel size 1.4µm x 1.4µm
Image area : 3673.6µm x 2738.4µm = 36.736mm X 27,384mm

First for 10m

100/10 = 10m

Field of view w = 2arctan(θ/2f)
2arctan (100/2*10)
2arctan(100/20)
arctan 5*2 = 10 degree

Focal length

F/X = D/W, so

F = X*D/W.
F = 36*100/10
F =360mm

Second for 20m

100/20 = 5m

Field of view w = 2arctan(θ/2f)
2arctan (100/2*5)
2arctan(100/10)
arctan 10*2 = 20 degree

Focal length

F/X = D/W, so

F = X*D/W.
F = 36*100/20
F =180mm

For S = 10m "big/wide/tall/long/size/fat/thin" (horizontal "wideness")

D = 100m "in the distance"

R = D/S = 100m/10m = 10.0 "ratio" (distance/subject-size "factor")

FL = R x W = 10 x 36mm = 360 mm lens exactly (or 1028.6mm)

AoV = 2 tan (D/(S/2)) = 2 tan (100/10/2)) = 2 tan (20) ... or ... from above

2 tan (2 x Ratio) = 2 tan (2 x 10) = 2 tan (20) = 4.4732° AoV "wide" (horizontal aov)

For S = 20m "big/wide/tall/long/size/fat/thin" (horizontal "wideness")

D = 100m "in the distance"

R = D/S = 100m/20m = 5.0 "ratio" (distance/subject-size "factor")

FL = R x W = 5 x 36mm = 180 mm lens exactly (or 1028.6mm)

AoV = 2 tan (D/(S/2)) = 2 tan (100/20/2)) = 2 tan (10) ... or ... from above

2 tan (2 x Ratio) = 2 tan (2 x 5) = 2 tan (10) = 1.296° AoV "wide" (horizontal aov)

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