CatMarley Loc: North Carolina

The horizontal FOV of a 50 mm lens is 39.6 degrees. therefore tan 39.6 = width/100 meters, tan39.6 = 0.83. 0.83 x 100 = width = 83 meters. (The actual width is the arctan function and is an arc, but my calculator doesn't have that, so tan is close enough) This is quite different from 70 meters. Maybe the navy was making sure you got a bit more than you needed.

davidf_logan Loc: Logan, Utah

Cat,

I think you have the formula incorrect. Plus, you need to use radians when calculating the Tangent. If I do the calculation then it is

Tan[Pi*39.6/(2*180)]= (w/2)/100 m.

where Pi/180 is the conversion from degrees to radians. Solving this for w gives you 72 m. The relevant angle is half the angular FOV (which is why there is a factor of 2 in the denominator in the argument of the Tan function) and the "opposite" side of the triangle calculated is actually 1/2 the width. See the link below and go to Derivation-of-the-angle-of view formula for the diagram.

https://en.wikipedia.org/wiki/Angle_of_view

EDIT: the "navy" formula relies on the fact that the triangles shown in the diagram in the link above are "Similar triangles." However, the relevant distance S2 is not necessarily the focal length but it is a good approximation to it. In the calculation of the width by the previous poster using the navy formula, 36mm (the horizontal dimension of the sensor) should have been used rather than 35mm and doing so also gives 72m.

Edit2: It looks like you calculated the Tan correctly but you missed a factor of "2" in two places (in the angle and in the width).

CatMarley Loc: North Carolina

I should have gone to Wickapedia!

davidf_logan Loc: Logan, Utah

What's interesting is that your formula gives an answer that is quite close to the actual value -- only about 15% off. That's because, for small arguments Tan[x] ~ x (as long as x is in radians). So the factors of 2 almost cancel themselves out for small-enough angles with the error being much worse for larger angles; 39.6 degrees is small enough that the error, while noticeable, isn't really that large. For an 18mm lens the error becomes essentially infinite!

User ID

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No need to know your trig. Similar
triangles work for everyone with no
need to know why or to speak math
jargon :-)

You need to know your sensor width.

When the FL equals the sensor width,
the FoV equals the subject distance.

Double the FL and you halve the FoV.

A 35mm on a FF has a 5ft FoV at 5ft
and a 105 sees a FoV of 5/3ft etc etc.

.

BebuLamar

Your error is that the 39.5 degrees angle of view isn't a right triangle and thus tangent function doesn't work correctly. If you want to use the tangent function then you have to use half the angle and then multiply by 2. however the other method which didn't use the tangent function but only the ratio of the film width/focal length = field width / distance works fine.

bleirer

On the calculator I mentioned you can enter the subject size, which might be the most important, and the distance for a certain sensor and it returns the required focal length to fill the frame with that subject. If the point is choosing lenses. Repeat link https://www.cambridgeincolour.com/tutorials/camera-lenses.htm

John Howard Loc: SW Florida and Blue Ridge Mountains of NC.

Thanks. Did not know that. I checked it out and it is a good tool. Just planing what I'll need on a trip.
Cheers,