Televid 77: Actual magnification? (1 Viewer)

[Etiennef]

I'm not really convinced that the comparison with a viewfinder is really applicable. The "scale" in the viewfinder does not necessarily correspond to the view in a scope.

If you take an example with an preview-screen instead of a viewfinder I think it becomes pretty apparent. In that case it would be like saying that a camera with a 3" screen has about twice the magnification of a camera with a 1,5" screen, even if the have exactly the same specs on all other parameters...

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A good way to test this would of course be if someone with one of the previously mentioned cameras could take a picture of an object of know size (for example a standard CD-case) at say 10 meters. Then it would be easy to compare and see if we are in the correct ballpark or not...

 

[JerryLogan]

I'm not really convinced that the comparison with a viewfinder is really applicable. The "scale" in the viewfinder does not necessarily correspond to the view in a scope.

If you take an example with an preview-screen instead of a viewfinder I think it becomes pretty apparent. In that case it would be like saying that a camera with a 3" screen has about twice the magnification of a camera with a 1,5" screen, even if the have exactly the same specs on all other parameters...

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A good way to test this would of course be if someone with one of the previously mentioned cameras could take a picture of an object of know size (for example a standard CD-case) at say 10 meters. Then it would be easy to compare and see if we are in the correct ballpark or not...

I am speaking of actual image sizes; from the time and land of film cameras where there was a measurable final image. So on the negative the 1200mm lens produced an image that was about half the size of that produced by the 40x scope. If you have access to an interchangeable lens camera it's easy to demonstrate even digitally by switching scope for lens. I assure you this is physics, not manipulation of viewfinders. A 1300mm lens equalivant will disappoint you if you're used to 44x images.

Best,
Jerry

 

[Etiennef]

Sure, you may very well be correct (of course).

But the issue for me right now is that basically you have just said that 26x "feels" like 1300mm when looking through a camera viewfinder. Indeed it's physics, and exactly due to this I personally would prefer a bit more proof than a subjective perception.
Especially since we now are not any more talking about a deviation of 10%; we are talking about nearly twice as much as a the commonly used formula would suggest. That is:

Scope mag * Camera focal length = Digiscoping focal length

My initial concern in the first post was that the above formula (at practical distances it turns out) was out by a few percent. What you now are saying is that the actual value is nearly 2x the calculated value...

 

[Etiennef]

Perhaps we are talking about different things.
Lets see if I can't explain what I'm talking about more precisely:

If I look at a ruler through my scope (40x magnification), at a distance of more or less exactly 10m, I get the attached image. In terms of markings on the ruler, this is equivalent to what I see with my eyes when looking "normally" through the scope. In other words, I don't see fewer, or additional, markings on the ruler with my own eyes compared to the camera (other than what's limited by the FOV).
(Sorry, the light is very bad right now and I just took a quick picture with my phone. It won't make any difference for this example anyway)

What I can fit into the scopes view along the ruler is basically 30 centimeters. For the sake of being able to read the numbers at the edge I added a small margin . If we would take this into account it would probably be closer to 31 centimeters, but that's not really super important right now.

Leicas specs for this scope+eyepiece say that the view at 1 km is 34 meters, so at this distance, which is 10 meters, it would mean 34 centimeters on the ruler.
As concluded earlier, the scope overshoots the magnification at shorter distances. Based on earlier experiments, about +10% on the specs gives the actual magnification at this distance (so this matches perfectly with the previous findings).

When I digiscope I zoom the camera to more or less the red square. I like to keep a bit of the black area in the frame for focusing reference.

If we measure the width of the square according to the ruler in the pic we end up at 26 centimeters, and after necessary cropping it would probably be more like just short of 25 centimeters.

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What I want to know is what focal length (35mm equivalent) on a super-zoom camera will give me the same view as the red area, or preferably marginally smaller due to the necessary cropping. To avoid confusion I added a picture to illustrate what the "final product" would be, provided that I don't crop more than necessary. This is represented by the second image.

Based on what I have concluded based on the commonly used formula, + my 10% corrections, it would be around 1300 mm, or at least close to that number...

If it in fact actually is 2200 mm, then something is way off with the frequently presented way to calculate this...

(Image quality and resolution and other factors will off course come into play also, but those aspects are outside the scope of this topic)

 

[Etiennef]

Well, after some investigation it seems like it's as easy as:

(sensor width * distance to object ) / with of frame = focal length

So 26cm wide at 10 meters turns out to be more or less 1380mm if the sensor (equivalent) is 36mm wide.

I've heard that manufacturers sometimes provide 35mm equivalents as literally 35mm sensor width (kind of like USB manufacturers give 1000 bytes per Kb, when it in binary computer-standards would be 1024). In that case it's about 1350mm.

I tested about half a dozen online-calculators as well, and they all seem to give the same result. So now I'm pretty confident what the actual corresponding 35mm equivalent focal length is on my digiscoping setup.