The information I have is that the camera used was The onboard cameras for the Apollo 8 mission were modified Hasselblad 500 EL cameras, with 80-millimeter and 250-millimeter Zeiss panacolor lenses.
The film was 70 mm sprocket film with gate dimensions (from what I can drag up) of 55 mm x 55 mm.
On the 1920 x 1920 pixel photo I have the earth image is 287 pixels wide, or 55 x (287/1920) = 8.22 mm wide.
If the 250 mm lens was used this makes the angular size of the earth 2 x atan((8.22/2)/250) = 1.88°
If we take the diameter of the earth as 7,918 miles, this makes the earth to moon distance of 241,728 miles.
@Rabinoz
You seem to know a fair bit about this kind of thing so I have a couple of questions for you. I don't want to obligate you with spending time on this if you don't want to. If you're too busy just disregard this post.
Can you tell me how far a camera would need to be from the earth in order to determine it's shape? Assume the best commercially available camera for the job. Do you know if FEers would accept digitally streamed images or would it need to be an old fashioned negative? And apparently there is some issue with the type of lense? Is there a type of lense which cannot be blamed for the roundness of the images produced at large distances?
Thanks for your time.
If you use a good 50 mm lens on a full frame 35 mm camera (or equivalent on smaller cameras) you will get little the distortion that so many FEers call "fish-eye lens" effect.
The sensor on a standard full frame 35 mm camera is about 36x24 mm (width x height), so the angluar field covered can be found from
2 x atan((half image dimenension)/ (lens focal length)).
For this standard 50 mm lens this makes the angular
Field
of
View is roughly 41° x 28°.
Now how far away depends on whether you want to show the
full disk or just
enough of the curvature to be convincing.
If you don't show the full disk, flat earthers will try to claim "so what, that just shows the "illuminated disk caused by the sun".
So to be really sure you are seeing the globe you need to be at a distance where the whole diameter of the earth (about 12,742 km) can be seen.
A bit more trig is needed (mind you bit of simple proportion would do it too). We need to find the distance for the half angle at the camera to be under 14°.
This can be worked out from
distance = (diameter/2)/sin(ang/2) or almost over 28,000 km - this is the distance from the earth centre. The
altitude is about 22,000 km.
Now the geostationary weather satellites are rather bit above this at an altitude of approximately 35,786 km, giving them view of all but even at this altitude they miss almost 9° of each polar region.
As far as I am concerned any altitude above about 100 km will show good curvature, the ISS is typically at about 400 km (it varies substantially), but the images from that leave no doubt - no wonder Flat Earthers deny its existence!
But if you want (almost) the lot
without using a wide angle lens you do need to be over 22,000 km. (I hope my tans and sins aren't confused, or I'll get my hide tanned for my sins!)
In my opinion, a more definitive test is to measure the dip angle from the local horizontal to the horizon. There are lots of references on the internet to this and it does not need such a high altitude to be convincing - a balloon or rocket with carefully aligned cameras could photograph the horizon in say four directions. Thid dip angle is about 3° at 10,000 m altitude and increases as the square root of the altitude above that.
The earth is huge!