Craziest question of the year, by far!Posted By Joel Elliott |
Fellas,
Alrighty, here's a good 'un. I was browsing the D&G website earlier this year and saw something that keeps me wondering. They've gotten up a line of long-focus refractor doublets, no news there, but now they can render them in rather long focal ratios. Even f/30!
So - let's imagine you had a way to mount up two 10 inch f/30 tubes in a giant binocular system. The spread would be about 550 inches or so, under the way I'd do it in my obviously deranged mind. This crazy idea has been dogging me all year. Math Guys: Is this wide enough to render a 3D image of the moon, parallax-wise?
I know that binoviewer owners report 3D like images all the time, but clearly this is an illusion. Good for them, and I mean that nicely. But seriously though, how wide a spread would it take to get a parallax rendered, genuine 3D image of something about 300,000 miles away?
Does anyone know the math involved in such a computation?
Nuts in NC,
J. Elliott
Alrighty, here's a good 'un. I was browsing the D&G website earlier this year and saw something that keeps me wondering. They've gotten up a line of long-focus refractor doublets, no news there, but now they can render them in rather long focal ratios. Even f/30!
So - let's imagine you had a way to mount up two 10 inch f/30 tubes in a giant binocular system. The spread would be about 550 inches or so, under the way I'd do it in my obviously deranged mind. This crazy idea has been dogging me all year. Math Guys: Is this wide enough to render a 3D image of the moon, parallax-wise?
I know that binoviewer owners report 3D like images all the time, but clearly this is an illusion. Good for them, and I mean that nicely. But seriously though, how wide a spread would it take to get a parallax rendered, genuine 3D image of something about 300,000 miles away?
Does anyone know the math involved in such a computation?
Nuts in NC,
J. Elliott