Timeline for Framing shots before purchasing cameras?
Current License: CC BY-SA 4.0
12 events
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| Feb 25, 2020 at 18:46 | comment | added | Rafael | @xiota, I updated the answer. | |
| Feb 25, 2020 at 18:40 | history | edited | Rafael | CC BY-SA 4.0 |
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| Feb 25, 2020 at 18:35 | history | edited | Rafael | CC BY-SA 4.0 |
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| Feb 25, 2020 at 18:34 | comment | added | Rafael | @MikeBrockington about the angles, the triangle must be a 90° triangle or this will not work. That is why I divided the triangle in the first place. Use the calculator I posted and try using A with double the length and you will see that gives you a different angle. | |
| Feb 25, 2020 at 18:29 | history | edited | Rafael | CC BY-SA 4.0 |
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| Feb 25, 2020 at 16:32 | comment | added | xiota | Use a card with a cut out the same size as the camera sensor. Distance from eye is focal length needed. Or can use crop factor for larger frame. No need for trigonometry calculations. | |
| Feb 25, 2020 at 12:48 | comment | added | Carsten S | @MikeBrockington, no, they don't. | |
| Feb 25, 2020 at 11:33 | comment | added | Carsten S | I understand that this works, but using trigonometry to convert a proportion to an angle and back again seems overkill to me. You can also get the focal length as B * sensor width / 2A. (Assuming that the focal point of the length will be where your eyes are now.) So in your example, the focal length would be the sensor width, 24mm for DX. | |
| Feb 25, 2020 at 3:03 | comment | added | Michael | Oh hey, now I know why purportedly movie makers in the 50s would hold out their arms and make the opposing sides of a rectangle with the squared thumb and index finger to get an idea of what the picture would look like when scouting out a location. | |
| Feb 23, 2020 at 23:01 | vote | accept | Ryan Leach | ||
| Feb 23, 2020 at 16:32 | history | edited | Rafael | CC BY-SA 4.0 |
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| Feb 23, 2020 at 16:24 | history | answered | Rafael | CC BY-SA 4.0 |