Timeline for How to produce a chasing-view in Mathematica?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Sep 20, 2014 at 22:47 | answer | added | Putterboy | timeline score: 2 | |
| Sep 19, 2014 at 18:21 | comment | added | Kuba |
pos2 = RotationMatrix[#, {0, 0, 1}].(1.1 {-0.00465, -1.6416, 0.64730}) + {pR Cos[#], pR Sin[#], 20 + Sin[#]} & and use it in options for graphics ViewVector -> {pos2[t - 1], pos2[t + .1]}, ViewAngle -> 1
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| Sep 19, 2014 at 18:14 | comment | added | kglr | See also: Slide 7 in Vitaly Kaurov's Mastering Dynamic Visualizations | |
| Sep 19, 2014 at 17:46 | answer | added | Apple | timeline score: 3 | |
| Sep 19, 2014 at 13:21 | comment | added | gst | Since you are moving in a circle you could of course use the the angular unit vector of your polar coordinates. | |
| Sep 19, 2014 at 13:08 | comment | added | gst | I guess a first step would be finding the tangential vector with respect to the motion and using it as "ViewPoint" in the Graphics3D. Tangent vector | |
| Sep 19, 2014 at 11:48 | comment | added | Sjoerd C. de Vries | And perhaps 14480 | |
| Sep 19, 2014 at 11:09 | history | edited | Putterboy | CC BY-SA 3.0 |
added 43 characters in body
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| Sep 19, 2014 at 10:49 | comment | added | Mr.Wizard | Related: (3528), (5649) | |
| Sep 19, 2014 at 10:48 | history | edited | Mr.Wizard | CC BY-SA 3.0 |
add direct import
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| Sep 19, 2014 at 10:45 | history | asked | Putterboy | CC BY-SA 3.0 |