Timeline for Projectile Motion - Arrow
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 23, 2011 at 13:52 | comment | added | Keeblebrox | @FxIII It's my pleasure :) | |
| Jun 23, 2011 at 13:19 | comment | added | FxIII | @Keeblebrox Thank you, I appreciate your corrections, your support is important to let everyone to contribute. | |
| S Jun 22, 2011 at 23:01 | history | suggested | Keeblebrox | CC BY-SA 3.0 | Spelling and grammar corrections |
| Jun 22, 2011 at 21:25 | review | Suggested edits | |||
| S Jun 22, 2011 at 23:01 | |||||
| Jun 20, 2011 at 15:18 | comment | added | FxIII | Anyway if you can relate the momentum to the speed given a feather setup, all can be computed thorough integration but i'm not sure you can have a closed form for the equations of motion (i.e. you can get an integration algorithm but not a parametric equation). | |
| Jun 20, 2011 at 15:13 | comment | added | FxIII | The problem is that the sole momentum is easy to compute is the one due the angle variation (you can see that deriving twice a parabola just a constant term remains). The other is caused by the spinning due the back feather. Here feather drags and friction are involved both converting kinetic energy into spinning, slowing down the arrow but adding gyroscopic effect. This influences the trajectory and is quite difficult to model | |
| Jun 20, 2011 at 15:10 | comment | added | FxIII | One can compute the moments of inertia with ease. These are two for rods, one for the rotation about its center of mass and the other for the rotation about the rod's axis. Superposition principle applies for moments of inertia so the arrow can be splitted into three part: feathers,body and tip. | |
| Jun 20, 2011 at 14:23 | comment | added | Dov | I think you mean: ![∂y/∂x = g/(2s²)·x+1][2] but in any case I think I recommended a better approach below. For one thing, you didn't explain about separating the x and y components, so this is hardcoded to an arbitrary 45 degree angle, with launchVelocity not being truly launchVelocity, but the component in both x and y | |
| Jun 20, 2011 at 13:01 | comment | added | Martin | Thank you Fxlll. Any idea where I could get the formulas which apply to the physic of an arrow? | |
| Jun 20, 2011 at 10:34 | history | edited | FxIII | CC BY-SA 3.0 | added 129 characters in body; added 6 characters in body |
| Jun 20, 2011 at 10:23 | history | edited | FxIII | CC BY-SA 3.0 | added 1 characters in body; added 1 characters in body |
| Jun 20, 2011 at 9:51 | history | answered | FxIII | CC BY-SA 3.0 |