Can aerodynamic trajectories ever be shaped like this?

  • #1

Main Question or Discussion Point

I have recently written a trajectory program which includes aerodynamic resistance, that is, only form drag at this point. Out of curiosity I put in some ludicrous inputs such as speeds above Mach 1 (I know that wave drag becomes significant at that speed and my calculations would be rendered inaccurate), but either way, I get my program to give a trajectory that has a shape like a sideways sigma (not the circle-like one) and that the "projectile" takes a near 90° turn downwards. My question is, can aerodynamic trajectories ever look like this (in theory, at least) or is this an artifact of shoddy programming. I do not want speculations as much as I want a clear, well supported answer. I'll try to eventually get a picture here.
 

Answers and Replies

  • #2
haruspex
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If you want validation of your equations/software, you'll need to post that too.
 
  • #3
Danger
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Welcome to PF, Elroy. (Please pass along my regards to Rosie and Astro. :smile:)
Having thrown, shot, and otherwise dispensed a significant number of projectiles in my day, I can honestly say that what you described can't occur on this planet. Maybe near a couple of neutron stars orbiting each other, but I doubt it.
As Haruspex said, however, you really need to provide a lot more information regarding how you arrived at your observations.
 
  • #4
Here is a link to the photo:

http://www.flickr.com/photos/83181851@N04/

You're right, a better question to ask would be about my actual equations. Now my equations are written in such a way that a computer can understand them, not a human; I'll do my best to explain. First I'll post the raw code, then I will explain.

objvely=(objvely+(vectgrav*tstep))
objvely=objvely-(tstep*compdragcoef*sqr(objvely)/objmass)
objvelx=objvelx-(tstep*compdragcoef*sqr(objvelx)/objmass)
newx=(objvelx*tstep)+oldx
newy=(objvely*tstep)+oldy

Line 1:[new velocity of the object in the y direction]=[object's y velocity] + [acceleration due to gravity (-9.8)] * [time frame of calculation, a pre-selected fraction of a second]

Line 2:[new velocity of the object in the y direction]=[object's y velocity] - [fraction of time] * [composite coefficient of drag, gotten from the drag formula] * [square of the object's velocity as in the drag formula] / [the mass of the object, changing a force to a corresponding acceleration]

Line 3: same as Line 2, except in the x direction

Line 4:[new x coordinate] = [object's x velocity] * [time frame] + [old x coordinate]

Line 5: see Line 4

Here it is in dimensional analysis:

Line 1: m/s = m/s + (m/s^2 * s)

Line 2: m/s = m/s - (s * (kg/m) * (m/s)^2) / kg OR (s * N) / kg

Note: kg/m comes from the air density and the surface area in the drag formula: (kg/m^3)*(m^2)

Hopefully I have presented my case a little better this time. I will post the software once I improve its usability.

Thanks
 
  • #5
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You cannot calculate y and x as independent variables. The square root in the drag formula should take the total velocity into account, and multiply the drag with the corresponding vector components for x and y velocity.
This should explain the (wrong) upwards curved trajectory and probably a part of the downwards movement, too.

"nearly horizontal -> close to vertical" as general shape can happen, you can get similar shapes with table tennis balls in reality. Even the upwards curvature is possible there, but this is related to spin.
 

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