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I am trying to figure out the air resistance that would act on a rocket that might reach Low Earth Orbit. First off I am only just entering high school so I have an extremely limited math background. I searched around quite a bit for an air resistance formula and eventually found this-

$$Drag \, Force = \frac{c_d \cdot \rho_{air} \cdot A \cdot v^2}{2}$$

Where $c_d$ is the drag coefficient and $A$ is the surface area of the object.

I plugged in my numbers and they essentially told me the drag force would nearly double the original force. Making it clear my rocket wouldn't be able to move. This cannot be. I fiddled with the numbers and found that the only way to squeeze a logical answer out of this formula is to make the surface area the tip of my rocket, something like .0001 meters. I couldn't trust this answer, however, because it varied by several billion newtons. Am I using this formula correctly.

Next, after I decided this formula wasn't helping me I looked further until I eventually came across a lot of talk about linear at quadratic drag. I looked into but I had no idea what it all meant. Is it possible to explain linear or quadratic drag to someone with a background in nothing more than Algebra so that I can calculate the drag on my rocket. If so I would really appreciate an explanation.

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  • $\begingroup$ The density of the atmosphere varies considerably between ground level and Low Earth Orbit level. That means drag will vary with height as well as speed. If you were not able to make any headway with the internet resources, it is unlikely you will do any better here. Probably best to wait a few years until your math skills are better, otherwise you are not going to understand what you are doing. $\endgroup$
    – sammy gerbil
    Aug 29, 2016 at 17:26
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    $\begingroup$ In addition to what sammy gerbil says, I suspect you are misinterpreting the surface area. The choice of area (total surface versus frontal cross section) depends on what drag coefficient you use. The drag coefficient is experimental, and specific to your individual rocket. Also, if the rocket isn't moving, there is no drag, so the rocket always moves first. The drag force would simply prevent it from moving faster. Bottom line: you have to actually fly the rocket to know what the drag coefficient is. $\endgroup$
    – Bill N
    Aug 29, 2016 at 19:27
  • $\begingroup$ Is there anyway to, relatively simply, determine the Drag Coefficient without actually putting my rocket through a wind tunnel. My rocket is purely theoretical. $\endgroup$
    – max
    Aug 29, 2016 at 21:14

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Is it possible to explain linear or quadratic drag to someone with a background in nothing more than Algebra so that I can calculate the drag on my rocket. If so I would really appreciate an explanation.

Sure. Consider a rocket which at a speed of 30 m/s has a drag force of 50 N. If the drag force is linear (as a function of speed), then at 40 m/s the drag force will be $$\left(\frac{40}{30}\right)^1 50\;\mathrm{N} = 66.7\;\mathrm{N}.$$ The first power next to the ratio indicates a linear relationship of velocity to drag.

If the drag is quadratically related to the speed, then the power will be a square (2) and the drag at 40 m/s would be calculated as $$\left(\frac{40}{30}\right)^2 50\;\mathrm{N} = 88.9\;\mathrm{N}.$$

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  • $\begingroup$ Ok, but I would have to calculate the drag force at, say 10 m/s, to increase it either linearly of quadratically, and to that I would have to use the formula I had in my question, correct? $\endgroup$
    – max
    Aug 29, 2016 at 21:09
  • $\begingroup$ Also how do I know whether my rocket has linear or quadratic drag, based on its acceleration? $\endgroup$
    – max
    Aug 29, 2016 at 21:11
  • $\begingroup$ Drag force is something which must be determined experimentally. From the experimental data you get both the values and the behavior (linear, quadratric, transitional from one to the other, or a combination -- the fins may be linear while the body is quadratic, or vice versa). One can use experiments that others have done to estimate your situation. $\endgroup$
    – Bill N
    Aug 31, 2016 at 14:36

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