# Hit a roadblock trying to calculate minimal velocity.

by Galorian
Tags: minimal, roadblock, velocity
 P: 9 Does anyone know how I can calculate the minimal initial velocity required to get to a certain altitude for a given object? I'm assuming an earthlike planet and a streamlined projectile with a maximum diameter of 8mm and a mass of roughly 40 grams and want it to reach an altitude of 55560 meters at the apex of its trajectory. If possible I would like a parametrized solution so that I could toy around with the variables (particularly air density and/or molar mass, and possibly projectile velocity at the desired altitude), though this isn't a must if it complicates matters overmuch. Personally I've reached a differential equation that even Wolfarm Alpha can't seem to solve, hence this thread...
 PF Gold P: 946 Welcome to PF! Perhaps you can present some more of the background for problem you are trying solve and any constraints that you've set for it? If nothing else, showing the work you've done so far, up to the point where you are stuck, will allow people here a much better opportunity to provide you with some relevant help. Modelling realistic ballistic trajectories (complete with aerodynamic interaction and all) is going to give rise to complicated solutions that usually are easiest to solve with numerical methods. In fact there most likely wont even be a closed solution, making numerical methods the only way.
P: 9
 Quote by Filip Larsen Welcome to PF! Perhaps you can present some more of the background for problem you are trying solve and any constraints that you've set for it? If nothing else, showing the work you've done so far, up to the point where you are stuck, will allow people here a much better opportunity to provide you with some relevant help. Modelling realistic ballistic trajectories (complete with aerodynamic interaction and all) is going to give rise to complicated solutions that usually are easiest to solve with numerical methods. In fact there most likely wont even be a closed solution, making numerical methods the only way.
Thanks!

Basically it's an 8mm depleted uranium spike who's mass is somewhere between 35.5 to 50.8 grams fired directly upwards and hitting a target that's 10 leagues (55.56km) up above the shooter. The incident accrued on a settled desert planet with breathable atmosphere and no mention of particularly high or low gravity, so I'm assuming earthlike conditions with little to no humidity and manageable temperatures.

The atmosphere of the planet was never mentioned to be out of the ordinary in terms of pressure and was sufficient to support a rather impressive atmospheric phenomenon (a stationary vortex 20 leagues tall and 4 leagues wide), so I don't think it's particularly thin and for the sake of a "low end" result I'm assuming earthlike atmospheric pressure.

P: 9

## Hit a roadblock trying to calculate minimal velocity.

Damn, almost forgot!

Here's the equation I ended up with that Wolfarm couldn't solve (at least not on trial version computing time limits):

y''=((A*(e^(-B*y)))*(C-x*y'')^2)-g

x being time, y being distance, A being AerodinamicCoefficient*CrossectionSurfaceArea*Pressure*AirMolarMass/(2*R*Temperature*ProjectileMass), B being AirMolarMass*g/(Kb*Temperature), g being the gravity acceleration and C being the initial velocity (which is what I'm trying to find).

I'm interested in getting the minimal initial velocity required for the feat and prefer that the uncertainty within the variables be erred on the side of caution (basically taking earthlike variable ranges and picking the values that'll give a lower result for initial velocity), thus making this calculation easier to defend as giving a reasonable lower estimate for the feat.

I'm far from infallible and the equations were derived from drag equations found in Wikipedia (air pressure I calculated using a formula we learned in classical physics), so feel free to point out any mistakes.
 PF Gold P: 946 You should not use v0 - a*t as an expression for speed, use y' instead. A vertical upwards, drag-only (i.e. no lift) trajectory in height $h$ can be modeled in a simple way using an exponential atmosphere, like you have done, and constant gravity, which gives $$\ddot{h} = - \frac{\rho_0 e^{-h/H}}{2\beta} \dot{h}^2 - g$$ with $h(0) = 0$ and $\dot{h}(0) = v_0$, where $g$ is the (constant) acceleration of gravity at the surface, $\rho_0$ is the air density at surface, H is the scale height (ref [1] uses H=6700 m for Earths atmosphere), and $\beta$ is the ballistic coefficient given by $$\beta = \frac{m}{C_D S}$$ with $C_D$ being the drag coefficient, $m$ the mass, and $S$ is the surface area. Note the unit of $\beta$ as defined above is mass per square-length, e.g. kg/m2 or lb/ft2. There are other definitions of $\beta$ that also include g where the unit is pressure, like N/m2. You may also want to search some references (like [2]) for a "good" value of the ballistic coefficient for your artillery piece. I believe that WolframAlpha is able to solve the above differential equation numerically, but it may require a "pro" subscription to do so. I was able to get a trajectory using my paid Android WolframAlpha app, but their normal website "ran out of time". Otherwise it should be fairly easy to solve it using some other ODE library or math tool. [1] Dynamics of Atmospheric Re-Entry, Frank J. Regan, AIAA 1993 [2] http://en.wikipedia.org/wiki/Ballistic_coefficient
 PF Gold P: 946 I forgot to mention, that in ref [1] above, Regan derives an approximate analytical solution to the "opposite" problem of a vertical downwards drag-only trajectory (i.e. the same equation as above, but with opposite sign for the gravity term) using Taylor expansion and shows a small program that tabulates speed as a function of height using the ten first terms of the expansion. If a similar derivation can be made for the "upward problem" you may be able to replace solving an ODE numerical with a "handful" of evaluations of a slightly more complex expression. Another thing to note is that Regan consistently use $\rho_0$ = 1.725 kg/m3, and not 1.225 kg/m3, in order to get a better fit of the exponential air density model over the height range 5 to 40 km relevant for a re-entry problem. Since you want to model an upward trajectory that passes 0 to 5 km with the largest speed, you may want to use a value somewhere between those two values. I guess that it should probably be closer to 1.225 kg/m3 than 1.725 kg/m3 since it is the lowest parts of the atmosphere that will have the most effect on your trajectory.
 P: 9 Thanks! I've been trying to get a solution out of Wolfarm and it directed me towards Mathematica. I got the trial version and I can't seem to get it to work (not surprising, given I've never used it before and I kind of went into the deep end with this one...). This is the equation I want it to solve- h''[t] = (-(1.1455*e^(-h[t]/6700))*(h'[t]^2)/1916.44) - 9.8 h[0]=0 and max h[t] = 55560. The first condition is fairly straightforward, but I'm not sure how to work the second one into the code for the program nor do I know how to tell the it to return the value of h'[0] as the solution... I've tried to go over a few tutorials but it's taking way too much time and I've got a test coming later this week which so I can't afford the time it'll take to learn this program at the moment... Any help would be appreciated.
P: 9
 Quote by Filip Larsen I forgot to mention, that in ref [1] above, Regan derives an approximate analytical solution to the "opposite" problem of a vertical downwards drag-only trajectory (i.e. the same equation as above, but with opposite sign for the gravity term) using Taylor expansion and shows a small program that tabulates speed as a function of height using the ten first terms of the expansion. If a similar derivation can be made for the "upward problem" you may be able to replace solving an ODE numerical with a "handful" of evaluations of a slightly more complex expression. Another thing to note is that Regan consistently use $\rho_0$ = 1.725 kg/m3, and not 1.225 kg/m3, in order to get a better fit of the exponential air density model over the height range 5 to 40 km relevant for a re-entry problem. Since you want to model an upward trajectory that passes 0 to 5 km with the largest speed, you may want to use a value somewhere between those two values. I guess that it should probably be closer to 1.225 kg/m3 than 1.725 kg/m3 since it is the lowest parts of the atmosphere that will have the most effect on your trajectory.
Didn't refresh the page for a while so you kind of ninja'd me :P
 PF Gold P: 946 I'm no Mathematica expert, but you should be able to solve it using NDSolve. Plugging the equations straight in (with a small change to make sure the drag force is always opposite the direction of movement so that the equation stay valid also when v turns negative), I get that the following inputs produces an altitude plot with a max height around 55 km: eqs := {h''[t] == -(\[Rho]/(2 \[Beta])) Exp[-h[t]/H] h'[t] Abs[h'[t]] - g, h[0] == 0, h'[0] == v} params := {\[Rho] -> 1.225, \[Beta] -> 3000.0, H -> 6700.0, g -> 9.8, v -> 3700} s = NDSolve[eqs /. params, h, {t, 0, 150}] Plot[Evaluate[{h[t], h'[t]} /. s], {t, 0, 150}, PlotStyle -> Automatic] This gives the following plot: As it is, you have to solve for each set of parameter values you want to try out. I have given it a few runs and ended up with an initial speed of 3700 m/s and a ballistic coefficient calculated from the values you gave in the first post using a drag coefficient of 0.25. I have no idea if a value of 3000 kg/m2 is too unrealistic for such a small projectile or not, though I suspect that is a bit high. Someone with better Mathematica skills could probably make Mathematica find the initial speed that result in a given maximum height without the need for any manual bisection of values and repeated runs. Attached Thumbnails
 P: 9 Thanks a bunch!

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