Adding curvature to accelerator path to change exit angle

G. Flanagan

Alna Space Program

July 13, 2011

Examine the path for the case of an accelerator that starts off horizontal, but then transitions to a curve so that the exit angle is at a specified value. The curve adds a normal acceleration component to the projectile, which is limited to a set value.

The formulation for the accelerator path is much easier for a constant projectile longitudinal velocity, so we’ll do this deviation first as a check of the steps and to get a clearer idea of the important parameters. The derivations will take advantage of small-angle assumptions.

For small angles, we use the relation that the local radius of curvature of the accelerator is approximately equal to , where w is the height of the path. Then, the radial acceleration is equal to , where v is the instantaneous velocity. If the normal acceleration limit is ay, then we can solve for the radius of curvature, and thus get a function for w with respect to x. x in this case runs from the start of the curved section of the accelerator, and goes to the exit.

At the beginning of the curved section, w=0, and the slope is equal to zero. This determines the constants. The function for the track height is

The slope at the exit.

By setting the slope equal to the specified value, exitAngle, we can solve for the length of the curved section.

The height at the exit is given by

Construct a function that returns the length of the curved section and the maximum height of the accelerator. I’ve include ax (the longitudinal acceleration in the argument list so that the list is identical to another function we’ll construct for an accelerating projectile.

Repeat the derivation from above, only change vexit to a variable velocity with a steady acceleration. The velocity as a function of x will be Sqrt[2 ax x]+v1, where v1 is the velocity at the beginning of the curved section. ax is the longitudinal acceleration of the projectile.

This time, we can’t trivially set the constants equal to zero. Also, the slope is indeterminate at x=0, so we have to take a limit. v1 can be determined if one knows the length of the curved section, xcurve. The function for the height is then

Attempt some additional simplifications.

And a stand-along function that simply returns the height as a function of the position x. Note that the fw expression will return a complex result for some values of ax. I crudely have taken the real part without further analysis. The problem is related to the complex branches chosen by the Log and ArcTanh functions. I suspect that by altering the default branches, the complex part will cancel, but I haven’t performed the analysis to verify.

The slope at the exit. Given the desired exit angle, and using a small angle assumption, we can solve for xcurve. However, the equation does not have a closed form answer, so a numerical method must be used.

Create a function by copying the two results and using FindRoot to determine xcurve.

The variable and constant functions should yield similar results if the longitudinal acceleration is small.

Height of exit as a function of the allowed normal acceleration. 10km/sec exit velocity

Look at height for a more modest exit velocity.

Consider an accelerator with a 200g normal acceleration limit. The curve section length is.

Graphic of curved section profile with 1:1 aspect ratio.

Total length of the accelerator.

Length of the straight portion.

Graphic of the entire accelerator. Green line simply gives a reference horizontal.

Consider an accelerator with a 100g normal acceleration limit. The curve section length is.

Total length of the accelerator.

Length of the straight portion.

Graphic of the entire accelerator. Green line simply gives a reference horizontal.

Consider an accelerator with a 200g normal acceleration limit. The curve section length is.

Total length of the accelerator.

Length of the straight portion.

Graphic of the entire accelerator. Green line simply gives a reference horizontal.

It’s going to hard to find terrain that would support the curve of a 20° exit angle.

If one is willing to design for high normal acceleration (on the order of 100-200 g’s), then the height change of the accelerator would be compatible with modest mountains with a lot of earth moving or tunneling. The high normal acceleration complicates the problems of supporting the projectile within the tube. For a straight gun, the sliding supports (or magnetic levitation system) only needs to provide for guidance and stability. For the curved system, there are considerable loads to be considered. There is an engineering trade-off between the complexity of using lift, versus the additional construction cost of adding a curve.

Although the calculated curves may be compatible with terrain, I don’t think it would be practical to build vertical structure to support the curve. The exit point is nearly the same height at the world trade center. That alone might be feasible, but building a series of supports over a 4.5 km path seems too ambitious.

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