LUNAR’clips 2002                        Volume 9, Number 2

Copyright © 2002 by LUNAR, All rights reserved.

How Far Can Your Lawn Dart Go?

An errant rocket led to the question of how to calculate the maximum "lawn dart" distance for a rocket, and here are insights into the problem:

I include a purely physical explanation, no equations, below all the math.

Horizontal distance d(x) is horizontal velocity v(x) * time of flight t(f).
For a vertical launch achieving a final velocity v(f), v(x) = 0, and t(f) = 2 * v(f) / a.
So d(x) = 2 * v(x) * v(f) / a. Since v(x) is 0, so is the whole term.

For a launch at 45 degrees, both v(x) and v(y) are v(f) * sqrt(2) / 2 (Pythagorean theorem).
We refine the formula for time of flight to reflect that we're not going straight up, so

t(f) = 2 * v(f) / a becomes t(f) = 2 * v(y) / a
So d(x) = v(x) * t(f)
= v(x) * 2 * v(y) / a
= (v(f) * sqrt(2) / 2) * 2 * (v(f) * sqrt(2) / 2) / a
= v(f) * v(f) * sqrt(2) * sqrt(2) * 2 / 2 * 2 * a
= v(f) * v(f) * 2 * 2 / 2 * 2 * a
= v(f) * v(f) / a.

To get the fraction you're talking about, we need to find the apogee for a vertical flight, d(y), and compare it to d(x). Easy. Only difference between apogee and downrange is that we average the velocity to get apogee, because we're slowing to a stop. We also have to cut the time of flight in half, because only the first half of the flight contributes to the apogee.

So d(y) = (v(y) / 2) * (t(f) / 2)
= (v(f) * sqrt(2) / 2) * (t(f) / 2)
= (v(f) * sqrt(2) / 2) * ((2 * v(y) / a) / 2)
= (v(f) * sqrt(2) / 2) * ((2 * (v(f) * sqrt(2) / 2) / a) / 2)
= v(f) * v(f) * sqrt(2) * sqrt(2) * 2 / 2 * 2 * 2 * a
= v(f) * v(f) * 2 * 2 / 2 * 2 * 2 * a
= v(f) * v(f) / 2 * a

The fraction is then (v(f) * v(f) / a) / (v(f) * v(f) / 2 * a)
= (v(f) * v(f) / a) * (2 * a / v(f) * v(f))
= (v(f) * v(f) / v(f) * v(f)) * (a / a) * 2
= 1 * 1 * 2
= 2.

So for a purely ballistic trajectory, your maximum possible distance - lawn dart radius, if you will - is exactly twice your apogee. This was a real nuisance trying to figure out safety for a CATS Prize shot, because any shot to 200 km had a lawn dart radius of 400 km, which from Black Rock included such sparsely populated areas as Boise, Reno, Eugene, Stockton, Modesto...

Without the equations, it goes like this. Say you leave the pad at 100 m/sec. You'll coast up for 10 seconds (I use 10 m/sec^2 for gravity instead of 9.8 because it makes the math so much easier), and fall back for another 10. That's 20 seconds time of flight. Your average velocity on the way up is 50 m/sec. Your time on the way up is 10 seconds. So, you went 500 meters. That's your apogee.

Now tilt the rail 45 degrees. Buy off the RSO and LCO, they're not easy but they're cheap. You still leave the rail at 100 m/sec, but now you have to divide it into horizontal and vertical components. Old Man Pythagoras says they're both the same, your total velocity divided by root 2. So you get about 70 m/sec each way. Pretty cool, how half of 100 is 70. Your time of flight goes down, it's only 14 seconds now, 7 seconds up, 7 seconds down. Average vertical velocity is now 35 m/sec, so your apogee is now...35 * 7 = 245 meters. Back to half of 500 is 250, Nature always makes you pay it back. But your average horizontal velocity is the same as your initial horizontal velocity, 70 m/sec, and you're going downrange both on the way up and on the way down, so you use the whole 14 seconds. 70 * 14 = 980. You go 980 meters downrange and scare the spit out of a skunk. Leave the rocket there, you don't want it back.

The 980 being less than 1000, as twice the apogee would be, is due to me rounding root 2 to 10/7. That's about 1% off; the error gets doubled because the rocket comes back down.

If it is not ballistic, i.e. there is wind drag, then the maximum distance angle can be higher. Increasing your time of flight can buy you more than increasing your horizontal velocity. In the artillery, our max range angle was about 47 degrees. Artillery shells aren't very draggy for their weight; they don't slow down much in flight.

If you include burn time in the calculation, instead of using an instantaneous velocity, your max angle could be down around 30 degrees, for a long burn high drag rocket.

The maximum distance reachable based on vertical trajectory apogee would be the one from 45 degrees; double the apogee. Add a fudge factor for very low T/W rockets that have high gravity losses. Better yet, don't fly very low T/W rockets.

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