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Ballista Equation

Date: 09/30/2002 at 22:46:10
From: Jessi
Subject: Geometry/Physics

Dear Dr. Math,

I'm writing a report about the ballista, an ancient weapon in the form 
of a giant crossbow, which fires bolts and grapeshot; see

   The Ballista
   http://members.lycos.nl/onager/ballista.html 

and I have to create an equation for how ballistas would accurately 
hit their targets. I have figured that since they used giant crossbow 
bolts, the process to solve for arc length would help find the 
distance between the weapon and its target, but I do not know how to 
add the velocity of the bolt and how many degrees they would have to 
adjust it up and down to hit its target. Any help would be nice.

Sincerely,
Jessi


Date: 10/01/2002 at 11:20:32
From: Doctor Ian
Subject: Re: Geometry/Physics

Hi Jessi,

Here's something that you can read to get you started thinking 
about the issues involved:

   Real Life Uses of Quadratic Equations
   http://mathforum.org/library/drmath/view/60810.html 

Note that it talks about holdover (aiming for a point a certain 
distance above the target) rather than angular correction.  

Let's say that a projectile leaves with an initial speed of V[i], at 
an angle (relative to the horizontal) of A degrees. For simplicity, 
let's also ignore air resistance, because once you start correcting 
for that, the real-world solution is to start collecting tables of 
information and using those to interpolate to nearby cases. 

(This is, in fact, what long-distance rifle shooters do. They have one 
table that compiles expected bullet drops at various distances, and 
another that compiles expected deflections for various wind speeds and 
distances. These are compiled empirically, rather than computed, 
because they need to account for things like bullet shape and 
rotation.)

Anyway, the motion can be decomposed into two separate motions, 
occurring simultaneously. One of the motions is horizontal, with a 
constant speed of 

  V[i]cos(A)

The other speed is vertical, with a speed given by 

  V[i]sin(A) - gt

where g is about 32 feet per second per second. The height of the 
projectile above the launch device is given by 

  h = V[i]sin(A)t - (1/2)gt^2

Let's say that the target and the launch device are at the same 
height. The projectile will rise for a while, and then fall back to 
the same height when h is zero:

        0 = V[i]sin(A)t - (1/2)gt^2

          = V[i]sin(A) - (1/2)gt

  (1/2)gt = V[i]sin(A) 
 
        t = 2V[i]sin(A)/g

This tells you the time of flight for the projectile, as a function of 
the initial velocity and angle. How far does it go during that time?  
The horizontal distance is given by

    d = V[i]cos(A)t

(This is just the old familar distance-equals-rate-times-time.)

We can solve this for t to get

    t = d/(V[i]cos(A))

And now we have two things that are equal to the same thing (t), so 
they must be equal to each other:

  2V[i]sin(A)/g = d/(V[i]cos(A))

Solving for d, we get

                  2V[i]sin(A)V[i]cos(A)
              d = ---------------------
                           g

                  2V[i]^2sin(A)cos(A)
                = ---------------------
                           g

Does that look reasonable?  Let's do some simple checks.

  1. Suppose the angle is 90 degrees. Then sin(A) is 
     always zero, so d is always zero. So you can't 
     shoot anything forward by shooting straight up.
     That makes sense.

  2. Suppose the angle is 0 degrees. Then cos(A) is 
     always zero, so d is always zero. This makes sense,
     since we assume that the launcher and target are
     at the same height; and the projectile will start
     dropping immediately under the influence of gravity,
     so it can never hit the target. 

  3. Suppose we take the derivative of d with respect to
     A (leaving V constant).  We get

       d/dA d = 2V[i]^2/g  d/dA sin(A)cos(A)

     Now, 

       d/dA sin(A)cos(A) = sin(A)*-sin(A) + cos(A)cos(A)

                         = cos^2(A) - sin^2(A)

     which is zero when sin(A) = cos(A), or A = 45 degrees.  
     So regardless of initial velocity, an angle of 45 degrees
     should give us the maximum possible distance. 

So you should probably go through the derivation again on your own to 
make sure that this formula is correct, but it seems pretty 
reasonable. 

Does this help? 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Basic Algebra
High School Trigonometry

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