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/
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