In the last post I described the draw force curves of compound and traditional bows. The compound bow was able to store more potential energy because the archer pulled the bow string for a longer distance at full draw weight. More stored energy in the bow resulted in more kinetic energy for the arrow and faster arrow velocity at the point of nock release. But what is happening to the arrow during the time from string release to nock release? Clearly the arrow starts with a velocity of zero and reaches full velocity as the arrow leaves the bow. The velocity profile during this cycle will determine the forces on the arrow and arrow rest, launch time, and ultimately the size of our arrow groups. We will begin with the draw curve from the last post. This figures shows the bow draw force as a function of draw distance for both compound and traditional bows. The x axis is distance from full draw, so during launch the arrow travels from left to right (0 to 19.5 inches in my example) In order for the arrow to change velocity, speed, it must accelerate. We know from Newton's equations of motion that force is equal to mass times acceleration. Therefore, if we know the arrow mass, we know the arrow acceleration at any point in the arrow flight. The arrow starts with a velocity of zero and increases speed until the arrow releases from the bow string. To calculate the increase in velocity with time, I started with the basic equation of motion relating the position of an object at time t (Xt), to the starting position (Xo), Velocity (V), acceleration (a), and time (t). To make my calculations simpler, I divide up the motion of the arrow and bow string into 1/2" intervals, and reorganized the equation in terms of delta X = 0.5". I solve this simple quadratic equation for time, with the traditional A, B, and C coefficients equal to: A = 1/2a, B = V, C = - (delta X) For my compound bow the solution for the first 0.5 inches is t= 0.004 s (4 milliseconds, ms) so it takes 4 ms to travel 0.5 inches. The velocity at the end of 4 ms is 21 ft/s. Now I repeat the calculation for the next 0.5 inch step using the new velocity and the acceleration computed from the bow draw curve. This process is repeated forty times to simulate the entire travel of the arrow in the bow. Doing the calculations in a spreadsheet like Excel makes quick work of the mathematics resulting in plots of arrow acceleration or arrow velocity versus time. (see attached Excel file: Arrow Speed and Acceleration.xlsx) The total time for an arrow to leave my compound bow is 16 milliseconds and it reaches a final velocity of 274 ft/s. This agrees with our computed arrow speed based on the stored energy in the bow, but now we can understand how long it takes the arrow to accelerate. The acceleration of the arrow starts off low due to the compound bow let off and then reaches a maximum value at 13 ms. In contrast, the traditional bow has the greatest draw weight and acceleration at full draw so the arrow accelerates quickly and leaves the bow after only 12 ms. The arrow launched from a traditional bow is in the air before the compound bow has reached peak acceleration. The final 274 ft/s velocity of the compound bow is achieved due to peak in acceleration from 12 to 15 ms, just before the arrow release. These velocity curves have interesting implications for arrow contact with bow rests and the susceptibility of the bow to move during launch. For these two examples, the compound bow is in contact with the arrow 25% longer than the conventional bow. For close targets, the arrow released from the "slower" traditional bow will get there first, but in reality the arrow from the faster compound bow will pass the arrow from the traditional bow after only five feet of flight and 19 ms of flight time. Slow and steady will not win this race! The next post will work through the details of calculating arrow trajectories from known arrow speeds and launch angles. If you want to understand how a compound bow works I recommend the post A Gear Head's Analysis of Compound Bow Function. The average velocity for each 0.5" distance step is simply the distances traveled, 0.5, divided by the computed time. This is the average velocity over the interval, not the final velocity at the end of the interval. Using a manipulation of the mean value theorem, it is possible to compute a better estimate for the velocity:Technical note for those that like the details: |