1. The problem statement, all variables and given/known data I am trying to calculate the exit velocity of an arrow from a bow. I am an archer and this is for my own personal knowledge so I need to keep the units meaningful in lbs, sec, mph, inches. The end velocity must me in mph. 2. Relevant equations The equation that I have found that best applies is: v = (eFx/m)^-2 v = The exit velocity. e = The efficiency (all force applied in the same direction). We can assume it will be 70% for most medieval bows. F= The force to draw the bow fully, commonly known as draw-weight. m= Mass of the arrow. x= This is the distance that the string will travel from its rest position to its fully drawn position. 3. The attempt at a solution e= 70 % f = 100 lbs m= .5 lbs x = 30 inches v= (.7x100 lbx30 inch /.5 inch) ^-2 3000 inch lbs / .5 inch .7 x6000 lbs 4200 lbs ^-2 5.66 lbs sq I don't understand what 5.66 lbs sq means.
Ok, in this post, I'll address your *method*, and in the next post, the calculation itself. The formula you gave was [tex] v = \left(\frac{eFx}{m}\right)^{-2} [/tex] I don't believe that this formula is correct, because the right-hand side doesn't have dimensions of velocity. I think that the exponent should be 1/2, not -2. Here's an outline of the basic physics: An object like a spring or a rubber band that has elastic properties can store elastic potential energy. Say it has some equilibrium length. If you try to stretch it by an amount x beyond it's equilibrium length, there is a law called Hooke's law that states that the spring or whatever will oppose being stretched with a restoring Force F = -kx. It's called a restoring force because it acts to oppose your stretching and to try to restore the spring back to it's equilibrium condition. You can see from the equation that the larger k is, the more force it takes to stretch the spring by the same amount. For this reason, k expresses the "stiffness" of the spring and is often called the spring constant. When you do work against this restoring force to stretch the spring, energy equivalent to that amount of work is stored in the spring as elastic potential energy. This elastic potential energy is given by: [tex] E_p = \frac{1}{2}kx^2 [/tex] When you let go of the spring (or in our case, the bow string), all of the energy that was stored in the string as elastic potential energy is converted into kinetic energy (energy of motion) and imparted to the arrow. The formula for kinetic energy is [tex] E_k = \frac{1}{2} mv^2 [/tex] Since the energy is just converted from one form to another, we have: [tex] E_p = E_k [/tex] [tex] kx^2 = mv^2 [/tex] [tex] v^2 = \frac{kx^2}{m} = \frac{(kx)x}{m} = \frac{Fx}{m} [/tex] [tex] v = \sqrt{\frac{Fx}{m}} = \left(\frac{Fx}{m}\right)^{\frac{1}{2}} [/tex] The extra factor of 'e' for the efficiency is to take into account the fact that the bow is not 100% efficient: not ALL of the elastic potential energy is converted into kinetic energy of the arrow. Also, the amount x you are using is not the amount by which the bow string is stretched. It is just the amount by which it is drawn back. I'm confident that this formula is correct. By dimensional analysis, the stuff in the square root sign has units of force x distance / mass = energy/mass = velocity squared.
Now that that's out of the way, let's have a look at the actual *calculation*. Your problem (over and above using the wrong formula) is that mass and force are two different physical quantities. As a result, you can't use the same units to measure them. This is where the inherent crappiness of the imperial system of units comes in. I think it's horrendous and wouldn't touch it with a ten-foot pole (yes, I see the irony) unless I absolutely had to. It's my understanding that the pound has always been a measure of force (or weight). The imperial unit for mass used to be the slug. However, I *think* that at some point, a unit called the pound mass (lb m) was defined to be the mass that an object would have to have in order to weigh one pound on Earth. Therefore, the unit of force is called the pound force (lb f) in order to make the distinction explicit. Now, if the lbf is really the weight of a 1 lbm mass on Earth, then their ratio should be the acceleration due to gravity, since F = ma. In imperial units, the acceleration due to gravity is (I had to look it up): 32.174 ft/s^2 I don't know how realistic your numbers are, but let's use them: F = 100 lbf m = 0.5 lbm x = (30/12) ft v = [(100 lbf)(30/12 ft) / (0.5 lbm)]^1/2 = [ (200 lbf/lbm)(30/12 ft) ]^(1/2) = [ 16 087 ft^{2}/s^{2} ]^(1/2) = 126.8 ft/s My trusty mac dashboard unit conversion widget says that this is equal to 86.48 mph. It seems too fast to me. Edit: I forgot the factor of e=0.7, but oh well.
I have to take a while to digest this. Thank you very much. The final answer should be somewhere in that area for 300 ft/s per radar detector. How does that convert? The numbers I provided as far as draw weight and drawlenght are realistic for a medieval bow. It should provide enough energy to pierce light armor at close range.
My assumptions about the true definition of the pound may be wrong. Again, I don't know much about this system of units and don't care to for the time being. Just as a sanity check, let me use the unit conversion widget to redo this problem in SI units: F = 100 lbf = 444.822162 N (newtons) x = 30 in = 0.762 m m = 0.5 lbm = 0.22679618 kg v = 38.65 m/s The trusty unit conversion widget says that this is equal to 86.48 mph (same answer). I think you need to take another look at your numbers. 100 lbs is a lot of force. Edit: Ok cool. I was still writing this when you posted your response.
A typical modern hunting bow has a 70 lb draw weight. The bows found on the Mary Rose from the 15th century have been determined to have draw-weights ranging from 100 lbs to 180 lbs. To determine the draw-weight a frame is used to hold the bow horizontal and weights are hung from the spring until the the string reaches a desired drawlength. When the sting reaches the desired length, the bower looks and sees how much weight he had to put on to get it there. I can draw a 100 lb and shot with some measure of accuracy. I wanted to explain how draw-weight is determined so we are on the same page. Can you explain why you are dividing the 30 inches by 12. Not clear on that. I am guessing you are relating it to the lb/ft draw-weight measure? The other thing I can't figure out is how the unit of seconds gets into the final velocity number. Could that efficiency number have some unit of time within it? Using the following formula: [tex]v = \sqrt{\frac{Fx}{m}}[/tex] 100 x30/.5= 6000 sqrt 6000= 77 What is this 77? Sorry, if I am going in circles.
I'm expressing it in feet. Sorry, I didn't show this explicitly. Again, it has to do with the relationship between pounds force and pounds mass: So the ratio in bold below becomes 32.174 ft/s^{2} No, it is dimensionless. The explanation for where the seconds come from is above. Well, it's the velocity, although the number is not correct, because you haven't included the pounds force/pounds mass conversion factor as discussed above.
Ok its starting to become a little more clear. The only part I need to understand now is (200 lbf/lbm). How do does the 100 lb become 200 lbf? I can see that some how it is multiplied by 2, but what is the reason for that? Sorry if you explained it already, but it is not clear to me. Thanks.
I'm still chewing on this :) Forgive me because I have been out of college for over 15 years already, I am trying to understand the following calculation: 200 x 2.5 =500 I don't see how you get 16,087.
I explained this in post #7...again, it's the conversion factor for units of mass to units of force (both unfortunately referred to as 'pounds'). 500 ft x 32.174 ft/s^{2} = 16 087 ft^{2}/s^{2}