Thinking about this a moment... DIY's may be a better representation for the ratio of effort put into drawing the bow to energy output. His method takes into account (energy loss from vibration, arrow rest contact, loss of energy from the extra weight of string accessories like a peep, loop, etc) that does not get transferred to the arrow. SE/PDF does not take any of that into account.
I don't see how peak weak (pounds) and KE (foot-pounds) can be compared. Your first sentence of your original post says: "That is the amount of energy you get out of the bow for all the energy you have to put into it." It takes more energy (effort) to draw a 70# PSE Full Throttle than is does to draw a 70# Mathews FX. But you say the energy in it is the same, 70#. You are pulling near 70# much longer on the FT than the FX, so more energy is needed to get past the peak weight and to full draw. OK, then KE divided by peak weight would be foot-pounds/pounds. The result comes out in feet. I can follow your way if you are comparing different arrows from the exact same bow. This would give me the difference in the amount of energy the different arrows received from that bow. Or maybe 60 & 70# peak weights of the same model bow. Just tying to understand. I am easily confused.
Mathews HTR 54 Pounds - 27" draw Easton Axis arrows at 392 grains 242 fps 51 Pounds of KE If I did this correct its at 1.06 Is that good? I know every deer I've shot I blew right through it. The axis arrows hit hard.
I think where the confusion comes in to play is because they aren't comparing efficiency and energy per say. I think you are right in your understanding of the study and some words are being used incorrectly in the explanation of the formula. Technically you are right that it would require more work to draw a bow that is 70 pounds versus a 60 pound bow at the same draw length. But the words you and I may read and understand are being used in a different way above. You do put energy into the bow when you draw it and it is stored as potential energy, that is the most possible energy that the bow will have. They are substituting draw weight here in the formula. We can easy calculate the energy the arrow has when it leaves the bow. Some energy is lost in the bow. By comparing the energy out divided by the peak pounds, you can get a ratio that is fairly consisted, whether it be an efficiency or not. Personally, I agree with you. It's not an efficiency, just a ratio since the units are fubar. The ratio is still some sort of measurement of the return from the bow. If there is a ratio of more than 1, that doesn't mean that the bow is creating energy. It means the ratio that corresponds to 100% efficiency may by 2 or 3. We don't know because we can't achieve it. I think the way to fix it would have to be multiplying the peak draw weight by a length, perhaps draw length. That would allow the units to work out but the ratio would decrease due the bottom number being roughly 2.5 feet draw length, creating a much smaller but maybe more true efficiency.
Sounds pretty straight forward to me and makes every bow comparable to one another. I think people here are trying to over think it. Its just a comparison or correlation using the data from your bow as you have it set up.
I also agree that it is not efficiency, but a ratio. I also worded my responses as such. However, there is a complete formula to calculate the efficiency of a bow. I will google it as I do not recall the whole procedure.
Courtesy of TAP, the inventor of "The Archery Program". "Bow efficiency is calculated in the following manner... One must measure the entire draw force curve and figure the amount of Stored energy the bow holds. This is done by measuring each inch of draw and weight. When each data point is logged they are added up then divided by the number of data points then by 12 to get ft/pounds. Now shoot an arrow and calculate the KE of the arrow. Divide the KE by the SE and WALLA you have efficiency. Oh...and one more thing. The efficiency changes about 5% over a range of arrow weight from 5 grains/pound to 10 grains/pound and then drops off on both ends." So basically what I posted earlier. It is Kinetic Energy / Stored Energy.
But since the average bowhunter may struggle to plot the entire force curve inch by inch due to the rigors involved, draw wgt/ke is doable and simple and gives you a ratio.
It may not fit a true definition of an efficiency ratio, and maybe shouldn't be called such. When all is said and done the only two factors that matter is: what you put in, and what you got out. I like it, I think it provides a great back bone for comparison. Nitrum30- 29", 300fps, 422g, KE 84.25, 1.27 ratio result. I haven't chronographed this bow but I know its faster than my CRX32 (299fps) was as I had to close up my pin gaps when i changed my sight over. ratio could possibly be better.
Sounds pretty straight forward because it came from the same IP address.... I knew you would miraculously start posting today.
I’m having a hard time grappling with the Physics of the formulas presented, so I googled an article located here: http://archeryreport.com/2011/02/bow-efficiency-care/ The article is dated 2011, but really how much has changed in bow designs since then. Please correct me if I’m wrong, but I have a few thoughts. 1. No machine is 100 percent efficient. This is a basic conservation of energy law. 2. The cams in a compound bow do not assist with drawing the bow, but rather assist (let down) with holding at full draw. Therefore, a 70 pound pull cannot ever release more than 70 lbs of energy. Efficiency is lost to hysteresis, string hardware and inefficient arrow weights. Update: As I hunted and pecked the keyboard “Western Extreme” aired on the Outdoor Channel and the topic of discussion related to new bows with efficiencies approaching 93%. They even discussed the RPM 360 as a model of efficiency. So while the formula mentioned in this thread may be incorrect, it may actually illustrate recent improvements in bow efficiencies. The error may be fairly linear, so there may be a way to back out a percentage and arrive at a close estimate.
I hadn't heard about this before but seems a great way to compare bows. Z7 extreme, 55# pull, 445 grain arrow, 238 FPS, 55.9 KE, gives 1.01 thanks for the education
Nice reference Jeffery. Folks, I hate to beat a dead horse, but you cannot get more energy out of a bow than you put into it. The formula is incorrect.
I have not read where anyone stated that they could? His formula is a ratio of Kenitic Energy divided by Peak Draw Force (KE/PDF). Sure he labeled it wrong as efficiency, but so what. His formula is incorrect as far as efficiency, but it would still suit comparing different bows KE output based on the peak draw weight just fine. I posted the correct formula to determine bow efficiency a few posts ago.
Didn't post to bag on your formula but to give an example of what you were trying to explain its very similar.
