Primitive Archer
Main Discussion Area => English Warbow => Topic started by: SuperCracker on October 22, 2008, 11:44:26 am
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Hi all. You guys have made some really nice bows here and I'm quite impressed.
Have any of you had the chance to chrono your bows? I'm jut curious what kind of arrow speeds you're getting.
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Hi,
I'm not at all intereseted in arrow speed. I'm more interested in cast, and most interested in cast with heavy arrows.
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well, they would be kind of related
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Shane, I've never had the opportunity to chrono any of mine, but I've seen most guys on here post their arrow speeds between 145 and 165 (feet per second I think). It is hard to get a self bow, unbacked, any faster that 165 from what I've seen.
I can't believe nobody has rung in yet to show off their arrow speeds. I'm sure it's coming though.
~~Papa Matt
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I believe I've heard of speeds approaching 200 fps with some selfbows. I think this came up as a thread a while back. Maybe do a search. From my perspective, it's not an issue with warbows for me. This would be more in the regular bows section. Medieval archers and bowyers had no way to measure arrow speed... cast was the important thing.
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If I use the rule: 100 fps + BDW*= average speed for an arrow that is 10 grains x BDW, most of my bows are shooting below average by about 10-15%. I'm closing the gap though, and I've made at lease one bow that shoots above average. I don't have data on all my bows, though. :-\
*BDW = bow draw weight
I think most of us don't keep records of arrow speed because we build for fun and sport....not for competition.
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Jack, that rule of thumb came up at the first mojam and is confusing a lot of folks, the way it works is 100 fps plus draw weight with all bow using the same 500 grain arrow. really a pretty useless formula. Most now just use a 10 grains per pound formula which would put all bows at the same standard. I think average is about 150 fps while 175 fps would be exceptional add about 10% to those speeds for laminated all wood bows, boo backed etc. Steve
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Badger...so 150 fps would be about average assuming a 50# bow @ 28" draw?
My bows draw between 20-22", so I use a chart in TBB that shows a loss of about 3 fps for every inch below 28". Is that about right? (I don't remember too good). :-\
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Jack, that sounds about right. Steve
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Badger, I'm going to bug you with another question, if I may?
Let's assume that a bow is 48" long, has a 22" draw, pulls 45#, and is a simple self bow. What would be the average speed? ;D
I'm thinking the average would be a lot less than 150 fps, right? Maybe 120 fps? It's my understanding that short bows with short draws generally underperform compared to longer bows with longer draws (with the same draw weight).
The reason I ask is because I'm trying to develop some formulas for bow speed and ideal bow mass for short bows. I'm also working on a spine chart for short arrows (under 24")
Thanks.
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Jack I will give you a simple formula that may be useful to you.
The formula is for figuring kinetic energy in an arrow and you can easily apply it to arrive at several bits of interesting info.
The formula V x V x M / 450240 = kinetic energy
example: 140fps X 140fps X 500 grain arrow, divided by 450240 =21.76 ft#'s
lets say you had a 40# bow that was storing 33# energy. (24" draw)
33# kinetic would be 100% efficent you have 21.76# so you have an efficiency of 2176/33=66% efficient.
The way you figure your stored energy is to measure the draw weight every 1" of your power stroke add all those numbers up and then divide it by 12.
Your stored energy is reffered to as sedpf ( stored energy per draw force) typicaly on a self bow you will be around 87% of your draw weight I think. Your efficiency will also usually run between 65% and 75% using 70% to estimate speed on a self bow will get you pretty close, draw length has little effect on efficiency but does affect stored energy quite a bit. See if this helps and if it is not clear feel free to ask again. Steve
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Ahh? Badger, I think the correct formula for kinetic energy is velocity squared times 1/2 the mass or VxVx(.5m)=Ek with Ek being kinetic energy..
I checked on wikipedia to make sure.
If I am missing something here, I appologise in advance.
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Maj, the 450240 we divide it by is a constant that is used where grains are fed in for mass, it takes care of that. I am not all that big on math, this is just a standard bowyers formula we have been using. It seems to work out acurately. Steve