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Benjamin H. Abbott




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PostPosted: Thu 08 Oct, 2015 1:13 am    Post subject:         Reply with quote

Certainly some of the Chinese-region military exams have heavy bows that were drawn only to demonstrate strength and not to actually shoot. I believe at the least the later Qing exams were like this. On the other hand, at least one much earlier exam required over thirty accurate shots at the test draw weight.
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Lafayette C Curtis




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PostPosted: Thu 08 Oct, 2015 1:55 am    Post subject:         Reply with quote

Other than the article in that link (which has a footnote to a contemporary French account of the examination procedures), there's also another article (also by Peter Dekker) with eyewitness testimony that later practitioners of Manchu archery continued to do strength training with heavier bows than the ones they really shot with: http://www.manchuarchery.org/interview-last-manchu-archer
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Timo Nieminen




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PostPosted: Sat 10 Oct, 2015 8:37 pm    Post subject:         Reply with quote

Dan Howard wrote:
Lafayette C Curtis wrote:
We should be really, really careful about taking the draw-weights cited in Chinese military examinations at face value. There's a fair amount of evidence that these examinations were done with special "strength bows" that weren't meant for shooting, and also that they were performed with a different draw (completely grasping the string in the fist) that would have been impractical for actual shooting. The result is that archers could handle a greater draw weight with the examination bow than they could actually handle in everyday military service, and the draw-weights of the bows they really shot with would have been somewhat lower than the strength bow they qualified with.

I'd like to read more about this. Do you have any cites?


For details of the archery sections of the examinations, the best source is Stephen Selby, Chinese Archery.

Some of the examination rules would have encouraged using stronger bows for the examination. For the mounted archery section of the Song examinations, the minimum draw weight allowed was 103lb, but shots with a 133lb bow would score 1.6 times more points (assuming the same accuracy).

The Qing examinations included drawing very strong bows. The standard was 156lb. These were drawn, without an arrow, holding the string in the fist. This was part of the strength test, not part of the archery test. Stronger and weaker bows were available, for extra points or fewer.

Korean examinations included stronger-than-normal bows (iirc, these were actually used to shoot at targets). I don't recall seeing figures for the draw weights.

"In addition to being efficient, all pole arms were quite nice to look at." - Cherney Berg, A hideous history of weapons, Collier 1963.
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Eirik R. F.




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PostPosted: Sun 08 Nov, 2015 2:02 pm    Post subject:         Reply with quote

Hello everyone. My first post. Let's see if I can contribute with something of interest.

I'm interested in the exact same questions as you, Cs. Norbert.

You talked about Matheus Bane's test and asked about -(kinetic energies) k = .04

Ok, let's take a look at it.

If we take something like Joe Gibbs and his world record we have the following information:

velocity, Vi = 64 m/s
bow weight, f = 175 lbs = 778,438 N
draw length, x = 32” = .8128m
efficiency, e = 0,7
mass arrow, m = 0,0637 kg
mass bow, mb = 1,43 kg
kinetic energies, k = 0,03125

I believethe way to calculate k is done as follows:
p/KE
or
(m*v/(0,5*m*v^2))

KE = 0,5*m*v^2
KE = 0,5 * 0,0637*64*64 = 130,457 Joule

p = 0,637*64 = 4,0768 Ns

k = p/KE
k = 4,0768/130,457 = 0,03125014372551875330568693132603

If we run the formula for velocity we get really close to the known speed.

Vi = sqrt(e * f * x / (m + k * mb))
Vi = sqrt(0.7 * 778,438 * 0.8128 / (0.0637 + 0.03125 * 1,43)) = 63,92 m/s

Now we can turn this around to find the weight of the bow if we want to.

63,92^2 = 4086,264

0.7 * 778,438 * 0.8128 = 442,9

442,9/4086,264 = 0.1083875

0,1083875 - 0,0637 = 0,0446875

0,0446875/0,3125 = 1,43

I want to do this with Mark Stretton's bow he used in one of his test because I don't know the physical weight of his bow. (Look at the bottom of this post). In the book he told us it was not his best bow, so let's assume the efficiency is 0,65.

KE = 113,76 J
velocity, Vi = 47,23 m/s
bow weight, f = 144 lbs = 640,5439 N
draw length, x = 32” = .8128m
efficiency, e = 0,65
mass arrow, m = 0,102 kg
mass bow, mb = ? kg
kinetic energies, k = 0,04234757

k = (47,23 * 0,102)/113,76 = 0,04234757

47,23^2 = 2230,6729

0,65*640,5439 * 0,8128 = 364,443857344

338,412153248 / 2230,6729 = 0,151708550925597383641501181101

0,151708550925597383641501181101 - 0,102 = 0,049708550925597383641501181101

0,049708550925597383641501181101 / 0,04234757 = 1,1738229826551413373069855271744 ~ 1,17 kg

The problem with this formula is that small inaccuracies give really big changes when you calculate.

----------------------------------------------------



I have a question myself.

In Alan Williams test he talks about 4 different armor values.
1. Low quality iron munition armor. Coefficient 0,5. R = 120-150 kJ/m2
2. Low carbon steel 0,3 %. Coefficient 0,75 R. = 180-210 kJ/m2
3. Medium carbon steel 0.6 %. Coefficient 1,1. R = 240 - 260 kJ/m2
4. Medium carbon steel 0,6, slack-quenched. Coefficient 1,5. R = >300kJ/m2

If you multiply these coefficients with the results of mild steel he tested (coefficient 1,0. R = 235 kJ/m2), you get the kinetic energy needed to penetrate these different armor qualities.
So we have for mild steel:
1 mm = 55 J
2 mm = 175 J
3 mm = 300 J
4 mm = 475 J

Let's look at 2 mm as an example.

175*0,75 = 131,25 J for quality 2, Low carbon steel R = 180-210 kJ/m2.

Now my question. What is the Vickers Plate Hardness (Hv) of these different armor qualities?

----------------------------------------------------------------

I have collected some information from different tests:

A report of the findings of the Defence Academy warbow trials Part 1 Summer 2005.
140 lbs bow, 87 gram arrow, 46 m/s = IE 92,046 J, Lozenge plate cutter, Ash, Mark Stretton.
140 lbs bow, 71 gram arrow, 46 m/s = IE 75,118 J, Long bodkin, Aspen?, Mark Stretton.
140 lbs bow, 70 gram arrow, 49,68 m/s = IE 86,383 J, Short bodkin, Aspen?, Mark Stretton.

The Secret of the English War bow.
144 lbs bow, 102 gram arrow, 47,23 m/s = IE 113,76 J, Lozenge plate cutter, Ash?, Mark Stretton.

The Great Warbow.
150 lbs bow, 95,9 gram arrow, 53 m/s = IE 134,69 J, MD 89,9 J, Long bodkin, Birch.
150 lbs bow, 86,6 gram arrow, 53,55 m/s = IE 124,167 J, MD 80,1 J, Target blunt, Ash.
150 lbs bow, 74,4 gram arrow, 57,8 m/s = IE 124,279 J, MD 75 J, Fluted bodkin, Aspen.
150 lbs bow, 57,8 gram arrow, 62,25 m/s = IE 111,989 J, MD 67,4 J, Small bodkin, Birch.
150 lbs bow, 53,6 gram arrow, 64,3 m/s = IE 110,804 J, MD 64,1 J, Small bodkin, birch.

Discovery Channel Documentary.
167 lbs bow, 85,04 gram arrow, 58,8 m/s = IE 147,01 J, Long Tudor bodkin, Simon Stanley.

Joe Gibbs World Record distance shooting with a "livery arrow".
175 lbs bow. 63,7 gram arrow, 64 m/s = IE 130,457 J, Small bodkin, Aspen, Joe Gibbs.

IE = initial kinetic energy.
MD = max distance.
J = joule.

There's a lot more, but than I have to look through my books.
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Cs. Norbert




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PostPosted: Wed 11 Nov, 2015 5:20 am    Post subject:         Reply with quote

Hi Eirik R. F. and thank you for contributing!
In the first part of your calculation I noticed that you interpreted k as a relation between the kinetic energy of the arrow and momentum, I don't know how it should be calculated, but if I'm not mistaking, the k factor. should represent the sum of the ke's of the moving parts of the bow. This is what I've found on the web: "k for medieval bows range from 0.03 and 0.07".
Your assumption regarding the efficiency of Joe's bow is close, but it can be estimated using for reference the 0.7 coef. for 9 grains/pound http://www.dryadbows.com/Defining%20Bow%20Performance%20Dryad.pdf , so the efficiency should be ~0.66 at 5.6 gr/lb. I don't know how you got the mass of the bow (you calculated it or asked Joe), I've asked Ian on youtube, but after a while he disabled the comments for that video. It's very important to have as many real values as we can (instead of estimated or calculated) to have a smaller tolerance for error in the final calculation/estimation, as you stated "small inaccuracies give really big changes when you calculate." so knowing the real weight of the bow would help a lot.
You gave this:

velocity, Vi = 64 m/s #known
bow weight, f = 175 lbs = 778,438 N #known
draw length, x = 32” = .8128m #known
efficiency, e = 0,7 #assumption
mass arrow, m = 0,0637 kg #known
mass bow, mb = 1,43 kg #either calculated with another calculated factor "k" or you know the real mass
kinetic energies, k = 0,03125 #calculated, but I don't know if it represents momentum/ke, I used 0.04
If using k as p/ke in Bane's calculations for the 110lb bow, we get a k factor of 0.036 instead of 0.04!

You might have 3 uncertain values in that equation (e, mb and k), if you estimate e using 0.7 for 9 gr/lb, you might get a more "accurate" estimation, if you know the real mass of the bow that leaves one uncertain value k and one roughly estimated e! If you use this calculation methods in A Karpowicz turkish recurves (where you have all the data for the equation), you will get k factors from ~0.007 to 0.04!

Now for the second part and your question, I don't know the VPH for the rest of the plates, except the tested 0.15C (0.1C) carbon mild steel which has a VPH (Hv) of 152 and the toughness 235 Kj/m2. I've searched for a relation between fracture toughness or toughness and VPH and I've found something, but I didn't have anymore time to spare for research, maybe you'll have more luck. I'm currently out of my country and I don't have the resources (laptop and a permanent internet connection) to find it. Williams extrapolated data by testing different plates of different quality and by quality (*,**,***) he meas the slag and carbon content. Those are all rough estimations and I believe that, the margin for error is very high in his estimations - see page 2 of this topic, the post containing: PART I - ESTIMATING LONGBOW PERFORMANCE and PART II - ESTIMATING ARMOUR EFFECTIVENESS!
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Eirik R. F.




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PostPosted: Tue 01 May, 2018 11:43 am    Post subject:         Reply with quote

C.S Norbert

Just forget about my previous post. It's incorrect.

I just want to make a quick comment on your post on page 2 about the test in Alan Williams' book, and in particular on your point about him being wrong about the fracture toughness of iron.

The qualities mentioned *, **, *** and **** have the following coefficients: 0.5, 075, 1.1 and 1.5.
* = 120 -150 kJ/m2
** = 180 -210 kJ/m2
*** = 240 -260 kJ/m2
**** = 300 kJ/m2 or more

Here's how he presumably calculated it.
120 / 235 = 0.51
180/235 = 0.76
260/235 = 1.1

1.5 * 235 = 352.5 kJ/m2
352.5/235 = 1.5

235 kJ/m2 is the fracture toughness of the mild steel he tested and serves as the reference.

The actual coefficients for the different qualities are as follows:
120 kJ/m2 = 0.510
150 kJ/m2 = 0.638
180 kJ/m2 = 0.765
210 kJ/m2 = 0.893
240 kJ/m2 = 1.021
260 kJ/m2 = 1.106
300 kJ/m2 = 1.276
320 kJ/m2 = 1.361 <-- This is the fracture toughness of 1050 steel he tested without heat-treatment.
352.5 kJ/m2 = 1.5

So you are correct that wrought iron have around 60 to 65 % of the fracture toughness of the mild steel. It's actually 63.8 %.


You made one mistake. You said 250 Joules was enough to just penetrate the 1050 steel. That's not correct. 250 Joules defeated the 1050 steel. Just over 300 Joules defeated the steel after heat-treatment. This is predictable with this formula:

55 * 2^1.6 * (460/235) = 326 Joules.

The steel had a Vickers Plate Hardness of 460, and this is very close to the fracture toughness denoted as kJ/m2. It is the manganese in mild steel that gives the steel a VPH of only 152 while the fracture toughness is 235. Fracture toughness matters, but VPH is a good substitute for it when dealing with medieval armor or modern carbon steel for that matter. We don't want to rip armor apart when an indentation test give us an adequate prediction.
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