Why did the the Ming crossbows become so weak compared to earlier Chinese ones?

[HackneyedScribe]

I don't think you understood what I was saying at all. The linear model I used is for calculating stored energy, and that is used to calculate efficiency. NOT velocity. Bichler calculated stored energy using the actual force draw curve, which is not linear. His way of calculating gives a higher stored energy but lower efficiency. If you don't believe me then just multiply the draw weight by powerstroke and divide by two(how we get stored energy by people here) and you will see that the resulting stored energy is significantly less than the stored energy Bichler stated.

1276 lbs * 14.76 inches / 2 = 9417 inch lbs = 1064 joules

That is less than the potential energy of 1277 joules Bichler stated.

Shooting at 488 Joules,

Efficiency using linear stored energy = 488 joules/1064 joules = 46%
Efficiency using actual stored energy = 488 joules/1277 joules = 38%

Andreas Bichler said:
I measured the draw weight in 25mm steps by a digital scale and generate a drawcurve. So its possible to calculate its potential energy from the area under the (not linear) curve.

That means he calculated the yellow area under the force draw curve:


Our equation used here for stored (potential) energy: Draw weight * powerstroke/2 . This is the equation used to calculate the area of a triangle. That means it excludes the yellow area. Ergo such an estimate tend to underestimate the force draw curve, but most efficiencies are calculated using this estimate because very few people have access to the actual force draw curve of bows, but information on draw weight and powerstroke is easy to come by. If you want to compare the efficiencies of different bows, then make sure they are either all calculated under the linear estimate of stored energy, or all calculated under the actual energy. Otherwise you would end up with inaccurate, not to mention unfair, results.

Ergo our calculated stored energy is smaller, and because efficiency is calculated by dividing by stored energy, we get a larger efficiency. Bichler's method gets lower efficiency but offsets this with calculating a higher stored energy. Our method (along with most peoples) calculates a lower stored energy, but offsets this with higher efficiency. Either way the amount of joules that gets transferred into the arrow is the same when given the same draw weight and powerstroke of the same bow. The rest of what you said stems from this misunderstanding.

[HackneyedScribe]

Also please show the formula used by Adam Karpowicz that you say could be used to calculate the dry fire speed of 101 meters per sec..

[Hurricane Six]

few people have access to the actual force draw curve of bows

But we can access it!. 1277 joules. 34% efficiency.

This is severely underestimating flight bows (high velocity) with large stacking and overestimating war bows (low velocity) with minimal stacking. I f we have the info, we should not pretend. It is what it is.

Also please show the formula used by Adam Karpowicz that you say could be used to calculate the dry fire speed of 101 meters per sec..

Simple log: y = -21.4ln(x) + 132.5 (rounded)

Based 100% off the points of Adam's composite bows

[HackneyedScribe]

But we can access it!. 1277 joules. 34% efficiency.

We have access to actual stored energy of ONE bow so far, but we didn't have access to other prods that were compared. If we did they would decrease in efficiency likely around the board..... compare like with like.

Simple log: y = -21.4ln(x) + 132.5 (rounded)

What does y stand for, what does x stand for, how did you get -21.4 and 132.5, and how did that get 101 m/s for dry fire speed? You should explain this equation.

[Hurricane Six]

We have access to actual stored energy of ONE bow so far, but we didn't have access to other prods that were compared. If we did they would decrease in efficiency likely around the board..... compare like with like.

And if you have another one, I would be glad to here it. Generalities do not nullify facts. 34% efficiency @ 58 m/s. FACT. Can we not agree on that?

If you found something you didn't like, you would judge it on one example. Would you not?

What does y stand for, what does x stand for, how did you get -21.4 and 132.5, and how did that get 101 m/s for dry fire speed? You should explain this equatio

n.

First, you want me to give you a formula, then you want me to explain it.

It is a simple logarithmic function. The numbers are based on Adam's graphs, which likely overestimated efficiency.

Here is a simpler equation: y = 94.688x2 - 160.67x + 10^2

X equals efficiency.

I have been told by the moderators not to use complex mathematics, but you asked. It is on you.

[HackneyedScribe]

And if you have another one, I would be glad to here it. Generalities do not nullify facts. 34% efficiency @ 58 m/s. FACT. Can we not agree on that?

If you found something you didn't like, you would judge it on one example. Would you not?

It's a fact that the efficiency is calculated using a more conservative method than that of other crossbows here, yes.

n.

First, you want me to give you a formula, then you want me to explain it. How could I explain a formula if I can't figure out the area of a triangle?[IMG]file:///C:\Users\DJ\AppData\Local\Temp\msohtmlclip1\01\clip_image001.gif[/IMG]

It is a simple logarithmic function. The numbers are based on Adam's graphs, which likely overestimated efficiency.

Here is a simpler equation: y = 94.688x2 - 160.67x + 10^2

X equals efficiency..

image don't work and now you still haven't explained what y means. How did you get 94.688, 160.67, and 10^2. How is that supposed to calculate dry fire speed of Bichler's bow?

[Hurricane Six]

It's a fact that the efficiency is calculated using a more conservative method than that of other crossbows here, yes.

image don't work and now you still haven't explained what y means. How did you get 94.688, 160.67, and 10^2. How is that supposed to calculate dry fire speed of Bichler's bow?

It is the equation of the trend line that fits the points best (34% at 58 m/s & 39% @ 53 m/s). If X is efficiency then Y must be m/s (velocity).



No insult intended. What is your level of educational attainment?

[HackneyedScribe]

It is the equation of the trend line that fits the points best (34% at 58 m/s & 39% @ 53 m/s). If X is efficiency then Y must be m/s (velocity).

How does that explain anything?

So far when I asked you to show how you estimated dry fire speed, you gave this equation without explanation:
y = -21.4ln(x) + 132.5 (rounded)

When I asked you to explain that equation, you gave this equation without explanation:
y = 94.688x2 - 160.67x + 10^2

You say y = velocity, and x equals efficiency. OK, how did you use that to get numbers 21.4, 132.5, 94.688, 160.67 and 10^2. Why are you logging efficiency? Where are these numbers coming from? How do they apply to Bichler's crossbow?

For example, here is how I calculated a 61 lb Han crossbow efficiency: http://historum.com/asian-history/13...4?postcount=29

It was using the equation:
Energy = "Projectile Energy in Joules"/"Potential Energy in Joules"

Potential Energy in Joules = 'draw weight in lbs' * 'powerstroke in inches' * 0.5
= 61 lbs * 19 inches *0.5
= 579.5 inch lbs
= 65.5 joules

Since he shot a 30.1 gram projectile that achieved 54 m/s, the energy of the projectile is calculated by:

Energy in joules = "projectile weight in kilograms" * "velocity in m/s" * "velocity in m/s" * 0.5
= .0301 * 54 * 54 *.5
= 44 Joules

Shooting at 44 Joules,
Efficiency = 44 joules/65.5 joules = 67%

A mere 61 lb crossbow shooting to 44 joules, not bad for a crossbow.

^That's something you can follow through. Every number is explained. Instead of asking me about my education level, I hope you do the same and explain the many numbers in your equation in order to get how you calculated the dry fire speed of Bichler's crossbow.

[Common Soldier]

I don't think you understood what I was saying at all. The linear model I used is for calculating stored energy, and that is used to calculate efficiency. NOT velocity. Bichler calculated stored energy using the actual force draw curve, which is not linear. His way of calculating gives a higher stored energy but lower efficiency. If you don't believe me then just multiply the draw weight by powerstroke and divide by two(how we get stored energy by people here) and you will see that the resulting stored energy is significantly less than the stored energy Bichler stated.

1276 lbs * 14.76 inches / 2 = 9417 inch lbs = 1064 joules

That is less than the potential energy of 1277 joules Bichler stated.

Shooting at 488 Joules,

Efficiency using linear stored energy = 488 joules/1064 joules = 46%
Efficiency using actual stored energy = 488 joules/1277 joules = 38%

Andreas Bichler said:
I measured the draw weight in 25mm steps by a digital scale and generate a drawcurve. So its possible to calculate its potential energy from the area under the (not linear) curve.

That means he calculated the yellow area under the force draw curve:


Our equation used here for stored (potential) energy: Draw weight * powerstroke/2 . This is the equation used to calculate the area of a triangle. That means it excludes the yellow area. Ergo such an estimate tend to underestimate the force draw curve, but most efficiencies are calculated using this estimate because very few people have access to the actual force draw curve of bows, but information on draw weight and powerstroke is easy to come by. If you want to compare the efficiencies of different bows, then make sure they are either all calculated under the linear estimate of stored energy, or all calculated under the actual energy. Otherwise you would end up with inaccurate, not to mention unfair, results.

Ergo our calculated stored energy is smaller, and because efficiency is calculated by dividing by stored energy, we get a larger efficiency. Bichler's method gets lower efficiency but offsets this with calculating a higher stored energy. Our method (along with most peoples) calculates a lower stored energy, but offsets this with higher efficiency. Either way the amount of joules that gets transferred into the arrow is the same when given the same draw weight and powerstroke of the same bow. The rest of what you said stems from this misunderstanding.

In Bilcher's video on his 1270 lbs great horn crossbow, I don't see where he list the powerstroke length, or the comment about how he measured the energy of his crosswbow. Can you point where he says the powerstroke length? Some of the German was hard to make out, but nothing seem to match the powerstroke length. You could derive a length from the energy and draw weight using the simple linear equation, but if you are saying the linear equation was invalid, then you won't get the right powerstroke length.





A

[HackneyedScribe]

In Bilcher's video on his 1270 lbs great horn crossbow, I don't see where he list the powerstroke length, or the comment about how he measured the energy of his crosswbow. Can you point where he says the powerstroke length? Some of the German was hard to make out, but nothing seem to match the powerstroke length. You could derive a length from the energy and draw weight using the simple linear equation, but if you are saying the linear equation was invalid, then you won't get the right powerstroke length.

It's in the comment section of his video here: https://www.youtube.com/watch?v=nY2untEwCnU

Where in reply to Peasant Mob, Andreas Bichler said:
"The powerstroke for this crossbow is 375mm".

In another reply he said:
I measured the draw weight in 25mm steps by a digital scale and generate a drawcurve. So its possible to calculate its potential energy from the area under the (not linear) curve.

375 mm is 14.76 inches.Usually I don't put any weight into comment sections of youtube videos. But here the comment was said by the guy who made the crossbow.

[Common Soldier]

It's in the comment section of his video here: https://www.youtube.com/watch?v=nY2untEwCnU

Where in reply to Peasant Mob, Andreas Bichler said:
"The powerstroke for this crossbow is 375mm".

In another reply he said:
I measured the draw weight in 25mm steps by a digital scale and generate a drawcurve. So its possible to calculate its potential energy from the area under the (not linear) curve.

375 mm is 14.76 inches.Usually I don't put any weight into comment sections of youtube videos. But here the comment was said by the guy who made the crossbow.

The video link you posted still doesn't give the crossbow powerstroke length, and Bilchesr doesn1t say a word in the video that I can hear. Are you thinking of a different link? If so, can you post that link? It isn't in the link you provided above. if you have a different link, please post it. None of the links I have from Bilcher include what you say.


(I was wrong about the 1270 lbs horn crossbow link, that one is written in English, but the powerstroke length was not listed, and he doesn't speak at all in it. I was thinking of his 1200 lbs crowwbow video, which was written in German.)

I

[HackneyedScribe]

The video link you posted still doesn't give the crossbow powerstroke length, and Bilchesr doesn1t say a word in the video that I can hear. Are you thinking of a different link? If so, can you post that link? It isn't in the link you provided above. if you have a different link, please post it. None of the links I have from Bilcher include what you say.


(I was wrong about the 1270 lbs horn crossbow link, that one is written in English, but the powerstroke length was not listed, and he doesn't speak at all in it. I was thinking of his 1200 lbs crowwbow video, which was written in German.)

I

As I said, it's in the comment section of the video, not the video itself. You can watch the video a million times and you won't see it because you would be looking in the wrong place. Look in the comment section. Go to the video link and instead of watching the video, scroll down.

[Common Soldier]

As I said, it's in the comment section of the video, not the video itself. You can watch the video a million times and you won't see it because you would be looking in the wrong place. Look in the comment section. Go to the video link and instead of watching the video, scroll down.

Can you provide a link to the comment section, as requested several times? The link you proviced did not include any comment section that I can see. Do you have another link with the comment section?

[HackneyedScribe]

Can you provide a link to the comment section, as requested several times? The link you proviced did not include any comment section that I can see. Do you have another link with the comment section?

The video link IS the link to the comment section. All you had to do was scroll down. I gave you the link, I told you to scroll down and NOT watch the video because the statement was made in the comment section, I told you the person Bichler was replying to. What more do you want, a picture? Somehow I already knew you were going to make me do this... :



^Compare that with post 90, the link shown in the picture is the link I provided. Everything I quoted from him was word for word. Powerstroke of 375 mm equals 14.76 inches.

[Common Soldier]

My mistake, I did see the comment section at the bottom, but I went through the 92 comments, and did not see where Bilcher said what the powerstroke length was. I assume that the commenter "Medieval Crossbow" was Andreas Bilcher replying to "Peasant Mob".

They did discuss the efficiencies of composite crossbow. Perhaps a section got deleted? Peasant Mob talked about a powerstroke of 18" using a liner power curve, but I don't see any explanation by Medieval Crossbow of describing how he obtained his energy calculation

[HackneyedScribe]

They did discuss the efficiencies of composite crossbow. Perhaps a section got deleted? Peasant Mob talked about a powerstroke of 18" using a liner power curve, but I don't see any explanation by Medieval Crossbow of describing how he obtained his energy calculation

The powerstroke of the crossbow is not 18'', 18'' is just a guess from a commentator who himself was dubious about the number. Actual powerstroke is 375 mm = 14.76 inches, as the creator of the crossbow himself said. Medieval Crossbow (Andreas Bichler) described how he obtained his energy calculation in the last statement of the picture I provided. If you don't understand that then you don't understand how the force draw curve works. So you'll need to learn how to read the force draw curve before trying to understand Bichler's statement.

They did discuss the efficiencies of composite crossbow. Perhaps a section got deleted?

Obviously, I can't fit every single comment into only one single picture. All you have to do, is go to the video link, go to Peasant Mob's comment, and scroll down to the end of that discussion, which isn't very long. It's very easy. It took me ten minutes to find the comment, take a snapshot, put it online (after resetting my password), and then respond to your post with said picture. All that was in ten minutes. It shouldn't take you that long to do much less.

[Hurricane Six]

Sorry for the delay:


That trend-line is generated by an 8 core processor. I forgot to update it did not consider all the points gernerates by the crossbow. The trendline is generated by a computer to best fit the data points that were input. Meaning, the velocity and efficiency are points on a graph. The computer generates a trendline to pass through these points. Turns out the polynomial is a perfect fit R2 = 1.00, even better than the logarithmic trend-line = .99

y = -4E-05x2 - 0.0042x + 0.7239
R² = 1

As you can see, top speed is 90 m/s not 101 m/s as I said earlier (trendline without all points gave false results).


Plug it in for yourself:
y = -4E-05(57.74)2 - 0.0042(57.74) + 0.7239 +|= 34% efficiency. PERFECT!


Dry fire speed = ~90 m/s

[HackneyedScribe]

Sorry for the delay:


That trend-line is generated by an 8 core processor. I forgot to update it did not consider all the points gernerates by the crossbow. The trendline is generated by a computer to best fit the data points that were input. Meaning, the velocity and efficiency are points on a graph. The computer generates a trendline to pass through these points. Turns out the polynomial is a perfect fit R2 = 1.00, even better than the logarithmic trend-line = .99

y = -4E-05x2 - 0.0042x + 0.7239
R² = 1

As you can see, top speed is 90 m/s not 101 m/s as I said earlier (trendline without all points gave false results).


Plug it in for yourself:
y = -4E-05(57.74)2 - 0.0042(57.74) + 0.7239 +|= 34% efficiency. PERFECT!


Dry fire speed = ~90 m/s

Why would you need an 8 core processor to calculate an equation like that? And previously you used these two equations:



So far when I asked you to show how you estimated dry fire speed, you gave this equation without explanation:
y = -21.4ln(x) + 132.5 (rounded)

When I asked you to explain that equation, you gave this equation without explanation:
y = 94.688x2 - 160.67x + 10^2

Now you say it's this:
y = -4E-05x2 - 0.0042x + 0.7239
y = -4E-05(57.74)2 - 0.0042(57.74) + 0.7239 +|= 34% efficiency.

The only number that matches with Bichler's number is the velocity, which is inputed into variable x, even though previously you said "If X is efficiency then Y must be m/s (velocity)", and prior to that you said "X equals efficiency". Now you are using x as velocity and y as efficiency. As I said, you really should show how you got -4E-05 and .0042 and .7239 along with the other numbers. What do they stand for? You're only showing the results, you haven't explained how the equation calculates efficiency, nor how you used that to get dry fire speed.

[Hurricane Six]

Why would you need an 8 core processor to calculate an equation like that?

You don't necessarily. You just need excel. However the computer that it was calculated on had 8 cores.

And previously you used these two equations:

As I said in my previous post, if you do not update the trend line formula when new data is input, it will give false equation, but a correct trendline. You always have to refresh.

even though previously you said "If X is efficiency then Y must be m/s (velocity)"

You can do it either way, If you want x to be efficiency or speed, you just switch the x and y axis inputs. You get a different formula, but it will generate the same results.

The only number that matches with Bichler's number is the velocity, As I said, you really should show how you got -4E-05 and .0042 and .7239 along with the other numbers. What do they stand for? You're only showing the results, you haven't explained how the equation calculates efficiency, nor how you used that to get dry fire speed.

X (velcoity in m/s) and Y (efficiency %) are variables. They will change and will fit Bilcher's numbers very well.. For example:

y = -4E-05(57.74)2 - 0.0042(57.74) + 0.7239 +|= 33.9% efficiency.

y = -4E-05(52.92)2 - 0.0042(52.92) + 0.7239; Y = 38.2% efficiency. ( You see, when you put in 57.92 m/s, you get 38.2%, spot on to Bilcher's numbers)

Now try: y = -4E-05(90)2 - 0.0042(90) + 0.7239; Y = 2.9% efficiency. (90 is the velocity in m/s)

AS you see, 90 m/s would give you an overall efficiency of 2.9% (effectively 0%). Input 92 m/s and you get 0% efficiency.

As I said, you really should show how you got -4E-05 and .0042 and .7239 along with the other numbers

Excel creates them. I tried to post the chart w/ the equation, but it didn't work, so I will use one from the internet.


https://i.stack.imgur.com/tgq8c.jpg


Notice the data points on the left. They are represented by the blue dots on the chart. The trendline and formula is computed by excel to best fit the data points. Notice the formula generated. Do you understand?


Let me simplify. The formula for a line is Y = mx + b. So lets say to have x = 30 and y = 10 and X = 0 & Y = 5. The formula would be Y = 0.1667x + 5. This is the formula of a simple line. When you put in 30 in the formula, y equals 10. When you put in 0, Y equals 5. Understand? The equation fits the data points (say velocity and efficiency).



To summarize: Heavy composite crossbows max out below 330 fps or 100 m/s in dry fire speed.

[Common Soldier]

You don't necessarily. You just need excel. However the computer that it was calculated on had 8 cores.



As I said in my previous post, if you do not update the trend line formula when new data is input, it will give false equation, but a correct trendline. You always have to refresh.



You can do it either way, If you want x to be efficiency or speed, you just switch the x and y axis inputs. You get a different formula, but it will generate the same results.




X (velcoity in m/s) and Y (efficiency %) are variables. They will change and will fit Bilcher's numbers very well.. For example:

y = -4E-05(57.74)2 - 0.0042(57.74) + 0.7239 +|= 33.9% efficiency.

y = -4E-05(52.92)2 - 0.0042(52.92) + 0.7239; Y = 38.2% efficiency. ( You see, when you put in 57.92 m/s, you get 38.2%, spot on to Bilcher's numbers)

Now try: y = -4E-05(90)2 - 0.0042(90) + 0.7239; Y = 2.9% efficiency. (90 is the velocity in m/s)

AS you see, 90 m/s would give you an overall efficiency of 2.9% (effectively 0%). Input 92 m/s and you get 0% efficiency.



Excel creates them. I tried to post the chart w/ the equation, but it didn't work, so I will use one from the internet.


https://i.stack.imgur.com/tgq8c.jpg


Notice the data points on the left. They are represented by the blue dots on the chart. The trendline and formula is computed by excel to best fit the data points. Notice the formula generated. Do you understand?


Let me simplify. The formula for a line is Y = mx + b. So lets say to have x = 30 and y = 10 and X = 0 & Y = 5. The formula would be Y = 0.1667x + 5. This is the formula of a simple line. When you put in 30 in the formula, y equals 10. When you put in 0, Y equals 5. Understand? The equation fits the data points (say velocity and efficiency).



To summarize: Heavy composite crossbows max out below 330 fps or 100 m/s in dry fire speed.

If the max dry fire speed max out below 330 ft/s, then that means the fastest possible bolt speed for a heavy crossbow is going to be less than that, no matter how light the bolt is.