# Thread: Where early muskets as powerful as multiple tonne trebuchets? A Comparison.

1. ## Re: Where early muskets as powerful as multiple tonne trebuchets? A Comparison.

Originally Posted by Hurricane Six
It is a quadratic equation T = v*y/g +/- Square root (V^2*y/g^2 - 2y/g). First you need to determine the vector component ratio and apply it to v (velocity).

T = time; V= velocity (to be adjusted with the y vector component based on launch angle); g = gravity; Y = height (input value needs to be negative).

I can walk you through it one step at a time, if you want. You seem to like equations.
Yes please, this looks similar to the time displacement equation but you multiplied by the y variable to the first section of the equation, multiplied by the square root of y in the first section of the square root, but left the last section of the square root well enough alone. Do you have a source explaining this particular equation you are using? It would be appreciated.

2. ## Re: Where early muskets as powerful as multiple tonne trebuchets? A Comparison.

It's a standard differential equation for coming up with acceleration, velocity, and position. Part of it anyway. You can find it in any physics book.

But still. Can people tell me what happens to a wall when hit with a bullet and what happens to a wall when hit with a projectile, and why we might use them for different things and how this conversation got started?

3. ## Re: Where early muskets as powerful as multiple tonne trebuchets? A Comparison.

Fine I'll tell you.

The musket ball has less area to strike the wall so it does less Work(physics term) on the wall and thus it does less damage to the wall. The Projectile is larger so it does more Work on the wall. Thus it does more damage to the wall. Do all the Energy(ironically is work, but you skip straight to Joules) calculations you want. But until you do more Work on what you hit, you will do less damage with things like muskets.

Also, it would help you if you included the right equations with the right equipment from the beginning to determine which one was more powerful with what it was designed to do.

4. ## Re: Where early muskets as powerful as multiple tonne trebuchets? A Comparison.

It's a standard differential equation for coming up with acceleration, velocity, and position. Part of it anyway. You can find it in any physics book.
I'm not seeing acceleration in the equation though, unless if you mean gravity but that's just the acceleration downward rather than overall acceleration of an object. Can you calculate velocity with just knowing the time it takes a flying object to hit the ground, when shooting in an arc without knowing the distance the object shot to? It's what Hurricane Six is doing but I want to see how.

Also I realize it's weird comparing a musket to a trebuchet. That's why I mentioned momentum over joules, but nobody mentioned further about that.

5. ## Re: Where early muskets as powerful as multiple tonne trebuchets? A Comparison.

Originally Posted by HackneyedScribe
I'm not seeing acceleration in the equation though.
Once it leaves the trebuchet the acceleration is gravity; down. 9.8 m/s^2

Can you calculate velocity with just knowing the time it takes a flying object to hit the ground, when shooting in an arc without knowing the distance the object shot to? It's what Hurricane Six is doing but I want to see how.
The trebuchets as such that we've studied them should be known quantities and highly tested even by today's standards and have some fairly good records for what the velocity leaving the trebuchet is depending on what you are doing with the trebuchet. Once you have that you are dealing with a fairly basic arc calculation.

At the top of the arc KE will be 0, PE will be a crapton. It starts moving down. When it hits something it doesn't really matter what its KE is. What you WANT to calculate is the transfer of KE from the projectile to something else. And the projectile transfers a helluva lot more than a musket bullet.

6. ## Re: Where early muskets as powerful as multiple tonne trebuchets? A Comparison.

Once it leaves the trebuchet the acceleration is gravity; down. 9.8 m/s^2
Yeah, I edited that part the moment you posted. But how would that equation allow you to calculate the velocity of an object while traveling in an arc, without knowing the distance the object is traveling?

Once you have that you are dealing with a fairly basic arc calculation.
My question is: Right now velocity is calculated as 31 m/s and the only variable I've seen been used is that it took about 5 seconds for the object to hit the ground, plus the launch angle. How is the 31 m/s velocity calculated based off of just the time it takes to hit the ground?

7. ## Re: Where early muskets as powerful as multiple tonne trebuchets? A Comparison.

Post 12 data. At least for his starting data. There's a lot of follow-on argument that makes you want to look at his work as much as his numbers and you wish there's a public spreadsheet to read over this.

But again, like I said. It's not about the energy carried by the projectile. But the energy transferred.

Here's your irony about trebuchets. If your projectile goes through and through you don't do a lot of comparative damage. If it stops it doesn't do a lot of damage. If it crashes down through the wall hitting a much larger area than it has any right to by its own diameter, you've got your launch velocity right.

Bullets. If it goes through and through, again, not too much comparative damage. If it stops, it hurts and can cause nasty infection, but surgery can handle things. If it stops and breaks up. Hell what if the bullet doesn't even penetrate...

90% of the calculations in this thread are about how the projectile flies through the air and how much energy they carry. Nothing is done to answer the question about how a particular projectile may transfer energy when it strikes. Which projectile do you want to hit the wall because it will do a good job breaking the wall? Which projectile do you want to fly over and hit the city because you lit it on fire but maybe not do so much damage to the wall?

8. ## Re: Where early muskets as powerful as multiple tonne trebuchets? A Comparison.

I've dug up an equation that I've forgotten about. It is v= tg/2sin(theta)
v = initial velocity
t = time
g = gravity (9.8)
theta = angle of shot

Because Hurricane Six set launch angle at 42 degrees and time at 5.25, I put these into the equation to get:
v =5.25(9.8)/2sin(42) =38.45 m/s.
I wonder how Hurricane Six used his different equation to get 31 m/s. I surmise because it's launched at a height, he took 1 second off of time to adjust for that, in which you'll get 31 m/s. However, it looks like he used another equation and I'm interested in this new equation. I know velocity and energy isn't everything there is to knowing the damage a trebuchet can do, but right now I'm more interested in the physics calculation anyways.

9. ## Re: Where early muskets as powerful as multiple tonne trebuchets? A Comparison.

Meh. Like I said. I'd like to see his work instead of just his results the way the thread his gone. But also like I said. He's asking the wrong question from the beginning.

10. ## Re: Where early muskets as powerful as multiple tonne trebuchets? A Comparison.

Originally Posted by Gaidin
Post 12 data. At least for his starting data. There's a lot of follow-on argument that makes you want to look at his work as much as his numbers and you wish there's a public spreadsheet to read over this.

But again, like I said. It's not about the energy carried by the projectile. But the energy transferred.

Here's your irony about trebuchets. If your projectile goes through and through you don't do a lot of comparative damage. If it stops it doesn't do a lot of damage. If it crashes down through the wall hitting a much larger area than it has any right to by its own diameter, you've got your launch velocity right.

Bullets. If it goes through and through, again, not too much comparative damage. If it stops, it hurts and can cause nasty infection, but surgery can handle things. If it stops and breaks up. Hell what if the bullet doesn't even penetrate...

90% of the calculations in this thread are about how the projectile flies through the air and how much energy they carry. Nothing is done to answer the question about how a particular projectile may transfer energy when it strikes. Which projectile do you want to hit the wall because it will do a good job breaking the wall? Which projectile do you want to fly over and hit the city because you lit it on fire but maybe not do so much damage to the wall?
True enough. I kinda chose not to even do this battle (work) on top of the others as well, I only implied it once maybe I think. Thing is that if you want to compare the kinetic energy of two weapons, it shouldn't be that at the start, but rather at impact. I decided to go that route and point out that the trebuchet round will keep most of the energy and always kill. The musket would lose said energy much faster. Then of course the area of impact plays a role. Which is why I mentioned cuirasses. Those tend to be very thin and wouldn't do much against modern firearms. But a musket bullet that is made of stone and has a larger area of impact is another story, especially with a gambeson that would absorb much of the blunt trauma. But obviously at normal ranges that musket would be a relatively sure kill. But those are so close that trebuchets weren't likely much used for those. The trebuchet would obviously also be a sure kill IF it hit. But the problem is obviously like you said (and I did earlier) they were used for very different tasks, with very different effects and so on.
Structural and "medical" damage inflicted on a target are of course a topic of their own. Though you got it right actually. Though to my knowledge, trebuchets never really got to that speed where they would simply rip through. Whatever wall they hit often fell apart immediately. Barring some implausible scenarios, I can't imagine a musket ever causing significant structural damage.

11. ## Re: Where early muskets as powerful as multiple tonne trebuchets? A Comparison.

it shouldn't be that at the start, but rather at impact.
That is fair enough. We can do that. However, we need to accurately measure the effects of drag and compare at equal distances.

But a musket bullet that is made of stone and has a larger area of impact is another story, especially with a gambeson that would absorb much of the blunt trauma.
Most were made of lead (which is soft) but very dense 11.3g/cc vs stone 2.5 g/cc.

Because Hurricane Six set launch angle at 42 degrees and time at 5.25, I put these into the equation to get:
v =5.25(9.8)/2sin(42) =38.45 m/s.
I wonder how Hurricane Six used his different equation to get 31 m/s. I surmise because it's launched at a height, he took 1 second off of time to adjust for that, in which you'll get 31 m/s. However, it looks like he used another equation and I'm interested in this new equation. I know velocity and energy isn't everything there is to knowing the damage a trebuchet can do, but right now I'm more interested in the physics calculation anyways.
Your equation is fin, but does not include launch height. Try this: http://www.convertalot.com/ballistic...alculator.html

Input known variables (launch height and angle) and guess and check launch velocity to arrive at the correct time of flight. Note that final velocity is is 38 m/s.

You can find the equation I used here: http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html

Scroll down near the end.

The additional height will theoretically yield greater energy at impact than at launch. a 40 lb projectile dropped from 90 ft. will yield 3,600 ft. lbs of energy (excluding drag). Kinetic plus potential energy will be equal to a launch velocity of 38 m/s at ground level.

I will update original post.

12. ## Re: Where early muskets as powerful as multiple tonne trebuchets? A Comparison.

Originally Posted by HackneyedScribe
I've dug up an equation that I've forgotten about. It is v= tg/2sin(theta)
v = initial velocity
t = time
g = gravity (9.8)
theta = angle of shot

Because Hurricane Six set launch angle at 42 degrees and time at 5.25, I put these into the equation to get:
v =5.25(9.8)/2sin(42) =38.45 m/s.
I wonder how Hurricane Six used his different equation to get 31 m/s. I surmise because it's launched at a height, he took 1 second off of time to adjust for that, in which you'll get 31 m/s. However, it looks like he used another equation and I'm interested in this new equation. I know velocity and energy isn't everything there is to knowing the damage a trebuchet can do, but right now I'm more interested in the physics calculation anyways.
That equation is valid only assumig no air drag. The angle for max range in a vacuum is 45 degrees, but less taking air drag into account. Since he gave 42 degrees instead of 45 degrees,his figures might be taking air drag into account which could explain his lower value of velocity.

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