Physics behind long lange siege engines

[Max Caysey]

So after thinking some more on the subject of long range artillery I gon interested in the physics behind it.

Now I know when magic is afoot, science goes out the window, but bear with.


So my heavy trebuchet can fire a 150 pound ball of (stone/metal) 14.400 ft., in one round. Now lets just assume it takes 6 seconds for it to land.

I have calculated the speed necessary for this to be 1646 MPH. (if its a flat trajectory) (don't know how to include an arch) So somewhat over Mach 2. (But that is without including drag)

Now if we take a L6 steel ball, a 150 lbs. ball would have a diameter of 10 inches (9,92 to be exact).

Now what I cant seem to figure out is the drag and thus what speed the ball has to start with and what speed it lands/impacts with for the ball to travel in an arch 14400ft. in 6 seconds.

When I know what speed it lands/impacts with I can calculate the energy.

Is there anyone here who can do this sort of math? I have tried using som calculators online, but the whole drag element is a bit too much for my math and googlefu.

Thanks!

[Baby Gary]

http://www.ucl.ac.uk/~zcapf71/Trebuchet%20coursework%20for%20website.pdf
(http://www.ucl.ac.uk/~zcapf71/Trebuchet%20coursework%20for%20website.pdf)
take a look at that, I don't have enough time at the moment to find a good answer. Also when I calculate the numbers stuff I am not going to take into account air resistance, just a heads up

[Baby Gary]

I just realized what you need to do

first you are trying to find the work (joules) that the ball does on the ground, right?

note: all my calculations will be rounded to the nearest hundredth

if so then this is what you do

work = force x distance
force = mass x acceleration

so that is how you would get the work, first you need to find the force, the mass is easy, you gave it to me (but it has to be in kg) 150 lbs = 68.04 kg. Acceleration is downward acceleration, which is our good friend gravity; that differers from planet to planet and D&D world to D&D world but I will use earth's gravity, 9.8 mps^2. multiply those (68.04*9.8) and you get 666.79 newtons of force that the ball hits the ground with.

now you need to know the distance that the ball hits the ground with. It reaches the apex of its flight at 3 secs (assuming that the entire thing takes 6 secs from fire to land) and we can find out how far it is away from the ground with gravity. with the formula d=1/2gt^2 we can figure this out

d=1/2 gt^2
d=1/2*9.8*3^2
d=1/2*9.8*9
d=44.1

with this we can figure out the answer

w=f*d
w=666.79*44.2
w=29405.44

[Max Caysey]

I just realized what you need to do

first you are trying to find the work (joules) that the ball does on the ground, right?

note: all my calculations will be rounded to the nearest hundredth

if so then this is what you do

work = force x distance
force = mass x acceleration

so that is how you would get the work, first you need to find the force, the mass is easy, you gave it to me (but it has to be in kg) 150 lbs = 68.04 kg. Acceleration is downward acceleration, which is our good friend gravity; that differers from planet to planet and D&D world to D&D world but I will use earth's gravity, 9.8 mps^2. multiply those (68.04*9.8) and you get 666.79 newtons of force that the ball hits the ground with.

now you need to know the distance that the ball hits the ground with. It reaches the apex of its flight at 3 secs (assuming that the entire thing takes 6 secs from fire to land) and we can find out how far it is away from the ground with gravity. with the formula d=1/2gt^2 we can figure this out

d=1/2 gt^2
d=1/2*9.8*3^2
d=1/2*9.8*9
d=44.1

with this we can figure out the answer

w=f*d
w=666.79*44.2
w=29405.44

Yes I', trying to figure out how many joules the ball would impact with.

So are you saying that the top of the ballistic curve is 44.2 something over the ground? And is this considering that the ball to travel the 14.400 ft. in 6 seconds has to travel an average speed around 1600 mph (without drag)? This would mean, I think, that it would not be landing only under the influence of gravity, it would still has a lot of energy left...

[Drakevarg]

According to Heroes of Battle, trebuchets have a maximum range of 1500 feet, much more manageable in 6 seconds.

Shooting any further than that and you've apparently built some kind of centrifugal railgun, and all bets are off.

[Max Caysey]

According to Heroes of Battle, trebuchets have a maximum range of 1500 feet, much more manageable in 6 seconds.

Shooting any further than that and you've apparently built some kind of centrifugal railgun, and all bets are off.

Apparently the Far Shot feat, the distance enchant as well as two spells increase the range increment. Thus giving a very high maximum range.

[Drakevarg]

Apparently the Far Shot feat, the distance enchant as well as two spells increase the range increment. Thus giving a very high maximum range.

Reasonably certain Far Shot doesn't apply to siege weapons, otherwise Rapid Shot would as well and suddenly you've got a semiautomatic catapult. I'm pretty sure they count as items/vehicles rather than weapons in the martial sense (Exotic Weapon Proficiency: Trebuchet is not a thing). As for the rest... magic. The math is "because we said so."

[Tvtyrant]

Apparently the Far Shot feat, the distance enchant as well as two spells increase the range increment. Thus giving a very high maximum range.
The assumption that D&D uses Earth physics instead of physkcs is the problem. In RL you can target something you do not have line of sight to, but no something with an interfering trajectory. So firing over a mountain is fine, but firing through it is impossible. In D&D you can hit anything with line of sight, so even if the trajectory would go through a low hanging beam it hits just fine and the weapon fires in a perfectly straight line.

[Max Caysey]

The assumption that D&D uses Earth physics instead of physkcs is the problem. In RL you can target something you do not have line of sight to, but no something with an interfering trajectory. So firing over a mountain is fine, but firing through it is impossible. In D&D you can hit anything with line of sight, so even if the trajectory would go through a low hanging beam it hits just fine and the weapon fires in a perfectly straight line.

Right...

I just wanted to, for fun, to try and see if I could calculate how much power there would be in a 150 lbs. stone/steel ball being flung from a trebuchet capable of flinging said ball 14400 ft. in 6 seconds. I found that it would have to have an average velocity of around 1600 mph. Which would imply the trajectory to be flat, the maximum range at a launch angle of 45 degree to be much farther that the 14400 ft. and the impact surely damaging more that 14d6. Like if it it damaged 14d6 per 1500 ft. which one could extrapolate from the damage of a standard heavy trebuchet in HoB well then a trebuchet hat could fire 14400 would damage 134d6...

So I just wanted to do the math for fun, but I found it to be quite difficult considering drag and ballistic curve and such.

[Baby Gary]

Yes I', trying to figure out how many joules the ball would impact with.

So are you saying that the top of the ballistic curve is 44.2 something over the ground? And is this considering that the ball to travel the 14.400 ft. in 6 seconds has to travel an average speed around 1600 mph (without drag)? This would mean, I think, that it would not be landing only under the influence of gravity, it would still has a lot of energy left...

I see your point but I disagree with it

first you only care about the vertical amount of energy that it would have, or so I got from what you said. When finding this out you don't care about the horizontal anything you just need to find the downward energy. Think about it if you throw a brick of a building it will still hit the ground at the same time if you just dropped it or gave it some horizontal acceleration.

also my reason why its peak is only 44.1 meters. you know that the entire flight takes only 6 seconds and that (assuming the flight path is a perfect parabola) the apex of the parabola is at 3 seconds. Also the only downwards force on the projectile is gravity. you know gravity is 9.8 meter/second^2 and that the ball goes from its highest point to the ground in only 3 seconds. just plug those numbers into the formulas that I used in my first post and you get what I got.

also one thing, all of my measurements will be in metric because thats how all the physics formulas work. just a heads up.

[edit] this is all assuming no air resistance and earth gravity (9.8 mps^2)

[Max Caysey]

I see your point but I disagree with it

first you only care about the vertical amount of energy that it would have, or so I got from what you said. When finding this out you don't care about the horizontal anything you just need to find the downward energy. Think about it if you throw a brick of a building it will still hit the ground at the same time if you just dropped it or gave it some horizontal acceleration.

also my reason why its peak is only 44.1 meters. you know that the entire flight takes only 6 seconds and that (assuming the flight path is a perfect parabola) the apex of the parabola is at 3 seconds. Also the only downwards force on the projectile is gravity. you know gravity is 9.8 meter/second^2 and that the ball goes from its highest point to the ground in only 3 seconds. just plug those numbers into the formulas that I used in my first post and you get what I got.

also one thing, all of my measurements will be in metric because thats how all the physics formulas work. just a heads up.

[edit] this is all assuming no air resistance and earth gravity (9.8 mps^2)

So are you saying if I drop a brick from 10 ft. 3m and trow a brick horizontally at 1600mph, 2600km/h, at exactly the same time, they will hit the ground at exactly the same time?

EDIT: Ok so I found THIS (http://www.shooterscalculator.com/ballistic-trajectory-chart.php?pl=%5BPreset+Name%5D&presets=&df=G1&bc=0.47&bw=1033968&vi=10000&zr=4800&sh=5&sa=0&ws=0&wa=90&ssb=on&cr=4800&ss=100&chartColumns=Range%7Eyd%3BRange%7Em%3BElevation%7E cm%3BElevation%7EMOA%3BElevation%7EMIL%3BWindage%7 Ecm%3BWindage%7EMOA%3BWindage%7EMIL%3BTime%7Es%3BE nergy%7EJ%3BVel%5Bx%2By%5D%7Em%2Fs&lbl=&submitst=+Create+Chart+), where I tried to type in the data. Weight of the ball, ballistic coefficient, and so forth, to try to get i to marry the 14400 in 6 seconds... One thing this calculator shows is how fast the initial velocity has to be, for the ball to reach 14000 in 6 seconds... More that 10000 ft/s it would seem!

EDIT: Looking at the chart as is, it seems that after 6 seconds the impact is 2 million joules. Would it be possible to survive that kind of impact to the torso?

Would you say these numbers looks right?

[JonU]

I see your point but I disagree with it

first you only care about the vertical amount of energy that it would have, or so I got from what you said. When finding this out you don't care about the horizontal anything you just need to find the downward energy. Think about it if you throw a brick of a building it will still hit the ground at the same time if you just dropped it or gave it some horizontal acceleration.

also my reason why its peak is only 44.1 meters. you know that the entire flight takes only 6 seconds and that (assuming the flight path is a perfect parabola) the apex of the parabola is at 3 seconds. Also the only downwards force on the projectile is gravity. you know gravity is 9.8 meter/second^2 and that the ball goes from its highest point to the ground in only 3 seconds. just plug those numbers into the formulas that I used in my first post and you get what I got.

also one thing, all of my measurements will be in metric because thats how all the physics formulas work. just a heads up.

[edit] this is all assuming no air resistance and earth gravity (9.8 mps^2)


You've over simplified how you got the overall height. In order to get the maximum height, you need to break the velocity into x and y vectors. Once you know those you can figure out the max height and overall distance covered. And the amount of work done would definitely have to include a horizontal component since it isn't traveling straight down but at an angle. I imagine the work done is actually higher than what you calculated.


So are you saying if I drop a brick from 10 ft. 3m and trow a brick horizontally at 1600mph, 2600km/h, at exactly the same time, they will hit the ground at exactly the same time?

EDIT: Looking at the chart as is, it seems that after 6 seconds the impact is 2 million joules. Would it be possible to survive that kind of impact to the torso?

Would you say these numbers looks right?

Your question about the bricks is 100% correct. If you dropped one brick and threw the other brick (actual velocity doesn't really matter) with a perfectly horizontal trajectory, they both hit the ground at the same time. Mythbuster actually covered this using bullets. I would also believe the 2 million joules would be pretty close to accurate.

https://youtu.be/tF_zv3TCT1U

[Necroticplague]

You've over simplified how you got the overall height. In order to get the maximum height, you need to break the velocity into x and y vectors. Once you know those you can figure out the max height and overall distance covered. And the amount of work done would definitely have to include a horizontal component since it isn't traveling straight down but at an angle. I imagine the work done is actually higher than what you calculated.
Well, to figure out the max height, you don't strictly need a horizontal component, just the vertical component's equation. The angle is already figured into the equation for the vertical component. IIRC, though it's been a few years, vertical component of velocity with respect to time is is gt+v*sin(theta) [assuming your g is negative; this basically breaks down to 'it's vertical velocity to start minus the decrease in vertical velocity that occurs due to gravity every second]. Set this equal to zero, you find the time when it's at it's maximum height (since, at that point, it's going horizontal). Plug the resultant t into the equation for the vertical displacement (.5g(t^2)+tv*sin(theta)+(initial height off ground)

Your question about the bricks is 100% correct. If you dropped one brick and threw the other brick (actual velocity doesn't really matter) with a perfectly horizontal trajectory, they both hit the ground at the same time.
Assuming that you're throwing on a flat plane. If you're throwing on a round one, the one thrown horizontally will land a bit later. Heck, if throw sufficiently fast horizontally on a round object, it might flat out never land.

[Yahzi]

So my heavy trebuchet can fire a 150 pound ball of (stone/metal) 14.400 ft., in one round.
What? No. Trebuchets had a range of 1000 ft, at best. Artillery that has its range measured in miles does not appear until WWI.

Edit: The Far Shot feat is incredibly stupid. Apparently you can take a feat that alters the laws of physics? And its explicitly not magical?

Far Shot should halve the range penalty for shooting, not impart extra energy to your projectiles (that somehow doesn't do extra damage).

[Anymage]

How much damage does the commoner rail gun do?

When you invoke magic and/or sketchy readings of RAW, trying to switch gears mid-discussion to RL physics creates a lot of broken things. This isn't a new discovery.

[Max Caysey]

What? No. Trebuchets had a range of 1000 ft, at best. Artillery that has its range measured in miles does not appear until WWI.

Edit: The Far Shot feat is incredibly stupid. Apparently you can take a feat that alters the laws of physics? And its explicitly not magical?

Far Shot should halve the range penalty for shooting, not impart extra energy to your projectiles (that somehow doesn't do extra damage).

Yes I know its silly, and as I've started with saying I know when magic is present physics goes away, but I still think its funny to see how screwed a siege engine gets when magic is added... like the fact that for it to actually reach 14400 ft. in 6 seconds the trebuchet would have to fire the ball at somthing like 11000 ft/s... Silly indeed.

Yes, its silly that damage is not increased... the regular trebuchet in HoB, would impact with about 114.000 joules at 1500 ft. My magic version would hit about 10 times as hard, but yes it doesn't deal 10 times as much, which it should, following its own logic (if one can even deduct any logic from the game at all).



How much damage does the commoner rail gun do?

When you invoke magic and/or sketchy readings of RAW, trying to switch gears mid-discussion to RL physics creates a lot of broken things. This isn't a new discovery.

I know... I just found the whole hyper sonic trebuchet really funny... And incidentally I am making an artillery unit for an upcoming game, so there were some use for it.

[Andreaz]

Yes I', trying to figure out how many joules the ball would impact with.

So are you saying that the top of the ballistic curve is 44.2 something over the ground? And is this considering that the ball to travel the 14.400 ft. in 6 seconds has to travel an average speed around 1600 mph (without drag)? This would mean, I think, that it would not be landing only under the influence of gravity, it would still has a lot of energy left...

If you just want the energy of the ball you can do it simpler than that with the same rounding assumptions.
Just calculate the velocity at the point of impact and use K = mv²/2.
Use the launch speed to find K, then add directly the potential energy from the difference "h" in height between launch and landing positions (U = mgh). You'll end up with the total energy being m (v²/2 + gh), with "gh" being positive if the landing point is lower than the launch point (this is true regardless of the Y axis being oriented up or down, g compensates for it). This is the maximum possible energy the projectile will have, and we're ignoring the dissipative phenomena.



The actual trajectory doesn't matter too much, you just need the launch angle to find the launch velocity. There are many ways to do it, but the simplest is based on the horizontal travel "d". d/6s = median horizontal velocity. Total velocity = horizontal velocity / cosine of launch angle to the ground. Unless your game is using the angle explicitly you'll just fiat it. Don't deviate too much from 45 degrees(pi/4 radians), that's the angle that throws the furthest.


And on drag... this is the realm of differential equations (in this case, one born from the F''(x) = aF'(x) relationship). Since Launches are almost always the same thing you might find a calculator or simplified equation for a given geometry, but I wouldn't count on it :\

[Max Caysey]

If you just want the energy of the ball you can do it simpler than that with the same rounding assumptions.
Just calculate the velocity at the point of impact and use K = mv²/2.
Use the launch speed to find K, then add directly the potential energy from the difference "h" in height between launch and landing positions (U = mgh). You'll end up with the total energy being m (v²/2 + gh), with "gh" being positive if the landing point is lower than the launch point (this is true regardless of the Y axis being oriented up or down, g compensates for it). This is the maximum possible energy the projectile will have, and we're ignoring the dissipative phenomena.



The actual trajectory doesn't matter too much, you just need the launch angle to find the launch velocity. There are many ways to do it, but the simplest is based on the horizontal travel "d". d/6s = median horizontal velocity. Total velocity = horizontal velocity / cosine of launch angle to the ground. Unless your game is using the angle explicitly you'll just fiat it. Don't deviate too much from 45 degrees(pi/4 radians), that's the angle that throws the furthest.


And on drag... this is the realm of differential equations (in this case, one born from the F''(x) = aF'(x) relationship). Since Launches are almost always the same thing you might find a calculator or simplified equation for a given geometry, but I wouldn't count on it :\

Thanks...

Its more math than I have had in a lot of years, but I'll give it a go! Feel free to plot in the necessary numbers to get a 150 lbs. stone ball 14400 ft. in 6 seconds. :smallbiggrin:

[InvisibleBison]

Edit: The Far Shot feat is incredibly stupid. Apparently you can take a feat that alters the laws of physics? And its explicitly not magical?

Far Shot should halve the range penalty for shooting, not impart extra energy to your projectiles (that somehow doesn't do extra damage).

Far Shot is a feat, feats are extraordinary abilities, extraordinary abilities can break the laws of physics. What's the problem?

[Kayblis]

The "hit in 6 seconds" thing is a game simplification. Thebuchets and Ballistas in real life can take far longer for its ammo to hit its target. Obviously, if you try to do math using a game simplification as absolute truth, you'll get stupidly incorrect values. Siege weapons are supposed to be used in game with an abstraction of time - it's more about "one shot every X rounds" than "the shot I fired this round MUST collide with the target this round". Anything using sieges shouldn't be quick enough or low-scale enough for that abstraction to make any sensible difference.