StarDestroyer.Net BBS

I have a question about the trap the Ewoks used to destroy one of the AT-STs -the trap with two giant redwood logs smashing together. What kind of force is created when two enormous logs like that slam together? Could that kind of force crush a real-world vehicle?

mass x height x force of gravity x angle gets you the amount of force applied.

We don't know the mass and we have to assume the force of gravity is the same, but I'm sure someone can work out a ballpark figure.

Let's assume the logs are ebony wood at its maximum density of 83lb/cubic foot.
Further assume the logs are 3 feet diameter and 40 feet long.
So with a volume of pi(r2)h, we have a mass per log of 1,131 pounds.
Assume the ropes are 50 feet long, and the arc of rotation is 90 degrees for simplicity.
Also assume one g or approx. 32.1 feet per second per second
Using the formula L=Iw (w=omega for angular velocity), we get:
L=(integral 0 to 50 r2dm)(|v|sintheta/|r|)
L=(2500*1131)(32.1*1*50)
L=4,538,137,500 pound-feet or, assuming the log faces are flat, 160,503,781 pounds per square inch. Per log. Total force of about 9 million pound-feet, over a 19 square foot area.

Oh shit, that looks awfully high. It's been too long since college physics and calculus. Surlethe, help? The Ewok attack looks plausible. Please prove me wrong.

OK, let's try Samuel's formula:

1131*50*32.1*1[sin 90]=1,815,255 pounds per log, 3.7 million pounds.

Now to look at the tensile strength of 304 grade stainless steel: 515 megapascals, or 74,675psi.

With a cross sectional impact area of 804 square inches per side, 304 stainless of 1 inch thickness can withstand 60,057,198 psi [EDIT] per side [/EDIT] uniformly applied before deformation.

Hallelujiah! Even using my original (probably f-d up) calcs, it looks like the "2 Logs=Walker Ki11" ROTJ brainbug is a non-starter.

Wouldn't the force of one log been enough to knock an AT-ST over?

Not that the movie's formula of 2logs + 1AT-ST = 1explosion wasn't a fun scene.

Samuel wrote:mass x height x force of gravity x angle gets you the amount of force applied.

We don't know the mass and we have to assume the force of gravity is the same, but I'm sure someone can work out a ballpark figure.

Last time I checked the force was a time derivative... maybe physics improved in the last five years. What you have is a gravitational potencial energy (mass x height x gravity acceleration) the angle I do not understand.

What's going on in this thread? Pound-feet is not a unit of force, nor is potential energy. If you want to know the force of collision, you need to take the time of collision into account. Take the change in the log's momentum (from a lot to zero), and divide it by how long the collision takes, and there's your (average) force.

For the best figures, the velocities and times should be measured, but if you want to approximate it, the maximum velocity of a pendulum can easily be calculated by a potential energy to kinetic energy conversion, and so would be
v = sqrt(2gh)
where h is the height the pendulum was lifted to.

So if we take Chocula's numbers; 513 kg dropped from a height of 15 meters (let's assume the "pendulum" here was at 90 degrees), and a collision time of 10 ms, we get
F = (sqrt(2 * 9.82 * 15) * 513) / 0.01
~= 881 000 N
per log.

Note that this is just the average force. The maximum force can be much higher than this. If we assume that the force-time graph looks like a triangle, it would be twice as big.

Bad math is bad. I assumed instantaneous stopping and failed to do the force conversions. Must. Review. Physics.

Dooey Jo wrote:What's going on in this thread? Pound-feet is not a unit of force, nor is potential energy. If you want to know the force of collision, you need to take the time of collision into account. Take the change in the log's momentum (from a lot to zero), and divide it by how long the collision takes, and there's your (average) force.

For the best figures, the velocities and times should be measured, but if you want to approximate it, the maximum velocity of a pendulum can easily be calculated by a potential energy to kinetic energy conversion, and so would be
v = sqrt(2gh)
where h is the height the pendulum was lifted to.

So if we take Chocula's numbers; 513 kg dropped from a height of 15 meters (let's assume the "pendulum" here was at 90 degrees), and a collision time of 10 ms, we get
F = (sqrt(2 * 9.82 * 15) * 513) / 0.01
~= 881 000 N
per log.

Note that this is just the average force. The maximum force can be much higher than this. If we assume that the force-time graph looks like a triangle, it would be twice as big.

So how flimsy would the armor on an AT-ST have to be for these logs to crush it?

If my calcs are anywhere in the ballpark, 1 inch of 20th century aircraft-grade stainless steel wouldn't deform if hit with the Trees Of Doom. Maybe 1/2 inch of mild steel equivalent is the lower bound, but since I've just shown that I've forgotten most of what I knew, I'll leave it for others to do the math.

Count Chocula wrote:Hallelujiah! Even using my original (probably f-d up) calcs, it looks like the "2 Logs=Walker Ki11" ROTJ brainbug is a non-starter.

Are you trying to suggest that it's impossible?

Unfortunately, we can't have that: Canon is canon. If it were much weaker than mild steel, that's just the way AT-STs are; perhaps their armour is some weird ceramic composite that resists blasters but snaps easily.

Not a bad explanation, though perhaps AT-STs are just fragile, mass produced shit that was never meant to survive serious force. They're scouts, after all, not tanks.

^ I'm not sure that's good enough, because it takes about four shots from AT-ST heavy blasters to take one out. It probably is cheapo armor for Star Wars tech, but still somewhat blaster-repellent.

Fair enough. But the fact that it breaks so easily when hit by a tree still suggest its a weak piece of crap, doesn't it? Surely Star Wars tech could make armor that could stop a blaster but is also impact resistant?

Actually, your refference to AT-ST firepower has got me thinking: has anyone done firepower calcs on a scout walker? Would it be possible to do them from the shots of walkers blasting trees on Endor, or are their any canon statements on firepower?

Sorry if the question is a bit of a highjack.

Edit: fixed typo.

Should we take into account that those logs smashed right into the grenade launchers on the sides of the walkers? Could that have contributed something to the damage done?

If you have the DVDs I suggest watching it frame by frame.

One frame the logs impact. The sides of the walker seem to bow inwards in roughly about a frame or so (might be a few frames). I am not sure if this is proof of "buckling" due to stress or just some give in the material as part of the design (or just because they are that flexible) The scene is obscured by an explosion. Once the epxloison fades, the walker head is crushed.

Make of that what yuo will. I'll comment more later.

Just for reference...

At 7:14, an AT-ST is shooting and hits the trees:



At 03:07 Chewbacca's AT-ST fires on the other and at at 4:05 the logs are used to strike the sides of an AT-ST:



Note that the ends of the logs are rounded and not flat, for what it is worth. Also, right after the log-smashing, another AT-ST is tripped up in a log-slide and its cockpit explodes as soon as it tumbles to the ground.

After NecronLord's observation, and reviewing the footage, you're right: canon wins. My mistake was assuming "big boxy scary looking thing with big guns and grenades" = "Armor Max."

Whatever the force involved was (I'll go with 880,000N) it did the job. I shoulda looked at the AT-STs as Humvees with TOWs, not as Bradley AFVs.

According to wookiepedia, there's an up-armored version of the AT-ST that acts more in the armored support role.

It's easy to confuse the AT-ST as the imperial equivalent to a light armored vehicle, since that's how pretty much every SW game portrays them, but most EU sources have them used as scout units for heavier hover-tanks or walkers.

Quick notes, don't have time to do the full maths:

Endor's gravity is, while near Earth's, at least somewhat lower. The hang-gliders that the Ewoks are riding around in aren't nearly big enough to produce enough lift to hold an Ewok and some big heavy rocks up. Either gravity's significantly lower, or the atmosphere is inexplicably much denser.

Getting back to the fragility of the AT-STs, when the one falls on its side after stepping on the logs that are rolled down to it, it blows up just as it hits the ground. Maybe those weapons on the side do indeed touch off whatever it is that is exploding inside.

It's interesting, the one Chewbacca destroys flares quite brilliantly (as if it is full of flash paper!) for a few moments (more so than the one felled by rolling logs), after what looks like a secondary explosion. The AT-STs power cells are to the rear, so perhaps he happened to strike it at just the right place which resulted in such a vigorous explosion. Note that some internal structure (turntable?) remains above the "hip" assembly after the explosion.

Terralthra wrote:Quick notes, don't have time to do the full maths:

Endor's gravity is, while near Earth's, at least somewhat lower. The hang-gliders that the Ewoks are riding around in aren't nearly big enough to produce enough lift to hold an Ewok and some big heavy rocks up. Either gravity's significantly lower, or the atmosphere is inexplicably much denser.

The gravity can't be significantly lower then the Earth's. There are too many examples of objects, and people falling at the same rate they would on Earth.

My knowledge of materials science is fairly insignificant, so bear with me if this is a stupid question, but is there any way that walker's weak armour could be chalked up to some assembly flaw or use of substandard materials in that particular vehicle that quality control somehow missed? The evidence shows that that particular AT-ST was that crappy; can we assume that it is also representative of its class overall? Can anyone think of any counter-examples of better performance in the EU?

as long as you don't you the density of the trees themselves, there's no point in making a mathematical calculation. they are alien trees after all, and could just as well be 4 or 5 times as dense as our standard earth tree^^

FSTargetDrone wrote:Getting back to the fragility of the AT-STs, when the one falls on its side after stepping on the logs that are rolled down to it, it blows up just as it hits the ground. Maybe those weapons on the side do indeed touch off whatever it is that is exploding inside.

That is the side with the grenade launcher, so it's possible if the grenades are preprimed to go off on impact.