Knockback on the moon?

[bigdamnhero]

I'm writing a sci-fi adventure set partly on the moon. `Tho it's a heroic game, I thought it'd be funny to use knockback to reflect people getting knocked around in the reduced gravity. It's a total space opera game, so I'm not too fussed about being realistic. But my logical brain got curious, and I tried to figure out how it might actually work in the real world.

 

I failed.

 

I found a study where they measured olympic boxers as punching with a force of 3427 Newtons (standard deviation of 811). Obviously that's not typical, but we'll take it as a heroic high-end. I remember F=ma from my "Physics for Poets & Lovers" course in college, but does that mean hitting a 100kg target would impart 342 m/s^2 acceleration to the target? That's 34G!? Even allowing for imperfect energy transfer, I'm clearly missing something; probably several things. Anyone else care to take a stab at it?

 

 

Thanks,

bigdamnhero

"Hell I dunno, if I wanted schoolin' I'd have gone to school."

[Cancer]

Re: Knockback on the moon?

 

That 34 G sounds right. I recall reading a technical journal paper saying that a solid Olympic-class jab to the jaw can induce an (instantaneous) acceleration for the target's head of 300g (yes, three hundred times Earth standard gravity). The head's smaller than the body by (very roughly) a factor of 10, so the 34g number seems OK.

[Jaxom]

Re: Knockback on the moon?

 

You could do some simple first-order approximations just using the ratio of gravities...

 

If your assumption is that on Earth knockback is "stopped" by friction with the ground then that friction is directly proportional to the gravity as I recall (been a while so that may be wrong). If I'm right there, then friction against the same surface on the moon would be less by the same factor as the gravity decrease, 1/6th (again, if memory serves). That'd mean that to first order, knockback should be about 6 times greater.

 

There are, of course, other factors that would be more important in some cases.... Lunar "soil" is a fine powder and as I recall it has lower coefficient of friction than, for example, concrete or asphault. Same problem would apply if you decided to have a fight on a glacier (or ski-slope in some cases)... Tim, since I mention this where you can find it, standard rule applies and you and Garry cannot use this on Houngan first. =P

 

Of course, in zero-G you get zero friction, obviously, so you would fly backwards until you hit something. This could lead to some bad situations if super-tank punches someone through the bulkhead... Eep.

 

Remember that the rule that breaking things while flying backwards would *not* change for gravity so you lose momentum normally while flying through furniture on the moon. 3" of "normal" knockback would mean 1", then through something for 1 BODY (or was it 1d6 BODY) and then another 1" back... On the moon that would equate to 6", same damage for flying through the chair and then another 6" knockback.

 

Generally speaking though, you will probably be a lot happier if you don't think a whole lot about this as it is going to change the flavor of things a *lot*. It'll become very much like some Manga where the opponents square up from 20" apart and then charge at each other, fly past one another (or apart anyway) turn and repeat. This will *strongly* favor anyone using ranged weapons (unless you apply something similar to recoil in which case you learn why people should grapple in zero-G combat).

[Basil]

Re: Knockback on the moon?

 

I'm writing a sci-fi adventure set partly on the moon. `Tho it's a heroic game, I thought it'd be funny to use knockback to reflect people getting knocked around in the reduced gravity. It's a total space opera game, so I'm not too fussed about being realistic. But my logical brain got curious, and I tried to figure out how it might actually work in the real world.

 

I failed.

 

I found a study where they measured olympic boxers as punching with a force of 3427 Newtons (standard deviation of 811). Obviously that's not typical, but we'll take it as a heroic high-end. I remember F=ma from my "Physics for Poets & Lovers" course in college, but does that mean hitting a 100kg target would impart 342 m/s^2 acceleration to the target? That's 34G!? Even allowing for imperfect energy transfer, I'm clearly missing something; probably several things. Anyone else care to take a stab at it?

3427 Newtons/100 kg = 34.27 m/s/s

 

You're off by a factor of 10.

 

BTW, that's about 3 1/2 g.

 

Oh, and don't forget that when hitting something like a human body, a lot of the energy goes to compressing the body and other losses, so the accelleration will be significantly lower than 3 1/2 g.

[bigdamnhero]

Re: Knockback on the moon?

 

3427 Newtons/100 kg = 34.27 m/s/s

 

You're off by a factor of 10.

 

BTW, that's about 3 1/2 g.

Um...well...when you put it like that... OK, now I feel REALLY dumb.

 

3.5Gs I can believe. Thanks, Basil!

 

Oh' date=' and don't forget that when hitting something like a human body, a lot of the energy goes to compressing the body and other losses, so the accelleration will be significantly lower than 3 1/2 g.[/quote']

And I know that "a lot" of the energy would be absorbed by the body, but I have no idea how much. 10%? 50%? 85.275%?

 

 

bigdamnhero

"...and they made a very satisfying thump when they hit the floor."

[bigdamnhero]

Re: Knockback on the moon?

 

If your assumption is that on Earth knockback is "stopped" by friction with the ground then that friction is directly proportional to the gravity as I recall (been a while so that may be wrong). If I'm right there, then friction against the same surface on the moon would be less by the same factor as the gravity decrease, 1/6th (again, if memory serves). That'd mean that to first order, knockback should be about 6 times greater.

Generally speaking though, you will probably be a lot happier if you don't think a whole lot about this as it is going to change the flavor of things a *lot*.

But this is, as I said, a Heroic game which means we're not normally using knockback at all. I'm just looking to see how closely using the knockback rules might simulate a fight in low gravity. In other words, I want to change the flavor of things; I'm just seeing if this is the direction I want to go.

 

Thanks for pointing out that knockback damage would NOT be increased by the low gravity. (And you remembered right, BTW: lunar gravity is 0.16 G.)

 

 

bigdamnhero

“I’m a busy man and I can’t be bothered to punch you at the moment. Here is my fist; kindly run towards it as fast as you can.â€

[Matt Frisbee]

Re: Knockback on the moon?

 

Don't forget two other issues in the problem -- event duration and inertia of the hit object. 3.5 G applied for 1/20th second on a mass of 75 kg won't move it very much.

 

Matt

[Manic Typist]

Re: Knockback on the moon?

 

To keep things simple and straight forward, I might recommend this that you divide Earth normal gravity by the Moon's (in this case .16), then take that number you get and use it to multiply all knockback distances.

 

I think that would be sexy. Just handwave everything else.

[SteelDoom]

Re: Knockback on the moon?

 

I wouldn't change the damage from knock back. While you would float further your mass hasn't changed and the force hitting you should give you the same velocity on the earth or the moon. It could get kinda complicated so I agree with the above poster about finding some kind of aproximation. Maybe one hex gets converted to two hexes or something like that.

[massey]

Re: Knockback on the moon?

 

You normally roll 2D6 to determine KB. On the moon, I'd have people roll 1D6. Roll the extra D6 to reduce damage from impacts, but not distance travelled.

 

Math? Bah. Math is for smart people. Hand-waving is good enough for me. It's space opera, you said, so real math figures are unimportant.

[Basil]

Re: Knockback on the moon?

 

To keep things simple and straight forward, I might recommend this that you divide Earth normal gravity by the Moon's (in this case .16), then take that number you get and use it to multiply all knockback distances.

 

I think that would be sexy. Just handwave everything else.

However, since the OP's game usually doesn't use knockback, that becomes 1/.16 * 0, which = 0.

 

I'd give all attacks a free "Does Knockback" and let it go at that. The KB rules don't model reality very well (being, deliberately, very cinematic), so any KB is fairly space-opera-y anyway.

 

If you want, you could give all attacks a fre "Double Knockback" and make things even wilder. IMO, that's about as far as I'd go, even in space-opera. YMMV.

[Manic Typist]

Re: Knockback on the moon?

 

Sounds good to me.

 

Just don't forget. Anything sufficient to cause KB, will cause KB onto BOTH parties unless one is properly braced.

 

Gotta remember, punching someone in space is gonna send you flying the same as them.

[BlackSword]

Re: Knockback on the moon?

 

However, since the OP's game usually doesn't use knockback, that becomes 1/.16 * 0, which = 0.

 

I'd give all attacks a free "Does Knockback" and let it go at that. The KB rules don't model reality very well (being, deliberately, very cinematic), so any KB is fairly space-opera-y anyway.

 

If you want, you could give all attacks a fre "Double Knockback" and make things even wilder. IMO, that's about as far as I'd go, even in space-opera. YMMV.

Grrr, Basil beat me to it.

 

Since the game normally doesn't use knockback (just knockdown), on the moon I would actually use the knockback rules (so there would be a difference between fighting Earth gravity vs Moon gravity). If you are going to add Mars into the mix, then Mars can use normal knockback, and on the moon either subtract a d6 or use Double Knockback.

 

Using normal knockback rules I would go with double knockback on the moon.

 

Handwave all the way.

[Jaxom]

Re: Knockback on the moon?

 

But this is' date=' as I said, a [u']Heroic[/u] game which means we're not normally using knockback at all. I'm just looking to see how closely using the knockback rules might simulate a fight in low gravity. In other words, I want to change the flavor of things; I'm just seeing if this is the direction I want to go.

 

Thanks for pointing out that knockback damage would NOT be increased by the low gravity. (And you remembered right, BTW: lunar gravity is 0.16 G.)

 

Well, my point wasn't so much to suggest using six times super-heroic knockback but to suggest the basic and appropriate physics that you might want to use. Considering normal scale of 1" equals 2 yds and then looking at a normal boxing match here you could assert that in Earth-norm a (strong) trained fighter can cause as much as 1" knockback if he hits an opponent off-guard then you'd want something that could cause as much as 6" knockback on the moon. I'd probably make it more than that because you can also easly lift them off the ground which eliminates friction all together.

 

If you wanted to resort to actual rules from the book I'd be tempted to institute normal super-heroic knockback since that is handy and gives numbers that aren't too wonky or to look at using some varient of the throwing rules based on STR and target mass (I don't happen to have these at hand but you could swipe from Ultimate Brick if you have it).

[Basil]

Re: Knockback on the moon?

 

Sounds good to me.

 

Just don't forget. Anything sufficient to cause KB, will cause KB onto BOTH parties unless one is properly braced.

 

Gotta remember, punching someone in space is gonna send you flying the same as them.

Well, the setting isn't "space", but the moon, so this isn't quite as much a problem. However, you do have a point; the puncher can suffer if not properly braced. I, personally, would say an unbraced puncher can do Knockdown to himself, and apply a little handwavium to say the Knockdown takes the form of the person landing up on the floor with his feet where they started and the rest of his body stretched out away from the (original location of the) person he hit. Adjusted for realism and dramatic sense, of course.

 

All that is YMMV, of course.

[ajackson]

Re: Knockback on the moon?

 

A realistic punch might be in effect a 1 kilogram impactor moving at 10 meters per second. Even on the moon, that's neglible motion. Essentially the only attacks which can realistically cause knockback are ones with the full weight of the body behind them, such as a tackle or slam in football.

 

Ignoring the realism of knockback for a moment, someone knocked back on a flat trajectory will go sqrt(1/0.16) = 2.5 times as far on the moon as on earth (and will then skid along the ground, possibly for a very long distance). If knocked back on an arching trajectory, they will go 6x as far.

 

On earth, knockback (flat) would be about 1 hex per 5 m/s velocity; knockback (45 degree arc) would be V^2/20 hexes.

[Manic Typist]

Re: Knockback on the moon?

 

Well' date=' the setting isn't "space", but the moon, so this isn't quite as much a problem. However, you do have a point; the puncher can suffer if not properly braced. I, personally, would say an unbraced puncher can do Knock[b']down[/b] to himself, and apply a little handwavium to say the Knockdown takes the form of the person landing up on the floor with his feet where they started and the rest of his body stretched out away from the (original location of the) person he hit. Adjusted for realism and dramatic sense, of course.

 

All that is YMMV, of course.

 

Yeah, but "The vacuum of moon" doesn't sound nearly as sexy as "the vacuum of space."

 

And I was being lazy in my answer, 'cause I knew you would know what I meant. Besides, they have to get to the moon somehow right? Maybe there's a fight along the way....

 

YMMV?

[bigdamnhero]

Re: Knockback on the moon?

 

Thanks for your help, all. I agree that using the KB rules as written is close enough to what I want to achieve. Give the characters some extra leaping and call it good. Or good & silly anyway.

 

But I was kinda curious what the “real†numbers might look like, if this were my usual hard-SF game instead of camp space opera. So I tried running a few numbers. Those allergic to math can skip the blue text. Those who feel like checking my assumptions and methodology, I’d be much obliged. I pulled a number of assumptions out of this air, so if anyone has any better ones please feel free to let me know. (I’m not a physicist, nor do I play one on TV.)

 

Olympic punch = 3400 N

Let’s assume that half of the force is absorbed by the elasticity of the target body

so 1700 N are transferred to acceleration.

Target body = 100 kg

a = f / m, so a = 1700 N / 100 kg = 17 m/s2

Let’s say the impact lasts 0.1s to make the math easy:

Initial Velocity VI = a * t = 17 m/s2 * 0.1s = 1.7m/s

Let’s say the blow was delivered at a 45-degree angle up and back.

So 1/2 of the velocity (0.85m/s) would be vertical and the other half is horizontal.

 

On Earth:

Looking at the vertical component first

0.85 m/s ÷ 9.8m/s2 = 0.087s for gravity to decel vertical velocity to 0 (ie – apex of move)

Total time airborne = 2 * 0.087 = 0.17s until tgt’s feet are back on the ground

Average velocity (vertical) is 0.85m/s * 1/2 = 0.43 m/s

0.43 m/s * 0.087s = 0.037 m = 3.7 cm vertical displacement, ie apex of move

Horizontal velocity stays fairly constant (ignoring air drag) until feet are back on the ground

0.85 m/s * 0.17s = 0.148 m = 14.8 cm horizontal displacement

So target is knocked backwards in a parabola, 3 or 4 centimenters up and 15 centimeters back, and is airborne for 1/5th of a second.

 

Not much, as you would expect. In reality, boxers don’t lift their opponents off the ground with every punch, which probably means that my assumptions were crap. (Or they all have KB Resistance!) Most likely more than half of the force is absorbed by the elasticity of the body, and the blow lasts less than 1/10s? Anyway, while I’ve never been hit hard enough to lift me completely off my feet, I’ve certainly been knocked backwards more than 15cm, so in game terms we’ll call it close enough.

 

Meanwhile, on the moon:

Gravity is 0.16G = 1.57 m/s2

Force, mass and starting velocity are all unchanged

0.85 m/s ÷ 1.57 m/s2 = 0.54s upward move

Total time airborne = 2 * 0.54 = 1.08s

Average velocity (vertical) is still 0.85m/s * 1/2 = 0.43 m/s

0.43 m/s * 0.54s = 0.23 m = 23 cm vertical displacement (apex of move)

Horizontal velocity stays fairly constant until feet are back on the ground

0.85 m/s * 1.08s = 0.92 m = 92 cm horizontal displacement

So target is knocked backwards in a parabola, 23 centimenters up and 92 centimeters back, and is airborne for just over a second.

 

Still only about 1/2" in scale. Kinda disappointing, really.

 

 

bigdamnhero

“There's no such thing as Silicon Heaven.â€

“Then where do all the calculators go?â€

[Basil]

Re: Knockback on the moon?

 

Yeah, but "The vacuum of moon" doesn't sound nearly as sexy as "the vacuum of space."

 

And I was being lazy in my answer, 'cause I knew you would know what I meant. Besides, they have to get to the moon somehow right? Maybe there's a fight along the way....

Yeah, fights in "zero-G" would get interesting.

 

YMMV?

Your Milage May Vary

IWO, you may have a different result/POV/answer/way of doing things.

[Jaxom]

Re: Knockback on the moon?

 

Well, the brief answers to your analysis are that different punches are more or less than 45 degrees. Also, you're imparting angular momentum to an object which has friction only at one end (i.e. a standing person) so if it is enough to knock you off balance you are going to step backwards to regain balance and perhaps stumble backwards depending on the actual reaction time. As a result, airborne travel is not the best approximation to compare to real-world stuff probably.

 

Likewise, a punch is not the best way of getting someone airborne and in motion... You'd do better to look for judo or aikido throws.

 

Of course, in a discussion about this thread at gaming last night it was also correctly pointed out that comicbook knockback which is suffcient to create a crater in the pavement way over there (on impact) would also impart the same amount of force to the guy making the punch which means tearing up whatever he is standing on. Probably better to decide what you think is appropriately cinematic and leave it at that...

[Guest HeroPink!]

Re: Knockback on the moon?

 

I saw the title, I had a "The Police" flashback.

 

"I hope my spine don't break

Knockback on the moon..."

[pinecone]

Re: Knockback on the moon?

 

I saw the title, I had a "The Police" flashback.

 

"I hope my spine don't break

Knockback on the moon..."

I got the same thing....

"An extra phase to land you'll take...knockback on the moon...."

[Fuzzy Gnome]

Re: Knockback on the moon?

 

I got the same thing....

"An extra phase to land you'll take...knockback on the moon...."

 

I took knockback on the Moon!

I got hit by a harpoon!

It's an RKA

But I don't weigh

Enough to be immune!

[bigdamnhero]

Re: Knockback on the moon?

 

I saw the title' date=' I had a "The Police" flashback. [/quote']

That's what I was thinking when I wrote it, but I couldn't think of any cool lyrics.

 

 

bigdamnhero

"He's not the first psycho to hire us nor the last. You think that's a commentary on us?"

[ajackson]

Re: Knockback on the moon?

 

Most likely more than half of the force is absorbed by the elasticity of the body' date=' and the blow lasts less than 1/10s?[/quote']

 

Dramatically under 1/10s. Let's assume that the punch starts out at 10 m/s velocity and slows down to 0m/s, for an average of 5 m/s.

 

Now, how far does the target have to move before the punch stops? For striking a punch meter, call it 5 cm.

 

So, the punch travelled 5cm between initial impact and stopping, at an average velocity of 5 m/s. Therefore, total time is 0.01 seconds.

 

Those numbers may not be especially accurate, but they're within the right order of magnitude.