lets maths...

[Alrik Vas]

got a question for you physics guys. a glitter boy weighs about 2 tons, is almost 12ft tall with human like proportions. the gun fires from the shoulder. when the recoil suppression is not engaged, it flies back 30ft almost instantly.

can we calculate what the actual muzzle velocity of the round is based off this information?

if its been done before, point me toward said thread.

[ShadowLogan]

Well the books state Mach 2 or 5 (depending on RMB or RUE) for the velocity of the slug, they also state the stock Glitterboy is 1.2tons fully loaded not 2 (not saying there might not be a 2 ton glitterboy somewhere, just not the the stock main book one).

Math wise I think it should look like this since we are talking about momentum really. We also would need to know what the mass of the fired round is, not to mention how long "near instantly" qualifies as in terms of a real number:

[round velocity] * [round mass] = [glitterboy mass] * ([glitterboy distance] / [glitterboy time])

[Alrik Vas]

1.2 tons, moved 30ft in 0.5 seconds.

I could be wrong, but I don't think we're not talking about momentum. I'm asking if the muzzle velocity of the round based on the above can be determined, not the printed speed of the round...I think Mach 5 is too slow. The Navy's got stuff that goes Mach 7 and produces under 32MJ of energy.

I got to thinking about it as at one point we figured out that the average laser rifle blast is 15MJ based on a 20 shot magazine (and if an eclip is still 1000 car batteries, or whatever it was) so taking those numbers I got to thinking that something like a Boom Gun, that can shred some light power armors in a single shot and can get most PA's with a good critical hit, has GOT to be producing way more juice than that.

but, disclaimer, I'm a knuklehead and don't necessarily know how it all works. That's why I posed the question.

[drewkitty ~..~]

To answer your specific question the answer is: Yes, it can be...sort of. The sort of is because the only thing we have a mass measurement for is the mass of the Glitterboy, and rest of the measurements would have to be assumptions. Sort of like answer the question "what is the friction coefficient between the ground and the GB when it is Knocked Down and knocked back from firing it's boom-gun while standing without it's stabilization system?"...I have no idea how to even WAG that answer.

Have people tried. Yes. But that was ages ago and is pretty sure whatever posts related to those threads have been deleted from the PB server. (talking about ten-twelve years ago.)

-----------------------------------
The little bit I did calculations for was pre- RUE and was on "what happens on the other end?" or "what are the side effects of getting shot by a boom-gun?"


BG Gun Pod (936 lb.):
Ranges: 11,000' Flechette/ 7000' Slug.
3d6x10 MD & SB flechette / 2d6x20 MD & knock back Slug.
payload--40 rounds.
Rate of Fire: single shots only
The gun pod was designed for Bots greater then 20', weighing greater then 10 tons. Due to the rifle like construction of the BG Gun pod, it looks much like an over sized Tommy Gun with the ammo stored in the forward circular mag. Flechette ammunition is the normal boom-Gun ammunition, while Slug ammunition is a single projectile.

A boom-gun slug looses virtually none of the kinetic energy that was used to launch in with in its total effective range, thus what ever is hit receives a knock back effect.
If hit by a Boom-gun slug:
♠ a 1.5 ton object will be knocked back 30',
♣ tanks have a 40% chance of being rolled over,
♥ Bots have a 50% chance of being Knock down,
♦ PA will be Knocked Down unless the pilot makes a 'roll with' roll, even if they make their 'Roll with' they will loose init and -1 APM.
❖ A man sized object will be thrown back 300'+2d6x10' and SDC/HP beings wearing MDC armor have a 70% chance of being KO'ed for 1d6 min, , man-sized MDC beings get a 60% of being KO'ed for 2d6 melees. For both SDC/HP and MDC beings saving vs 58% (using the coma bonuses) means being KO'ed for only 1d4 melees.

[Lenwen]

A boom-gun slug looses virtually none of the kinetic energy that was used to launch in with in its total effective range, thus what ever is hit receives a knock back effect.
If hit by a Boom-gun slug:
♠ a 1.5 ton object will be knocked back 30',
♣ tanks have a 40% chance of being rolled over,
♥ Bots have a 50% chance of being Knock down,
♦ PA will be Knocked Down unless the pilot makes a 'roll with' roll, even if they make their 'Roll with' they will loose init and -1 APM.
❖ A man sized object will be thrown back 300'+2d6x10' and SDC/HP beings wearing MDC armor have a 70% chance of being KO'ed for 1d6 min, , man-sized MDC beings get a 60% of being KO'ed for 2d6 melees. For both SDC/HP and MDC beings saving vs 58% (using the coma bonuses) means being KO'ed for only 1d4 melees.

I like this (as being a logical course of action) so much I will now be utilizing it in my games.

[Nekira Sudacne]

got a question for you physics guys. a glitter boy weighs about 2 tons, is almost 12ft tall with human like proportions. the gun fires from the shoulder. when the recoil suppression is not engaged, it flies back 30ft almost instantly.

can we calculate what the actual muzzle velocity of the round is based off this information?

if its been done before, point me toward said thread.

You are discounting the weight of the pilot in that. that can add another 1/10 of a ton on average. If you want to get litearl, a metamorphasis dragon climbed into a glitter boy would increase it's weight by 30 tons, which could arguably negate the knockback and need for pylons altogeather

[Thinyser]

Its pretty straight forward if you know the projectile's mass which we don't but we can make several calculations to get an idea.

If you assume the 1.2 ton unit gets tossed back 30' in .5 sec thats 60'/sec as an impulse velocity. To make this math easier it is best to convert to metric as joules are simply newton-meters. So that said 1.2 tons is 1088.62 Kg (lets assume there is some mass from other gear like small arms and eclips and junk that would make this an even 1100) 60' is 18.28m (call that 18.25) So we are moving 1100kg at 18.25m/s.

Here is a calculator http://www.1728.org/energy.htm
Which says that ends up at 1.8318e x10^5 Joules (which is 183180 Joules for those that don't understand scientific notation)

Using that calculator to solve for velocity assuming some various values for the projectile mass (since I don't know what it is) here is what you get.
.1Kg = 1914m/s or 4281 mph (which is 6.48 MACH) this is totally unrealistic as each of the 200 flechettes would weigh only .5grams which is half of what a large paperclip weighs.
.5Kg = 855.99m/s or 1914 mph (which is 2.9 MACH) 500g/200 flechettes= 2.5 grams each (also not realistic to do MD as they would not carry their momentum for the 2 mile range)
1Kg = 605.28m/s or 1354 mph (which is 2.05 MACH) 5g per flechette (this is the weight of a U.S. Nickel) and even going that fast its not gonna deal MD
2Kg = 428m/s or 957 mph (which is 1.45 MACH) 10g per flechette, now we're talking about something that "might" be able to carry its momentum 2 miles but it would only be going at most the starting 428m/s which will yield at most 915 joules or 675 ft/lbs of energy per flechette to the target that is comparable energy to a .357 magnum revolver and that is assuming it did not lose any velocity to drag. Likely it would be much lower energy delivered to target per flechette.

So as you can see assuming the mass of the 200 flechettes is about 1kg the speed they would be shot is about MACH 2. BUT the flechettes are too light to carry their energy downrange.

This is assuming that all the recoil is directly converted to movement in the Glitterboy and none is lost as heat/sound/friction (I assume it gets thrown to the ground and skids a few feet at the end). If energy is lost (which we know some will be) then there was more energy used to move it back those same 30 feet and that means more energy was imparted to the projectile since the reaction must be equal yet opposite. This means that there could be a few thousand more joules used to fire the boomgun and a somewhat faster projectile. MAYBE even as much as double? I don't know how much overall might be lost even assuming more than double (say 400,000 joules) would send a 2kg downrange at 632.46m/s or 1415 mph (about 2.5 mach) and each 10g flechette would have 2000 joules or 1475 ft/lbs of energy (about the same as a 7.62x39 round) which is again not even gonna do a singe MD per flechette.

[Alrik Vas]

well...not even 1md a slug, but there's 200 and max damage is 180.
also, would the shape of the suit produce enough drag to cut even more energy off?

Edit: okay, okay. So, same meters per second, but let's make the GB weigh what it should, considering it's 11ft of P0W4R 4RM0R! 5400kg. Everything else is the same, but the round weighs 5kg.

I fooled around with the calculator you posted the link for, but the result was in Greek. (9.2408e+5)

[Thinyser]

well...not even 1md a slug, but there's 200 and max damage is 180.

if you would allow 200 separate shooters wielding AK-47's to each fire one round (capable of a mere 6d6 SDC using Armor Piercing rounds) simultaneously and do 3d6x10 MD then it makes sense...

if you treat them as the RAW indicate (any attack that does less than 100 SDC just bounces off harmlessly) then 200 separate slugs be they from a boomgun or 200 different barrels fired simultaneously would not do MD.

also, would the shape of the suit produce enough drag to cut even more energy off?

At the slow speed of 18.25m/s air friction (drag) is not much of an issue. So not a significant amount.

EDIT: the notation should be 9.2408x10^5, which would move the . to the right 5 spots so it would be 924080 joules. A very significant improvement.

[drewkitty ~..~]

A boom-gun slug looses virtually none of the kinetic energy that was used to launch in with in its total effective range, thus what ever is hit receives a knock back effect.
If hit by a Boom-gun slug:
♠ a 1.5 ton object will be knocked back 30',
♣ tanks have a 40% chance of being rolled over,
♥ Bots have a 50% chance of being Knock down,
♦ PA will be Knocked Down unless the pilot makes a 'roll with' roll, even if they make their 'Roll with' they will loose init and -1 APM.
❖ A man sized object will be thrown back 300'+2d6x10' and SDC/HP beings wearing MDC armor have a 70% chance of being KO'ed for 1d6 min, , man-sized MDC beings get a 60% of being KO'ed for 2d6 melees. For both SDC/HP and MDC beings saving vs 58% (using the coma bonuses) means being KO'ed for only 1d4 melees.

I like this (as being a logical course of action) so much I will now be utilizing it in my games.

I would also figure these would be the case if something got shot point blank with the normal BG rounds because all (or nearly all) the flechettes would hit the target.

[Alrik Vas]

A very significant improvement.

but still slower than a turd out of Prosek's Imperial Exit...for an anti-armor round.

[ShadowLogan]

I could be wrong, but I don't think we're not talking about momentum. I'm asking if the muzzle velocity of the round based on the above can be determined, not the printed speed of the round...I think Mach 5 is too slow. The Navy's got stuff that goes Mach 7 and produces under 32MJ of energy.

We are talking about momentum though. That is basically what recoil is, the momentum the gun platform receives from firing the projectile. Which due to conservation must equal the energy the gun platform imparts to the projectile. So if the USN railgun fires a 32MJ projectile, the railgun (and the platform that fires it) also receives 32MJ of energy in the opposite direction. Now because the Railgun/platform is much more massive than the projectile fired it will have a lower velocity imparted to it.

[Thinyser]

A very significant improvement.

but still slower than a turd out of Prosek's Imperial Exit...for an anti-armor round.

Well lemme check on that.
So assuming the armor really weighs in at 5400kg and gets tossed back at 18.25m/s that gives us 8.9927e+5 Joules (call that 900,000)
Channel that into a 5kg group of 200 flechettes and they will be going 600m/s or 1324 mph or 2.03 MACH... BUT each flechette now weighs a respectable 25 grams or for those that think in grains for projectiles this is 385 gr. with 4500 joules or 3320 ft/lbs of energy each. This is pretty close to a .308 Win-mag rifle round but there are 200 of them.

Now make some very "fast and loose" assumptions you could say that it actually take twice as much energy to knock back the armor due to thermal/sonic/ (and mostly) frictional losses. so you have 1.8 million joules propelling that 5kg of projectiles. They would be going at 848.53m/s or 1898 mph (2.87 MACH) and each flechette would have 9000 Joules or 6638 ft/lbs of energy. More than a .338 Lapua but less than a .50 BMG.

[Alrik Vas]

what's funny about this, is based of the Mach 5 speed and if we assume the GB travels 18.25 m/s if the recoil system is disengaged...it means the suit should weigh around 31 tons.

[Thinyser]

what's funny about this, is based of the Mach 5 speed and if we assume the GB travels 18.25 m/s if the recoil system is disengaged...it means the suit should weigh around 31 tons.

[Alrik Vas]

but, lets look at it another way.

so, according to the 1,000 car battery figure for e-clips, a single shot of a 30 round mag is 15MJ. that's about 3d6MD. the boom gun does 10x that, so a single shot delivers 150MJ?

I ran that and it came up with like a 109kg round that travels at mach 5. haha...so, really, the 1100kg power armor would fly back at somewhere around 300mph...but, Brain, where are we going to get a 109kg round? O.o

that's like shooting black holes out of a Smart Car at 5x the speed of sound.

seems bad.

[ShadowLogan]

o, according to the 1,000 car battery figure for e-clips, a single shot of a 30 round mag is 15MJ. that's about 3d6MD. the boom gun does 10x that, so a single shot delivers 150MJ?

There may be a problems with using a car battery though. 1st there is not a standard car battery to work off of. 2nd, energy ratings for car batteries IIRC are a bit different in how they present it. 3rd due to changing technology, what year should we be looking at "car battery" to derive specs. 4th what materials are involved in the car battery.

[Alrik Vas]

it was an average with some assumptions made. flatline did the work on it, I believe. wish I could cite the thread.

the 1000 car batteries statement is from a book, probably RMB/UE or SB1.

[Thinyser]

but, lets look at it another way.

so, according to the 1,000 car battery figure for e-clips, a single shot of a 30 round mag is 15MJ. that's about 3d6MD. the boom gun does 10x that, so a single shot delivers 150MJ?

I ran that and it came up with like a 109kg round that travels at mach 5. haha...so, really, the 1100kg power armor would fly back at somewhere around 300mph...but, Brain, where are we going to get a 109kg round? O.o

that's like shooting black holes out of a Smart Car at 5x the speed of sound.

seems bad.

This is why we don't do the maths for these things because it all breaks down to garbage...

[ShadowLogan]

it was an average with some assumptions made. flatline did the work on it, I believe. wish I could cite the thread.

the 1000 car batteries statement is from a book, probably RMB/UE or SB1.

IIRC it was CB1r I've seen it in (10,000 car batteries = 10 eclips, with that being equivalent to 1 gigawatt under the Energy Absorption power), but Psycape has the Power Leech listing 2MD worth of energy in a car battery, 200MD in a standard Eclip, 300 for a long Eclip. That works out to 100-150 car batteries being in an Eclip (depending on type) instead of 1,000 listed in CB1r.

This is why we don't do the maths for these things because it all breaks down to garbage...

Don't even need to do the math for things, some of their statements are way off (Mach 5 to reach earth orbit for example, though if we treat it as some game system term for miles per second it works). Palladium doesn't do scale well either.

[Alrik Vas]

yeah, isn't the shuttle, err....wasn't the shuttle makin' bacon at like 17000mph?

bit of a difference.

though some of their megadamage guesswork isn't as terrible as it sounds, at least in comparison with itself. as a for instance, by the calculator, an M1A1's 120mm (I forget the current designation) by it's mass and velocity is delivering 12MJ, which is penetrating an obscene amount of armor at an extreme range.

if you continue with the car battery idea, it's less than a single shot from a Wilks 457, by almost a third. they put that tank damage lower by a d6 I believe, but in the realm of mega damage, it's really not that far off.

[flatline]

it was an average with some assumptions made. flatline did the work on it, I believe. wish I could cite the thread.

the 1000 car batteries statement is from a book, probably RMB/UE or SB1.

As I recall, if you replace "gigawatt" with "gigajoule" in the Energy Absorption description, then the comparison to car batteries works pretty well if you make your comparison against car batteries from the 1970s or 1980s (which seems apt considering when CB1 was written).

However, I hate trusting my memory, so I'll hunt around and see if I can find that thread.

--flatline

[ShadowLogan]

yeah, isn't the shuttle, err....wasn't the shuttle makin' bacon at like 17000mph?

bit of a difference.

though some of their megadamage guesswork isn't as terrible as it sounds, at least in comparison with itself. as a for instance, by the calculator, an M1A1's 120mm (I forget the current designation) by it's mass and velocity is delivering 12MJ, which is penetrating an obscene amount of armor at an extreme range.

if you continue with the car battery idea, it's less than a single shot from a Wilks 457, by almost a third. they put that tank damage lower by a d6 I believe, but in the realm of mega damage, it's really not that far off.

For the Shuttle yes that sounds about right. Though you'd only be able to cook the bacon at certain points in the flight.

This occurred to me in the past that when the projectiles hit one can view them as a collision. By the rules if you apply the collision damage (1d8 or 2d4 depending on main book per 10mph which does not consider mass/weight) and use that with muzzle velocities for projectile weapons the assigned stats break down.

M-16 does 4d6 SDC per RMB, wikipedia lists a muzzle velocity of 3,110ft per second (I've seen a slightly higher one here), putting everything into the same units, the M-16 should do 212d8 SDC or 2d8MD (truncated).

An AK-47 does 4d6SDC per RMB, wikipedia lists a muzzle velocity of 2330ft per second, that works out to 158d8SDC or 1d8MD (truncation rounding).

The .50cal does 7d6 per round per RMB (a weapon on the M1A1 tank), wikipedia lists a muzzle velocity of 2910ft per second, that works out to 198d8 or 1d8MD (truncation rounding, though if we round to nearest hundred for MD it would be 2d8)

The 120mm cannon on the M1A1 tank has an upper muzzle velocity (per wikipedia) of 1750ft/sec or 119d8 or 1d8MD (truncation rounding). Off hand I'm not sure what the damage is listed at.

Now I will admit that if we are talking in terms of actual kinetic energy things will be different, but the RAW don't really follow KE given the generic damage for collision only cares about speed (ie muzzle velocity), not the mass/weight involved. This is probably also breaking the spirit of RAW since the table could be seen as applying to human size objects or vehicles and not small projectiles, but it really isn't clarified to be that way. Still it shows how arbitrary the assigned numbers really can turn out to be.

[Alrik Vas]

I believe they listed the M1A1's standard penetrator at 1d6x100 SDC, or 1d6MD. If it falls under Armor Piercing (which...it should...you know, considering...) then a high strike roll nets double damage, which could hole another tank in a single hit...which is something that seems fairly accurate.

of course, 1d6x100 SDC seems low still, if you take other things into consideration...like the fact that Palladium has listed the grenade from an M203 as 1d4x100 before...granted, a direct hit will gib a human, and I think that's what they were going for, but the 2d4x10 they list it at now does it almost as well, haha.

Okay, back to boom sticks. So say we want to count it as a collision...the GB's round goes mach 5 by the book, so it's like 380d8? Yeah, gotta figure out how to include mass in that, for hilarious results.

[Mlp7029]

First off Palladium math rarely adds up. The people writing and editing are artists not engineers or scientists. The velocity listed for the Boom Gun round is wrong if you agree with the published damage values that say it does more damage than a 120mm tank gun. The boom gun round is smaller than a 120mm APFSDS round therefore to do more damage logically it must be moving faster. Yet the tank round does at least 5200 fps / 1,127 fps for Mach 1 = 4.6 to 5 depending on the source you look at. The boom gun round has to be doing more than Mach 5 just to have the same kinetic energy as the 120mm APFSDS round. In reality the boom gun round is not a AP round it is an anti-personnel round. Simple fix is to give the boom gun round an area of effect maybe 20 or 30 ft. Or change the boon gun round to a sabot round. Point is the boom gun round falls short of the kinetic energy of a modern tank round while supposedly being much more powerful.

[Subjugator]

Don't forget the weight of the ammunition.

1,000 rounds of it.

[Hotrod]

got a question for you physics guys. a glitter boy weighs about 2 tons, is almost 12ft tall with human like proportions. the gun fires from the shoulder. when the recoil suppression is not engaged, it flies back 30ft almost instantly.

can we calculate what the actual muzzle velocity of the round is based off this information?

if its been done before, point me toward said thread.

We'll begin assuming the center of mass for a GB is 6 feet=1.83 meters.
When fired without pylons/rockets, the GB goes flying back. It'll also rotate, since the gun is up around its head level. I'll assume that it hits the ground flat. I'll model the GB as a ballistic projectile 6 feet off the ground with a flat initial trajectory.

The initial vertical velocity is zero. The time the GB takes to fall (6 feet) is the same as the time it takes to travel 30 feet horizontally.

1.83 meters = .5*the accelleration due to gravity (9.81 meters/second^2)*(the time it takes to fall)^2

Solving for the time yields .611 seconds.

Dividing 30 feet by .611 seconds yields a recoil velocity of about 15 meters per second. That's how fast the GB is flying back.

The momentum of the GB is 15 m/s * 2 tons. Divide this momentum by the mass of the projectiles in a single GB gun round, and you'll have a fair approximation of the GB's muzzle velocity.

[Alrik Vas]

yeah, I fiddled with the numbers and for a 3kg solid penetrator to produce 150MJ of energy, the round would have to travel at around 17000mph.

I wonder how much friction heat would be in the barrel, melting point of tungsten is about 6000F.

[Hotrod]

For a 3kg slug, I get a muzzle velocity of around 20,000 miles per hour and a kinetic energy of around 120 MJ. That's Mach 28 or so, which is slightly faster than the low earth orbit speed at the equator. Boom indeed!

I think it's safe to say that such a weapon would require some exotic (sci fi) materials to work.

[Alrik Vas]

Golden Age tech for the win? lol