Glitterboy Boomgun and some Math Results

[ShadowLogan]

This topic is in part inspired by an offhand statement Killer Cyborg make in a previous thread that comes down to the Math concerning flying a Glitterboy to the Moon using its boomgun (Shemarrians are also included in the comment, but...)

Glitterboy Facts from the RPG:
-Fully Loaded Mass of 1.2tons (no mass breakdown, assumed to include the mass of the ammo, weapons, and lifesupport, but not the pilot) or 1,200kg
-Round leaves the barrel at Mach 5 (RUE/WB22) or 1715m/s (google search result for a calculator Mach to meters per second)
-Without the Pylons and Thrusters the Glitterboy would be thrown backward 9.1m (30ft)
-A DecendedGlitterboy Pilot (human) at level 15 is assumed with a total of 15attack/actions per melee with RCE, Boxing, HTH: Assassin (extra attack over other options). This works out to one action being equal to 1second duration. A lower level could be assumed, but since the time it actually takes isn't going to change between levels for things measured in "actions", this allows one to establish actual "time" more closely that don't come from being a "slow" or "inexperienced" pilot.
-Depending on which book one uses WB22 or RUE the Ammo of the USA-G10 is 100 rounds or 1,000rounds plus a 40/400 spare ammo it can carry separate bag. Both values will be considered
-RUE pg69 the length of the Boomgun is 2.1m (7ft)

Question 1. What is the mass of a Boom Gun Round? Sure we can look at the illustrations and attempt to make guess on the material used to determine the mass and such via volume. But there might be a simpler way to work out the mass of a projectile and by extension the fully loaded mass breakdown using the Delta-V Rocket Equation.

Using the Muzzle Velocity of the Railgun in place of Exhaust Velocity and using the recoil distance (assuming it happens in 1second) as the Delta-V and solving for the post mass and taking the difference yields up an individual round mass of...:
-WB22/RUE USA-G10 is ~6.35kg per round
-RMB USA-G10 is ~15.81kg per round (for those interested, this is due to the slower velocity of the round previously stated and rectoned in WB22)

This is fairly close to using conservation of momentum in recoil calculations (p=mass * velocity), or even Force = mass * acceleration. The slight difference is accounted for that using these at a basic level and not subtracting out the round from the GB fully loaded mass which the Delta-V equation does.

If the Recoil Velocity imparted IS LESS than 9.1m/s the mass of the round can come down. We know how far the suit can be thrown, but we do not know how long it takes (it take place within 1 action, which is a variable measaure of time, but as established above 15apm may be the maximum w/o resorting to tricks to get it up higher).

Question 2. What does the Boom Gun look like when used for space propulsion?
IF the Boom Gun fires in space (orbit) it can act as a means of propulsion due to Newton's Laws if one disables the recoil suppression system (cut the jets and the pylons have nothing to sink into). Assuming the distance covered from being thrown backward is 1second, that means firing the BG would increase the suits velocity by 9.1m/s each time. This however is misleading because you aren't properly accounting for the change in mass after each firing which would allow an even bigger increase in velocity after each firing and would not be how it works. If we plug the mass of the round from Question 1 into the Delta-V equation proper for the entire payload (main bin only, the extra ammo drum will of course add velocity) of ammo being used:
-WB22 standard = ~1,291m/s (~1.3kps) for just the main bin ammo (assuming the "fully loaded" mass DOES includes the main bin ammo and NOT the extra, and ignoring the mass of the pilot)
-RUE standard = ~3,154m/s (~3.2kps) for just the main bin ammo (this assumes the "fully loaded" mass DOES NOT include the main bin ammo as it is over 6metric tons alone and is added in later nor the mass of the pilot, and again NO extra ammo).
-RMB standard = ~576.6m/s for just the main bin ammo (assumptions and such are as the RUE standard)

For comparison to some real spacecraft chemical propulsion systems can offer a maximum change (or budget) in velocity of:
-the space shuttle's OMS only offers 305m/s (its SRBs and external tank fed engines combine to provide over 8kps, but those systems aren't available in orbit)
-Gemini has a total of 323m/s
-Apollo CSM 2,804m/s
-Apollo Lunar Lander has 4,690m/s total (combined Ascent and Decent stages, 2.2 and 2.4 respectively IIRC)
-3rd Stage of the Saturn V injecting the Apollo/Lander stack to the Moon impart 3,041.6m/s (it actually has more, but some is used in getting into Low Earth Orbit)

So the question is could a Glitter Boy uses its Boom Gun to fly to the Moon? The answer based on the Saturn V's 3rd stage performance:
-WB22 standard could NOT fly an Apollo style trajectory if it started in Low Earth orbit
-RMB standard could NOT fly an Apollo style trajectory if it started in Low Earth orbit
-RUE standard YES it could fly an Apollo style trajectory to the Moon starting in LEO, but won't be able to return without being able to reload its ammo bin, and it might not even be able to break to enter Lunar orbit or alter its trajectory (sending it off into deep space)
-RUE standard could also fly from Lunar Orbit to Earth on an Apollo style trajectory, but will have perform an aero-breaking maneuver to enter LEO (or re-entry the atmosphere, which it really isn't designed to do)

IINM Apollo did not fly a minimum energy trajectory to the Moon, but it was close which does provide a bit of margin to work with. Off hand the WB22 standard would still not work though, but the RUE standard increases its chance.

Question 3: What about the Man Portable Railgun in Mutants In Orbit (pg89)?
This weapon system might be functional with a bit of work. It gives us enough information, though we have to make a few assumptions. Recoil is listed at about 10ft/3m per second (likely due to the varying mass of an individual), but it doesn't tell us how fast the round is going (we can actually work that out though). We will need to assume the mass of the person firing the weapon (likely in a spacesuit) and assume the mass of the ammo drum is the actual mass of all the ammo instead of ammo + drum.

Using conservation of momentum equation, and assuming a 180lb (~81kg) person in a CAF Light Body Armor (DB2, MiO does not list mass of space suits, its a bit heavier than the DB2 spacesuit, but since we are firing a railgun it likely means the person expected to enter combat so I went with that), assuming the ammo drum mass is all the ammo (for simplicity) we end up with a muzzle velocity of 313.96m/s (for reference Mach 1 is 343m/s) for a total Delta-V of 65.89m/s with these assumptions.

A lighter individual, or a lighter spacesuit, or more ammunition could allow for more velocity, but at the same time heaver options also exist. Now weather the encountered recoil here can be applied to other railguns is likely highly debatable.

Just a quick note: IINM most Railguns don't tell us their recoil velocity (or distance) or the velicty of the round when fired or even the mass of the ammunition itself. So I have not considered any of them because of this (exception was the Triax T-550 GB variant, but still I had to assume the recoil information was still valid at 9.1. Doing the math for other GB variants that can use the Boomgun was also problematic in that I would have to subtract out systems and potential payload and add back in the stock masses.)

[Natasha]

-Fully Loaded Mass of 1.2tons (no mass breakdown, assumed to include the mass of the ammo, weapons, and lifesupport, but not the pilot) or 1,200kg

...

-Depending on which book one uses WB22 or RUE the Ammo of the USA-G10 is 100 rounds or 1,000rounds plus a 40/400 spare ammo it can carry separate bag. Both values will be considered

...

-WB22/RUE USA-G10 is ~6.35kg per round
-RMB USA-G10 is ~15.81kg per round (for those interested, this is due to the slower velocity of the round previously stated and rectoned in WB22)

If I understand these numbers, then 1000 * 6.35 and 100 * 15.81 should be << 1200 ?

[ShadowLogan]

-Fully Loaded Mass of 1.2tons (no mass breakdown, assumed to include the mass of the ammo, weapons, and lifesupport, but not the pilot) or 1,200kg

...

-Depending on which book one uses WB22 or RUE the Ammo of the USA-G10 is 100 rounds or 1,000rounds plus a 40/400 spare ammo it can carry separate bag. Both values will be considered

...

-WB22/RUE USA-G10 is ~6.35kg per round
-RMB USA-G10 is ~15.81kg per round (for those interested, this is due to the slower velocity of the round previously stated and rectoned in WB22)

If I understand these numbers, then 1000 * 6.35 and 100 * 15.81 should be << 1200 ?

They Should BE (BUT ARE NOT) given the text in the Glitterboy G10 says a fully loaded GB should have a mass of 1.2tons. I think a reasonable person would interpret the book text of fully loaded to include ammo.

I did mention that in order to use these values later I had to assume the fully loaded mass did not include ammo based on the above calculation (though I did not feed it back in later).

[Sohisohi]

They Should BE (BUT ARE NOT) given the text in the Glitterboy G10 says a fully loaded GB should have a mass of 1.2tons. I think a reasonable person would interpret the book text of fully loaded to include ammo.

Mate, people here don't even agree that the Boom Gun is considered part of the Glitter Boy; can't wait for the replies.

Anyway, both how gravity reacts on an object and the density of the atmosphere change significantly. This is why you'll never get the same ballistics shooting 90° upwards as opposed to shooting 180° horizontal. But, that's RL and this is Rifts; gg no re, going to rocket jump more often.

[Natasha]

-Fully Loaded Mass of 1.2tons (no mass breakdown, assumed to include the mass of the ammo, weapons, and lifesupport, but not the pilot) or 1,200kg

...

-Depending on which book one uses WB22 or RUE the Ammo of the USA-G10 is 100 rounds or 1,000rounds plus a 40/400 spare ammo it can carry separate bag. Both values will be considered

...

-WB22/RUE USA-G10 is ~6.35kg per round
-RMB USA-G10 is ~15.81kg per round (for those interested, this is due to the slower velocity of the round previously stated and rectoned in WB22)

If I understand these numbers, then 1000 * 6.35 and 100 * 15.81 should be << 1200 ?

They Should BE (BUT ARE NOT) given the text in the Glitterboy G10 says a fully loaded GB should have a mass of 1.2tons. I think a reasonable person would interpret the book text of fully loaded to include ammo.

I did mention that in order to use these values later I had to assume the fully loaded mass did not include ammo based on the above calculation (though I did not feed it back in later).

I'm trying to understand your method. You seem to use the 1.2 ton stat to show the 1.2 ton stat is ridiculous, and then use it anyway for your mass ratios in the ∆v equation. Then you seem to have computed either 100 or 1000 ∆v's depending on which boom gun you're considering. I don't know if that's what you did. I think if you showed your maths, it all would be clearer to me. I simply don't understand how you arrived at the results you did. I apologise for my poorly worded original question.

As to the question if it can be done, consider that there are different approaches to the problem. There have been plenty of folks looking into how relatively small ∆v's can get there. Such missions require months rather than days, though. They are considerably more complicated scenarios than the relatively simple Hohmann transfer from a parking orbit (and for such missions from launch pads > 30 degrees of latitude, there are two launch opportunities per day).

Using a Hohmann transfer from a 300 km parking orbit to, say, 384400 km your propulsion system will need to supply a ∆v of about 3 km/s. A fraction of this is required for lunar injection, but it's noticeably smaller than the ∆v required to enter the transfer orbit. The assumptions are that the orbits are circular so as to eliminate the variable flight path angles and velocities and that the orbits are coplanar so as to eliminate the required plane change, which wastes ∆v since it does no work on the spaceship.

[ShadowLogan]

I used the 1.2tons loaded in both cases yes. When calculating the mass of the round via the Delta-V equation I assumed the mass of all the ammunition was included in that figure. I cross checked that with a balanced equation for momentum and Force between the two objects, but could not account for the mass change easily but as the resulting numbers where close I just went with it as the difference could be explained by not properly accounting for the mass change. Now it is possible to lower the mass of the individual rounds if the assumption about the recoil distance requires more time.

Now the second section I agree can be a bit dicey since I had to work with a different assumption on what constituted "fully loaded" in order to make the math work in some specific cases. I noted what assumptions where in play. I admit I don't like it, but it is what it is in order to make the math work. Which probably shows that the numbers Palladium uses at times may be just numbers pulled at random.

Yes I know it is possible to fly a slower trajectory to the Moon from LEO. Which is why I described it as an Apollo Style trajectory. IIRC the first Lunar flyby (Soviet) flew a much more energetic trajectory to reach the Moon in ~24hrs. Though one probably should take into consideration the life support capacity of the GB suit, which would push for the higher energy trajectory requirements (which it can't fly) if one uses the RMB/RUE endurance instead of MiO endurance (which is greater) which can allow the Apollo Style trajectory but not much slower.

[Natasha]

It takes an object just under a second to fall 3.1 metres near the surface of Earth. I'd say your 1 second knock back time is an easy sell for glitterboys if their guns are shoulder-fired.
But, yea, it seems many numbers in Palladium can be replaced with words like "big", "bigger", and "biggest"

[Hotrod]

A while back, I calculated that the actual muzzle velocity of a boom gun would have to be around Mach 28.

[Armorlord]

We've had mounting evidence for years that conventional physics does not scale linearly with MDC materials and energies. From it the amount of energy to fire and charge MD vs SD shots, kinetic energy not transferring to occupants of a sealed MDC environments, only two special weapons having noteworthy recoil, ranged damage drop-off, etc.
It was with the New Generation book that well demonstrated difference was moved from supported conjecture to printed fact.
MDC materials have exponentially more damage potential at speed than SDC materials. To the point that an Abrams tank moving a supersonic speeds and a MDC motorcycle going down the highway have the deliver equivalent destructive force on impact.

Going by the numbers stated there, I would say that Boom Gun based propulsion would be more inefficient than the mass and velocity would indicate with conventional physics. A similar setup with SDC materials would be more useful for thrust, though more prone to wear and likely take a very different launch mechanism.

[ShadowLogan]

I did consider this scenario of using comic book physics (or palladium physics) results but left it out because I knew it was way off. However the payload without any extra ammo bins would provide velocities of 910m/s (WB22/RMB) or 9,100m/s (RUE). Though if the recoil distance is doubled the velocities imparted also doubles (this also assumes it takes 1second to cover the recoil distance).

[Hotrod]

Given the discrepancies in ammunition mass and muzzle velocity, I'll base my math off the GB itself.

When fired without pylons/rockets, the GB goes flying back 30 feet. 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.

I'll also assume that the center of mass for a GB is 6 feet=1.83 meters.

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.

Multiply those 15 meters per second by the total payload of the GB (100), and you get a fair approximation of its total available delta v when it uses its gun as a rocket engine. Note that this also assumes that the gun is fired in line with the center of mass, or else a lot of the kinetic energy from each shot will go into rotation.

[ShadowLogan]

@Hotrod.

That is more "comic book" math in terms of approximating the Delta-V total. 15m/s from each 100round equals 1.5km/s they way you are suggesting.

If we use 15m/s as the recoil velocity from one round (and plug that in as the goal for a Delta-V from firing one round), the mass of which would have to be approx. 10.45kg for each projectile. Using WB22 velocity (Mach 5) and payload (100). The calculated Delta-V is 3,510.01m/s OR 3.5km/s. This is because as you expend rounds, the mass of the loaded GB gets lighter, and thus each round can impart more velocity than the previous round.

[Hotrod]

@Hotrod.

That is more "comic book" math in terms of approximating the Delta-V total. 15m/s from each 100round equals 1.5km/s they way you are suggesting.

If we use 15m/s as the recoil velocity from one round (and plug that in as the goal for a Delta-V from firing one round), the mass of which would have to be approx. 10.45kg for each projectile. Using WB22 velocity (Mach 5) and payload (100). The calculated Delta-V is 3,510.01m/s OR 3.5km/s. This is because as you expend rounds, the mass of the loaded GB gets lighter, and thus each round can impart more velocity than the previous round.

I wouldn't call it "comic book math." Your focus is math based on information on the propellant, while my focus is math based on a test flight. Per Von Braun's maxim, "A test is worth a thousand expert opinions," I'm inclined to prefer my own method. In any case, both of our calculated delta-V approaches are on the same order of magnitude, which I regard as a minor miracle considering the source data.

I'll gladly admit that I'm assuming that the unloaded mass is pretty close to the fully loaded mass of the GB, but I think that's a reasonable approximation. I'm not going for high precision on this calculation since the source data is clearly inconsistent; same-order-of-magnitude is enough for me.

[lather]

I guess the main problem with saying a RUE GB round has mass of 6+ kg is that the armor, when fully loaded, is missing a few thousand kilograms.

[ShadowLogan]

I put "comic book" in quotes because it's the approach you might see in comic books.

And the math I used for Delta-V is the proper formula. So I was properly accounting for the mass of the rounds (as best I could), where your approach seemed to not consider the mass of the round or the changing mass of the glitterboy from firing the rounds over time. Granted Palladium doesn't seem to consider that either with the recoil.

And 2kps difference might seem close, but if you look above when I figured the Delta-V using 9.1m/s recoil it was a lot closer (910m/s just adding up the rounds "comic book" style vs 1291m/s vs Delta-V equation) with a difference of only 400m/s (or 0.4kps).

[lather]

I put "comic book" in quotes because it's the approach you might see in comic books.

And the math I used for Delta-V is the proper formula. So I was properly accounting for the mass of the rounds (as best I could), where your approach seemed to not consider the mass of the round or the changing mass of the glitterboy from firing the rounds over time. Granted Palladium doesn't seem to consider that either with the recoil.

And 2kps difference might seem close, but if you look above when I figured the Delta-V using 9.1m/s recoil it was a lot closer (910m/s just adding up the rounds "comic book" style vs 1291m/s vs Delta-V equation) with a difference of only 400m/s (or 0.4kps).

Sure, 6.35 kg is the correct answer for the RUE GB, but it's a physically absurd result because if true the fully loaded GB cannot have the given mass. I don't have any problem with fudging numbers to make the SCIENCE! a little more science and imposing this or that assumption, but I don't know that we can say the mass has been accounted for.

[ShadowLogan]

@ lather.
Yes RUE GB payload can result in some absurd results. I make mention of it in my initial post (in order for the math to work I have to treat the 1200kg "loaded" mass of the GB as not including ammo at times that are specifically called out).

While some GBs can result in absurd results, some (like the WB22 version) do not.

@ In General
While preparing a reply on this topic a stray thought came to me. What if the stated recoil distance is regarded as the average value, and what the impact would be above. I only did this using the WB22 stats, not RUE (payload) or RMB (lower velocity) in relation to the Delta-V equation. These are the results:
Assuming an Average 9.1m/s recoil velocity:
-each round has a mass of 4.94kg
-the total delta-v is ~910m/s (rounding to nearest whole unit)
-the smallest imparted velocity was ~7.1m/s and the largest imparted velocity was ~12m/s. I calculated the Delta-V for each round (all 100), and the average is 9.1m/s. Incidentally if you add up each rounds velocity this way you still get ~910m/s

Assuming an Average 15m/s recoil velocity:
-each round has a mass of ~7kg
-the total delta-V is ~1501m/s (using the 7kg figure per round and not 6.35kg in previous post)
-the smallest imparted velocity was ~10m/s and the largest imparted velocity was ~23.8m/s. I calculated the Delta-V for each round (all 100), and the average is ~15m/s. Incidentally if you add up each rounds velocity this way you still get ~1501m/s

I'm not sure what Hotrod considers close to the mass for a fully loaded vs empty GB, but @0.1kg mass per round and Mach 5 exit velocity the recoil imparted (.14m/s). So in order for the stated recoil and velocity of the round to work, the mass of the round has to be several kg.

[lather]

@ lather.
Yes RUE GB payload can result in some absurd results. I make mention of it in my initial post (in order for the math to work I have to treat the 1200kg "loaded" mass of the GB as not including ammo at times that are specifically called out).

While some GBs can result in absurd results, some (like the WB22 version) do not.

@ In General
While preparing a reply on this topic a stray thought came to me. What if the stated recoil distance is regarded as the average value, and what the impact would be above. I only did this using the WB22 stats, not RUE (payload) or RMB (lower velocity) in relation to the Delta-V equation. These are the results:
Assuming an Average 9.1m/s recoil velocity:
-each round has a mass of 4.94kg
-the total delta-v is ~910m/s (rounding to nearest whole unit)
-the smallest imparted velocity was ~7.1m/s and the largest imparted velocity was ~12m/s. I calculated the Delta-V for each round (all 100), and the average is 9.1m/s. Incidentally if you add up each rounds velocity this way you still get ~910m/s

Assuming an Average 15m/s recoil velocity:
-each round has a mass of ~7kg
-the total delta-V is ~1501m/s (using the 7kg figure per round and not 6.35kg in previous post)
-the smallest imparted velocity was ~10m/s and the largest imparted velocity was ~23.8m/s. I calculated the Delta-V for each round (all 100), and the average is ~15m/s. Incidentally if you add up each rounds velocity this way you still get ~1501m/s

I'm not sure what Hotrod considers close to the mass for a fully loaded vs empty GB, but @0.1kg mass per round and Mach 5 exit velocity the recoil imparted (.14m/s). So in order for the stated recoil and velocity of the round to work, the mass of the round has to be several kg..

Your original result for the WB22 version was 15.81kg per round which is well over 1.2 tons as well. Neither result is workable.
The revised mass per round in this post, regardless of what I think about their derivation, are indeed workable.

[ShadowLogan]

@later
No, my original result of ~15.81kg was for the Rifts Main Book version of the GB, RUE and WB22 are ~6.35kg. The only (relevant) difference between the Rifts Main Book Version and the WB22 version is the velocity of the round when fired (Mach 2 originally, now Mach 5), which makes a lot of difference in the math.

I know that the RMB and RUE versions result in the ammo mass being considered separate from the "fully loaded mass", that is mentioned in my original post when it is done this way in calculating the final delta-V figure.

[lather]

@later
No, my original result of ~15.81kg was for the Rifts Main Book version of the GB, RUE and WB22 are ~6.35kg. The only (relevant) difference between the Rifts Main Book Version and the WB22 version is the velocity of the round when fired (Mach 2 originally, now Mach 5), which makes a lot of difference in the math.

I know that the RMB and RUE versions result in the ammo mass being considered separate from the "fully loaded mass", that is mentioned in my original post when it is done this way in calculating the final delta-V figure.

My bad. I totally misread the OP

[ShadowLogan]

The Mass Ratio is Not constant though. With each firing of the Boomgun the unit will get lighter and a changing mass ratio. When I did the "average" version of this, I did for each of the 100 rounds (in WB22/RMB) at the specified velocity as they where fired, that is why the Min and Max D-V figures exist. I had to go with extremly light projectiles to get the Avg/Min/Max to be constant, but the resulting recoil velocity was no where near the list value.

[Natasha]

The Mass Ratio is Not constant though. With each firing of the Boomgun the unit will get lighter and a changing mass ratio. When I did the "average" version of this, I did for each of the 100 rounds (in WB22/RMB) at the specified velocity as they where fired, that is why the Min and Max D-V figures exist. I had to go with extremly light projectiles to get the Avg/Min/Max to be constant, but the resulting recoil velocity was no where near the list value.

I deleted the post because I did not like the wording about the mass. But I believe it is correct that the mass ratio is constant. If the initial mass of the entire system is 1200 kg and we say a round is 5 kg, then ln(1200/1195) = ln(1195/1190) = ln(1190/1185) = ...

[ShadowLogan]

The Mass Ratio is Not constant though. With each firing of the Boomgun the unit will get lighter and a changing mass ratio. When I did the "average" version of this, I did for each of the 100 rounds (in WB22/RMB) at the specified velocity as they where fired, that is why the Min and Max D-V figures exist. I had to go with extremly light projectiles to get the Avg/Min/Max to be constant, but the resulting recoil velocity was no where near the list value.

I deleted the post because I did not like the wording about the mass. But I believe it is correct that the mass ratio is constant. If the initial mass of the entire system is 1200 kg and we say a round is 5 kg, then ln(1200/1195) = ln(1195/1190) = ln(1190/1185) = ...

The Ratio is not. If you do the math the ratio for each round being 5kg and 1200kg fully loaded mass:
-Round 1: 0.00418
-Round 2: 0.00419
-Round 3: 0.00421
-Round 4: 0.00423
-Round 9: 0.00430
-Round 100: 0.539

I know I skipped Rounds 5-7 and 9-99, but if your premise was correct they should all have a ratio of 0.00418, and it is clear they are not. It might require taking it out to enough digits to actually see (which would explain getting a Avg/Min/Max to appear constant since the spreedsheet's default is two digits).