Question about RW Rotary MG and AC - Power Consumption

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Christopher_Perkins
04/19/08 04:14 PM
67.166.179.22

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You can find information about the carious calibres and rates of fire for to the various rotary Machine Guns and the rotary cannon, but what I cannot find is the rate of draw in megajoules tor the electricly powered variants, Does anyone have a good resource for this?

The primary use of this innformation for me would be to determine how large a barrery pack woudl be needed or determining how large the integrated battery pack would be in the RPG for these weapons.

the types that I know about are the
5.56 mm XM214 (Electro-Mechanical)
7.62 mm M134 (Electro-Mechanical)
7.62 mm XM133 (Gass Operated Prototype)
12.7 mm GAU-19/A (GE Cal 50 - 3 & 6 barrel) (Electro-Mechaical)
15 mm T45 (Variant of the M61) (Electro-Mechanical)
20 mm M197 (Electro-Mechanical - 3 Barreled)
20 mm M61 (Vulcan)(Electro-Mechanical - 6 Barreled)
23 mm GSh-6-23 (Gas operated- 6 Barreled)
25 mm GAU-12/U (Equalizer - 5 Barrel)(Hydraulic, Electric, Pneumatic)
30 mm GAU-13/A (Electro-Mechanical)(4 Barreled Variant of the GAU-8A)
30 mm GAU-8/A (Avenger) (7 Barreled - Hydrolic driven electric motor??)
Christopher Robin Perkins

It is my opinion that all statements should be questioned, digested, disected, tasted, and then either spit out or adopted... RHIP is not a god given shield
Karagin
04/20/08 12:04 PM
24.26.220.4

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Try asking Cray or jump over to the Tanker's Forum and ask them.
Karagin

Given time and plenty of paper, a philosopher can prove anything.
Christopher_Perkins
04/20/08 12:37 PM
24.125.204.221

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whats the link to the tankers forum?
Christopher Robin Perkins

It is my opinion that all statements should be questioned, digested, disected, tasted, and then either spit out or adopted... RHIP is not a god given shield
Karagin
04/20/08 06:01 PM
24.26.220.4

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http://63.99.108.76/forums

One bit of advice...90% of the folks on that forum have been in the military and they know what they are talking about when it comes to the weapons and vehicle system.
Karagin

Given time and plenty of paper, a philosopher can prove anything.
CrayModerator
04/24/08 05:32 PM
147.160.136.10

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Quote:

You can find information about the carious calibres and rates of fire for to the various rotary Machine Guns and the rotary cannon, but what I cannot find is the rate of draw in megajoules tor the electricly powered variants, Does anyone have a good resource for this?




You won't find a value in megajoules because you're looking for the wrong term. Joules are a measure of work accomplished or energy released, but it tells you nothing about how fast the work was accomplished or the energy was released. After all, a miniature 1/16th horsepower toy car electric motor and a 1200-horsepower racecar engine both release megajoules - the toy engine just needs about 6 hours to accomplish a megajoule of work, while the racecar engine needs 1.116 seconds.

The terms you want are watts (1 watt = 1 joule per second), kilowatts, or horsepower (1 horsepower = 747 joules per second).

According to this internet forum, with no supporting references, the GAU-8 uses a 77-horsepower motor.
http://www.strategypage.com/militaryforums/2-12468.aspx

Wikipedia says the GAU-12 25mm uses a 15hp/11kW motor.
http://en.wikipedia.org/wiki/GAU-12_Equalizer

This website says the Apache's M230 uses a 6.5-horsepower motor.
http://www.canit.se/~griffon/aviation/text/akandata.htm

So, the smaller guns use motors of various sorts (pneumatic, hydraulic, electric) of a few horsepower, while apparently the GAU-8 (which has to overcome the inertia of a LOT of barrel weight) uses about 50kW.

The power demand is almost certainly to get the barrels rotating up to speed in a short time. When you want to accomplish a lot of work (moving several hundreds pounds) a short period (a fraction of a second), you're using a lot of power.
Mike Miller, Materials Engineer

Disclaimer: Anything stated in this post is unofficial and non-canon unless directly quoted from a published book. Random internet musings of a BattleTech writer are not canon.
Askhati
04/26/08 04:19 PM
168.209.97.42

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Slightly off-topic, but the whole thing of minguns got me thinking: the AC-130 Spectre gunships, as in use by the USAF, fire a 105mm howitzer at an angle perpendicular to that of the flight direction. Does anyone know where I could get some pilot feedback on the effect that firing the howitzer has on the flight characteristics of the aircraft?
Evolve or DIE!
CrayModerator
04/26/08 05:12 PM
68.205.198.74

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Well, I can give you a quick approximation.

105mm shells are about 15kg and fly at a max of about 1000m/s.

An empty (unfueled, no payload) C130 is 40 metric tons.

Ignoring recoil compensation systems and sideways aerodynamic drag, the plane would surge sideways from recoil at 0.375m/s (0.8mph) when it was completely empty of ammo, fuel, and any payload. At maximum takeoff weight, that would be halved. And recoil compensation should drastically reduce the kick.
Mike Miller, Materials Engineer

Disclaimer: Anything stated in this post is unofficial and non-canon unless directly quoted from a published book. Random internet musings of a BattleTech writer are not canon.
Christopher_Perkins
04/27/08 06:18 AM
76.120.212.36

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Quote:

Quote:

You can find information about the carious calibres and rates of fire for to the various rotary Machine Guns and the rotary cannon, but what I cannot find is the rate of draw in megajoules tor the electricly powered variants, Does anyone have a good resource for this?




You won't find a value in megajoules because you're looking for the wrong term.


Won't be the first time... [expression of disgust}...

thanks

lot of that I was seeing but unable to understand...

hours and hours... gaaa

Quote:


Joules are a measure of work accomplished or energy released, but it tells you nothing about how fast the work was accomplished or the energy was released. After all, a miniature 1/16th horsepower toy car electric motor and a 1200-horsepower racecar engine both release megajoules - the toy engine just needs about 6 hours to accomplish a megajoule of work, while the racecar engine needs 1.116 seconds.

The terms you want are watts (1 watt = 1 joule per second), kilowatts, or horsepower (1 horsepower = 747 joules per second).




So
1.000 MJ / Sec is 1 MW or 1000 kW?
0.747 MJ / Sec is 1 HP?

Quote:


According to this internet forum, with no supporting references, the GAU-8 uses a 77-horsepower motor.
http://www.strategypage.com/militaryforums/2-12468.aspx




77 HP * 0.747 mj/sec = 57 MJ (per second?)

[Quote]
Wikipedia says the GAU-12 25mm uses a 15hp/11kW motor.
http://en.wikipedia.org/wiki/GAU-12_Equalizer




15 hp * 0.747 MJ/sec = 11.205 MJ (per second?)
11 kW / Sec = 0.011 MJ per second?
confused... different units per second...

Quote:


This website says the Apache's M230 uses a 6.5-horsepower motor.
http://www.canit.se/~griffon/aviation/text/akandata.htm




6.5 hp * 0.747 MJ/Sec = 4.8555 MJ (per second)

Quote:


So, the smaller guns use motors of various sorts (pneumatic, hydraulic, electric) of a few horsepower, while apparently the GAU-8 (which has to overcome the inertia of a LOT of barrel weight) uses about 50kW.




Wasn't there a russian 30 mm that was gas driven?

50 kW / sec = 0.05 MW / Sec = 0.05 MJ

Quote:


The power demand is almost certainly to get the barrels rotating up to speed in a short time. When you want to accomplish a lot of work (moving several hundreds pounds) a short period (a fraction of a second), you're using a lot of power.




yup...

little out of it with vicodin (hateful stuff) so am very much in "fun with cancellations" mode


Assuming 1 power point is 3.5 MJ (i know youve never been convinced, but run with it as the assumption - just trying to get my math module to work past the riniging in my ears ... btw, its always been MJ... i may have typed it mJ but always meant MJ)

so
GAU-8/A 30 mm = 50 kW = 0.05 MW /1 sec = 0.05 MJ / 3.5 MJ = 0.014285714285714285714285714285714 power points per second at max ROF of 58 Shells per second.
so 0.000246305418719211 power points per shot?

that cant be right... not for 7 barrels

Apachies M230 , 30 mm, 10 shots per second
6.5 hp * 0.747 MJ/Sec = 4.8555 MJ (per second) / 3.5
1.3872857142857142857142857142857 power points per burst
0.13872857142857142857142857142857 power points per Shot
thats more like it... what did i do wrong with the GAU-8/A

humm, and this is even a single barreled AC?
nothing ever easy, yes?


GAU-12 25mm
15 hp * 0.747 MJ/sec = 11.205 MJ / 3.5 MJ = Power Points
3.2014285714285714285714285714286 Power points

just realized that fall guy could probably have been doing 3.5 MJ / 5 Seconds

i am so out of it to try again right now... how close am I?
Christopher Robin Perkins

It is my opinion that all statements should be questioned, digested, disected, tasted, and then either spit out or adopted... RHIP is not a god given shield
CrayModerator
04/27/08 05:25 PM
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Quote:

Quote:

Joules are a measure of work accomplished or energy released, but it tells you nothing about how fast the work was accomplished or the energy was released. After all, a miniature 1/16th horsepower toy car electric motor and a 1200-horsepower racecar engine both release megajoules - the toy engine just needs about 6 hours to accomplish a megajoule of work, while the racecar engine needs 1.116 seconds.

The terms you want are watts (1 watt = 1 joule per second), kilowatts, or horsepower (1 horsepower = 747 joules per second).




So
1.000 MJ / Sec is 1 MW or 1000 kW?




Yes.

Quote:

0.747 MJ / Sec is 1 HP?




No. That would be 1000 horsepower. 1 horsepower = 747 watts.

Quote:


Quote:

According to this internet forum, with no supporting references, the GAU-8 uses a 77-horsepower motor.
http://www.strategypage.com/militaryforums/2-12468.aspx




77 HP * 0.747 mj/sec = 57 MJ (per second?)




That's incorrect, for the reason given above. The correct conversion is:

77 HP * 747W/HP = 57,519W = 57,519J/sec = 0.057519MJ/sec.

So after running this for 17.385 seconds, you'd use 1 Megajoule of energy powering the 77 horsepower motor.

Quote:

Assuming 1 power point is 3.5 MJ (i know youve never been convinced




That's a tiny quantity of energy so, no, I'm not convinced. It's far too little to accomplish the cited feats of BT energy weapons.

Quote:

but run with it as the assumption - just trying to get my math module to work past the riniging in my ears ... btw, its always been MJ... i may have typed it mJ but always meant MJ)

so
GAU-8/A 30 mm = 50 kW = 0.05 MW /1 sec = 0.05 MJ / 3.5 MJ =
0.014285714285714285714285714285714 power points per second at max ROF of 58 Shells per second.




That's correct, 1/7th of your under-sized power points per second.

Quote:

so 0.000246305418719211 power points per shot?




Very simplistically, yes. Realistically, no. The gun's motor only delivers peak horsepower to get the gun spinning. Once the gun is rotating at the desired speed, you'll only need to resist friction in the gun mechanism and meet the trivial demands of feeding in ammo - if you need 10kW to keep the gun firing, I'd be surprised. (This is the same reason a car gets worse mileage in-city than moving fast on the highway: it takes a lot of power to accelerate and not much to sustain a car moving against friction.)

So after the first, 0.2-0.4 seconds, power demands for the gun might be 5-10kW, or 0.005-0.01MJ/second... approximately 0.003 of your power points per second, almost irrelevant of the number of bullets being fired.

Quote:


that cant be right... not for 7 barrels




You're looking at it in the wrong way. The gun isn't eating energy to force bullets out the barrel; the gunpowder is doing that. The gun needs a moderate amount of horsepower to get the barrels spinning, and then a little juice to keep it spinning and keep the ammo belt flowing. The act of firing a bullet doesn't add up to much on that scale.

Quote:

Apachies M230 , 30 mm, 10 shots per second
6.5 hp * 0.747 MJ/Sec = 4.8555 MJ (per second) / 3.5
1.3872857142857142857142857142857 power points per burst
0.13872857142857142857142857142857 power points per Shot
thats more like it... what did i do wrong with the GAU-8/A




Your calculation with the GAU-8 was fine. You exaggerated the value of a horsepower by a factor of 1000 in the case of the M230.
Mike Miller, Materials Engineer

Disclaimer: Anything stated in this post is unofficial and non-canon unless directly quoted from a published book. Random internet musings of a BattleTech writer are not canon.
Christopher_Perkins
05/01/08 02:07 AM
67.172.211.128

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Quote:


Quote:

Assuming 1 power point is 3.5 MJ (i know youve never been convinced




That's a tiny quantity of energy so, no, I'm not convinced. It's far too little to accomplish the cited feats of BT energy weapons.




Cited by Whom? Not me.

Fallguy has erroneously linked the Energy Consumption of the Infantry Support Weapons with the Damage of the equivelent BattleField Class Weapons because the Writer of CBT (RPG) Companion's BACR did not do his Due Dillagance in ensuring that the Powerd Armour Variants (and their better power packs and larger Ammo Loads) of the Infantry Support Weapons would be able to do the BattleField Damage that they are stated to do in the Canon when BattleTech RPG 3rd Edition Stats are fed into the Infantry Platoon Creation Rules weapons conversion system

The Support Class Laser does 0.89 Damage per shot for 10 Power points.
The Small Laser (BA) and Small Laser (Vehicle) variants of the Support Class Laser do 3 Damage for an indeterminate amount of Power points.

Thats a multiple of 3.3707865168539325842696629213483 or 6 ap 27d6 when compared with the 6 ap 8d6 cited for the infantry Variant.

Taken the Easy way, the Support Infantry Class weapon fires for 0.01 Seconds delivering enough power to the target to laze a 1 cm diameter hole though 96 mm of RHAe (1985) where the front armour on a vehicle - (said homogenous armour plate massing 111 kg) out of 10 power points (35 MJ) and the BattleArmour & Vehicle Class Weapon fires for 0.03 Seconds and delivers enough power to the target to laze a 1 cm diameter hole through 108 mm of RHAe (374.63 kg) out of 34 power points (119 MJ)

35 MJ / .01 Sec = 3,500 MW
119 MJ / 0.0338 Sec = 3,500 MW (Same Laser, Prolonged Use)
119 MJ / 0.01 Sec = 11,900 MW (Possibly Same Laser, Dialed up Power Consumption, or Different Laser)

column of lazed materiel for Support Laser is 10 mm * 96 mm of WWII Spec rolled Homogenous Steel
10*10*96= 9600 cubic mm =

column of lazed materiel for Small Laser is 10 mm * 108 mm of WWII Spec rolled Homogenous Steel
10*10*108= 10,800 cubic mm = ?

If that is more energy required than provided (or an efficiency greater than 40%), then it was a 1 mm column instead of 10 mm... but i got one of them to work under the 40% maximum laser effeciency mark.

I keep forgetting the mass that you gave me for Rolled Homogenous Steel ... something like .7 g for a cubic cm? or was that kg?

and then there was te energy needed to convert the state of that mass in MJ?

i couldnt remember which f the sheets I put that in so I am reinventing the wheel again... at least I have the Armour More Spelled out...

Quote:

Quote:

0.747 MJ / Sec is 1 HP?




No. That would be 1000 horsepower. 1 horsepower = 747 watts.




1 HP
how many MJ / 1 Sec?
how many MJ / .1 Sec?
how many MJ / .01 Sec?

Quote:

Quote:

so 0.000246305418719211 power points per shot?




Very simplistically, yes. Realistically, no. The gun's motor only delivers peak horsepower to get the gun spinning. Once the gun is rotating at the desired speed, you'll only need to resist friction in the gun mechanism and meet the trivial demands of feeding in ammo - if you need 10kW to keep the gun firing, I'd be surprised. (This is the same reason a car gets worse mileage in-city than moving fast on the highway: it takes a lot of power to accelerate and not much to sustain a car moving against friction.)




Realisticly yes then...

Multiple power consumption rates then for each ROF

to spin up to that ROF from dead stop in a single second
to spin up to that ROF from a lower ROF (say LMG to MG)
to spin down to that ROF from a Higher ROF (say HMG to MG)
to maintain the spin (LMG, MMG or HMG / 1x, 2x, 3x, 4x, 5x, or 6x) for 1.5 seconds
to maintain the spin (LMG, MMG or HMG / 1x, 2x, 3x, 4x, 5x, or 6x) for 2.5 seconds
to maintain the spin (LMG, MMG or HMG / 1x, 2x, 3x, 4x, 5x, or 6x) for 4 seconds
to maintain the spin (LMG, MMG or HMG / 1x, 2x, 3x, 4x, 5x, or 6x) for 5 seconds
to maintain the spin (LMG, MMG or HMG / 1x, 2x, 3x, 4x, 5x, or 6x) for 9 seconds
to maintain the spin (LMG, MMG or HMG / 1x, 2x, 3x, 4x, 5x, or 6x) for 10 seconds
(1.5, 4, & 9 Seconds are Turn length - Fire Second)

While good power efficiency doctrine would be to spin up the Rotary MGs at the beginning of contact and to de rev at the end of contact, some guns will spin up for each trigger pull rather than maintaining a rotation. Sort of like the first stage of the trigger gets the gun spinning, and the second stage of the trigger fires.

Combat is more like driving in traffic when it comes to power consumption


Quote:


So after the first, 0.2-0.4 seconds, power demands for the gun might be 5-10kW, or 0.005-0.01MJ/second... approximately 0.003 of your power points per second, almost irrelevant of the number of bullets being fired.




you say almost...

so how many bullets variation would increase or decrease the power consumption by an amount that would round to 3.5 MJ/Second? Double, Tripple, Quadtouple?{?} Quintuple? Hextuple?

Quote:


that cant be right... not for 7 barrels




You're looking at it in the wrong way. The gun isn't eating energy to force bullets out the barrel; the gunpowder is doing that. The gun needs a moderate amount of horsepower to get the barrels spinning, and then a little juice to keep it spinning and keep the ammo belt flowing. The act of firing a bullet doesn't add up to much on that scale.




Not looking at the power to fire the bullet, looking at the energy to get 7 Barrels of a robust structure and calibre spinning at a rate necessary to fire a certain number of rounds in the first second of fire... do not know enoough of the fireing pattern for battletech MG whither or not the Doctrine is to keep the barrels spinning for 100 seconds of a battle when the gun actually fires for only 1 or 2 Seconds. Light Autocannon, Autocannon, Ultra Autocannon, and Rotary Autocannon probably maintain the 1x Spin rate for the entire battle, but with the ranges of LMG, MMG and LMG its unlikely.


Quote:


Quote:

Apachies M230 , 30 mm, 10 shots per second
6.5 hp * 0.747 MJ/Sec = 4.8555 MJ (per second) / 3.5
1.3872857142857142857142857142857 power points per burst
0.13872857142857142857142857142857 power points per Shot
thats more like it... what did i do wrong with the GAU-8/A




Your calculation with the GAU-8 was fine. You exaggerated the value of a horsepower by a factor of 1000 in the case of the M230.




if its not in a spreadsheet, i just don't get it most of the time...
Christopher Robin Perkins

It is my opinion that all statements should be questioned, digested, disected, tasted, and then either spit out or adopted... RHIP is not a god given shield
CrayModerator
05/01/08 10:00 AM
147.160.136.10

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Quote:

Quote:


Quote:

Assuming 1 power point is 3.5 MJ (i know youve never been convinced




That's a tiny quantity of energy so, no, I'm not convinced. It's far too little to accomplish the cited feats of BT energy weapons.




Cited by Whom?




Cited by the rules (armor mass lost per point of damage) and physics. You don't ablate (heat to thousands of degrees and burn off) 62.5kg of magical armor with 3.5MJ. You couldn't even boil 62.5kg of water with 3.5MJ. 3.5MJ wouldn't even boil a 2-liter bottle of water; it'd be just about enough to warm 1 liter (1 kilogram) of water from room temperature to boiling, and then power the vaporization process.

Quote:

column of lazed materiel for Support Laser is 10 mm * 96 mm of WWII Spec rolled Homogenous Steel
10*10*96= 9600 cubic mm =




That's a spurious comparison twice over. Primarily it's off-base since BattleTech armor loss entails blasting off entire "points" of ablative armor, which are 62.5kg per point for standard 'Mech armor. You need to look at the energy required to ablate 62.5kg of armor, not drill little holes.

Secondarily, the amount of energy needed to melt a thin hole in steel plate is different (far smaller) than the total energy put into the process. High power (and I do mean "very fast energy delivery") lasers will create surface plasmas that are excellent for soaking up laser light but, unfortunately, not so good at drilling into further armor. You may waste over 90% of the beam literally burning off the paint job.

Quote:

I keep forgetting the mass that you gave me for Rolled Homogenous Steel ... something like .7 g for a cubic cm? or was that kg?




7.8 grams per cubic centimeter, 7800kg per cubic meter.
Mike Miller, Materials Engineer

Disclaimer: Anything stated in this post is unofficial and non-canon unless directly quoted from a published book. Random internet musings of a BattleTech writer are not canon.
Christopher_Perkins
05/12/08 03:38 AM
24.125.106.130

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Quote:

Cited by the rules (armor mass lost per point of damage) and physics. You don't ablate (heat to thousands of degrees and burn off) 62.5kg of magical armor with 3.5MJ.




3.5 MJ * 7 = is the power Requirement per shot for a Support Pulse Laser's 5 AP, 5d6
5 AP 5d6 (n/S n/I) = 0.46 BT
so... 24.5 MJ to Render Combat Ineffective 62.5*0.46=24.5 kg of Barrier 10 Tech D "Standard" Ferro-Ceramic Armour, perhapse 0.46 mm thick
This is Equivelent in BattleTech to 0.66 points of Tech B Bar 7 Armour, or what appears to me to be 57.5 kg of Steel Armour ... and note this is Penetrate the Plate of Steel Armour, not -totally- ablate it
humm, this is about 10 mm thick if i am at all close in my estimates

A Support Pulse Laser Fires 15 Pulses
that is 367.5 MJ to Render Combat Ineffective 62.5*0.86=53.75 kg of Barrier 10 Tech D "Standard" Ferro-Ceramic Armour, perhapse 0.86 mm
This is Equivelent in BattleTech to 1.23 points of Tech B Bar 7 Armour, or what appears to me to be 107.5 kg of Steel Armour ... this is about 16 mm thick if i am at all close in my estimates

A Small Pulse Laser with the same power requirements, ap and damage rating as a Support Pulse Laser would fire 102 shots in a single Burst to do 3 BattleTech Damage
102 Shots * 7 power * 3.5 MJ = 2,499 MJ
3 BattleTech Damage is 62.5*3 = 187.5 kg of Tech D Bar 10 Standard "Ferro-Ceramic Armour", perhapse 3 mm thick

3 BattleTech Damage of Bar 10 Tech D armour is roughly equivelent to 4.29 points of Tech B Bar 7 Armour, or what appears to me to be 375 kg of Steel Armour ... this is about 60 mm thick IIAAACIME

so, while the canon in CBT RPG 3e Companion may imply that 3.5 MJ is enough to ablate 1 point of Armour in the CBT RPG 3e Companion, that is exactly the king of Bug that i have been screaming about in that book for years... LosTech & Combat Operations completely disagree with it
The Real damage rules published in Combat Operations clearly indicate that its closer to 427.33 MJ to render ineffective 62.5 kg of Tech D Bar 10 Armour...
This is roughly Equivelent to 125 kg of Steel (1.43 pts Tech B Bar 7) or possibly 20 mm thick

so, 1 cm * 1 cm * 3.14 beam * 2 cm = 6.28 cm*cm*cm
6.28 cm*cm*cm * 7.89 kg/(cm*cm*cm) = 49.5492 kg of steel actually being burned through out of a 125 kg plate of steel armour... or 39.6% damage to a plate being enough to render it ors de combat

How does 427.33 MJ stack up against 49.5492 kg of 1985 spec Rolled Homogenous Steel Equivelent Armour?


and if one takes into consideration the 833 MJ required to damage 3 points... thats even more power
6 cm * 1 cm * 1 cm * 3.14 = 18.84 cm^3 * 7.89 kg / cm^3 = 148.6476 kg out of 375 kg that was direcly affected in order to render the armour plate ineffective...
or 39.6% of the plate being damaged to render it ineffective

granted if one assumes 10 cm Lasers, this gets problematic... but if you dial it in to 10 mm (1 cm), works, mostly... and it is in the canon that a laser weapon is capable of dialing out to weld and dialing in to damage

also, the power requirements being so astronimical is also an easy way of explaining the heat for a laser... if a small pulse laser does 2 heat and uses 833 MJ, then the heat points could be calculated off the Power Draw of the weapon... and thus the HS Required if the unit is drawing off a Fusion Engine rather than a battery pack
Christopher Robin Perkins

It is my opinion that all statements should be questioned, digested, disected, tasted, and then either spit out or adopted... RHIP is not a god given shield
CrayModerator
05/12/08 08:16 AM
147.160.136.10

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Quote:

Quote:

Cited by the rules (armor mass lost per point of damage) and physics. You don't ablate (heat to thousands of degrees and burn off) 62.5kg of magical armor with 3.5MJ.




3.5 MJ * 7 = is the power Requirement per shot for a Support Pulse Laser's 5 AP, 5d6
5 AP 5d6 (n/S n/I) = 0.46 BT
so... 24.5 MJ to Render Combat Ineffective 62.5*0.46=24.5 kg of Barrier 10 Tech D "Standard" Ferro-Ceramic Armour, perhapse 0.46 mm thick
This is Equivelent in BattleTech to 0.66 points of Tech B Bar 7 Armour, or what appears to me to be 57.5 kg of Steel Armour ... and note this is Penetrate the Plate of Steel Armour, not -totally- ablate it
humm, this is about 10 mm thick if i am at all close in my estimates




Sorry, the fluff doesn't leave a lot of room for reinterpretation. BT lasers largely melt off the armor they destroy. BT lasers aren't satisfied with poking small holes in it. This is why I say that it may takes hundreds of megajoules to account for the damage of one small laser shot.
Mike Miller, Materials Engineer

Disclaimer: Anything stated in this post is unofficial and non-canon unless directly quoted from a published book. Random internet musings of a BattleTech writer are not canon.
Christopher_Perkins
05/14/08 03:52 AM
24.125.106.130

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Quote:

Sorry, the fluff doesn't leave a lot of room for reinterpretation. BT lasers largely melt off the armor they destroy. BT lasers aren't satisfied with poking small holes in it. This is why I say that it may takes hundreds of megajoules to account for the damage of one small laser shot.




Since when did BattleTech Fluff deal with Steel Armour?

I can only think of once, and that was more along the lines of
"This gun would normally have penetrated 300 mm of steel armour but was instead stopped by 30 mm of the new armour"

Actually, the armour is always described as being cut into, blasted off, but only once has it ever been described as vaporizing the armor without a trace... as Stackpoles description of the vaporization of an elemental with a Machine Gun, a Medium Laser and a Large Laser a point at variance with the body of the canon, I call it Hyperbole rather than Veritas

even then
11 Points of BT Armour
2 points being degraded by being hit with a burst of 63 M2 Shells (assume 80% hit rate, in scattered locations)
5 points being Degraded by being hit with 227.5 MJ
and 4 points being degraded by being hit with 361.67 MJ
and any 4 Points remaining BT damage (180.835 MJ) being expended against the Elemental and the ground

Anyway... What does the RPG 3 edition "vs characters" rules say for an Elemental Suit & Elemental being hit with
a 63 shot burst from AP 5 * 5d6
AP 7 * 39d6 @ 227 MJ
and AP 7 * 62d6 @ 361.67 MJ
Dead
Mostly Dead?
Flambe?

heck
5 BT @ (AP 7 * 12 d6) = 42 Shots * 20 Power/Shot * 3.5 MJ/Power = 2,940 MJ
8 BT @ (AP 7 * 12 d6) = 78 Shots * 20 Power/Shot * 3.5 MJ/Power = 5,460 MJ
if instead of increasing the D6 & Power Proportionate we simply prolong the shot from
1.55 BT @ (AP 7 * 12 d6) = 1 Shot * 20 Power/Shot * 3.5 MJ/Power = 70 MJ

so, again... the canon doesnt really deal with the amount of energy used to do BATTLETECH scale damage... We can only Infer it from the Equivelent Infantry weapons that do less damage...

Hopefully the upcomming iteration of the RPG Rules finally gives the generic stats for BattleField Weapons
Christopher Robin Perkins

It is my opinion that all statements should be questioned, digested, disected, tasted, and then either spit out or adopted... RHIP is not a god given shield
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