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Old 10-22-2013, 02:22 AM   #11
Anthony
 
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Default Re: [Spaceships] Conventional and EM Gun ranges in atmosphere?

Quote:
Originally Posted by vicky_molokh View Post
I'm not sure, that would imply that an object will drop its kinetic energy to zero at the same distance regardless of the kinetic energy it had. How would that happen?
Kinetic energy is proportional to v^2. Drag is proportional to v^2. Loss of energy from drag is equal to (drag force) * (distance). Thus, dE/dd = kE. This is a well known differential equation, and means E ~= e^(kd) (where E is energy, d is distance, and e is the base of the natural logarithm) and half-energy range is ln(2) / k (note: doing this quick by memory, so some signs may be off).
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Old 10-22-2013, 02:39 AM   #12
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Default Re: [Spaceships] Conventional and EM Gun ranges in atmosphere?

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Originally Posted by Anthony View Post
Kinetic energy is proportional to v^2. Drag is proportional to v^2. Loss of energy from drag is equal to (drag force) * (distance). Thus, dE/dd = kE. This is a well known differential equation, and means E ~= e^(kd) (where E is energy, d is distance, and e is the base of the natural logarithm) and half-energy range is ln(2) / k (note: doing this quick by memory, so some signs may be off).
I have trouble parsing that - my math is very rusty (the equation-related parts, anyway). Sorry.
What is 'k'?

What I was trying to say was that more powerful throws do result in greater range, so if range is proportional to calibre only, something seems missing.
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Old 10-22-2013, 03:00 AM   #13
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Default Re: [Spaceships] Conventional and EM Gun ranges in atmosphere?

IDHMBWM
Are all those quoted damage values for the Spaceships versions including the minimum multiplier for their velocity; IIRC *2 for EM and *5 for Grav?

And some advice I was given from the forum may apply.
A rough estimate, for gaming:
• Muzzle Energy x 4 gives
• Muzzle Velocity x 2 (mv equates to me^0.5)
• Damage x 2.05 (ME^0.52)
• 1/2D range x 1.4 (velocity increase ^ 0.5)
• Max Range x 2 (mv increase = range increase)
Acc +1* (treat this as per Match Grade loads and only add an increase if the original weapon is Acc 4+)

Or if I remembered those damage multiples correctly (*2 & *5):
EM get +40% 1/2D range, Grav get +120%
EM max *2, Grav max *5.
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Old 10-22-2013, 03:11 AM   #14
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Default Re: [Spaceships] Conventional and EM Gun ranges in atmosphere?

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Originally Posted by jacobmuller View Post
IDHMBWM
Are all those quoted damage values for the Spaceships versions including the minimum multiplier for their velocity; IIRC *2 for EM and *5 for Grav?
×1 for conventional, ×2 for EM. Didn't look much at grav, as it isn't in my setting.
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Old 10-22-2013, 07:43 AM   #15
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Default Re: [Spaceships] Conventional and EM Gun ranges in atmosphere?

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Originally Posted by vicky_molokh View Post
What I was trying to say was that more powerful throws do result in greater range, so if range is proportional to calibre only, something seems missing.
I'm not sure about extremely slow object but for things like various fastball pitches or different pressure loads in guns I believe what you're seeing is just an increased distance traveled before the projectile hits the ground, not the effects of any absolute limit on its range.
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Old 10-22-2013, 09:30 AM   #16
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Default Re: [Spaceships] Conventional and EM Gun ranges in atmosphere?

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Originally Posted by vicky_molokh View Post
Hmm. A 2½cm conventional slug from SS does 3d(2) dDamage, i.e. 30d normal damage.

Again 2cm VRF EM Gun does 60d, i.e. your guns seem to be better.

Just saying.
SS conventional shells appear to be 50% propellant, and the projectile itself loses about 50% of its mass (based on damage) before hitting the target. EM just lacks the propellant. This second loss is likely in the form of reaction mass, decoys, etc. The mass given for Gauss/Grav shells corresponds to a 2:1 ratio of lead or steel (I calculated masses for the unguided rounds of my setting by figuring out the mass of lead/steel of a sphere of the right caliber, then doubling it; my numbers were almost exactly what you see in SS, so I went with those instead). So, use the ballistics spreadsheet with a 2:1 ratio and then halve damage (to account for mass loss). Assume the increased range from using a rocket and the decreased range from weighing less for part of the travel time cancel out.
That, or just use Anthony's quick-and-dirty rule.
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Old 10-22-2013, 10:48 AM   #17
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Default Re: [Spaceships] Conventional and EM Gun ranges in atmosphere?

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Originally Posted by vicky_molokh View Post
I have trouble parsing that - my math is very rusty (the equation-related parts, anyway). Sorry.
What is 'k'?

What I was trying to say was that more powerful throws do result in greater range, so if range is proportional to calibre only, something seems missing.
k is an unspecified constant. Half damage range is strictly a function of projectile sectional density and drag coefficient. Actual maximum range is a rather complex function, because it involves both atmospheric drag and gravity, but is pretty heavily dominated by drag unless dealing with projectiles that are quite large or quite slow (at least, by bullet terms. A thrown rock is both large and very slow by those terms).
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