[Spaceships] Conventional and EM Gun ranges in atmosphere?

[vicky_molokh]

Originally Posted by Anthony

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.

[jacobmuller]

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.

[vicky_molokh]

Originally Posted by jacobmuller

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.

[lexington]

Originally Posted by vicky_molokh

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.

[Anthony]

Originally Posted by vicky_molokh

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).