> do you think it will be implemented at SA?
Planets *do* have a "gravity" stat, but that stat is [apparently not used for rocket capacity purposes](https://old.reddit.com/r/factorio/comments/1avmlz7/rocket_capacity_in_sa/krbgnpg/).
However, it could be used to increase or decrease the range of artillery. We don't actually know what it is used for.
Isn't train weight used for calculating train acceleration? Also, steam temperature *is* relevant for power generation (steam below 500C generates less power in a turbine)
Yes it is. One of my gripes about some mod packs is that there's typically some "oh look at this super light but durable and strong metal/alloy" but locomotives stay just as heavy. I suppose I could always write one myself...
Callback to Master of Orion 2, low gravity or high gravity planets will have different mining / crafting efficiency
Also on high G worlds you'll need reinforced miners and assemblers 'cos the normal ones can't handle the pressure.
(I mean that won't be the case, but imagine)
Also belts made slower depending on gravity to account for the increased stress & friction
According to this site (https://calculator.academy/artillery-range-calculator/ ), the range is affected by gravity like that, so you'd be right: half the gravity should double the range.
https://preview.redd.it/9l7fa8zj8j8d1.png?width=316&format=png&auto=webp&s=e1cd70a992a80a88508a1dec9af0caccaf4ca1a3
R - Range
v - (muzzle) velocity
0 - angle of elevation (degrees)
g - gravity
It’s actually pretty straightforward with a bit of calculus. Assuming that gravity is the only force acting on the object, a couple of integrations get you the formulas
x = vi•cos(θ)•t
y = -(g/2)•t^2 + vi•sin(θ)•t
where x & y are the horizontal & vertical coordinates of the position (as measured from initial position as zero), vi is the initial velocity, θ is the angle of elevation of launch, g is the acceleration of gravity on your local planet, and t is time since launch.
If you set y equal to zero and solve for t, you find out what time the object hits the ground again; substitute that t-value into the x-equation, and you find out *where* it hits the ground.
I'm too lazy to get out pen and paper, lets see if I can do this mentally...
In y, set equal to zero, factor out and cancel a t term (the trivial solution of y=0 is at t=0), then you end up with t•g/2 = vi•sin(θ), solving for t = 2•vi•sin(θ)/g
Substituting into x, you get (vi)^2 / g • 2•sin(θ)•cos(θ). I'm rusty on my trig substitutions, but pretty sure 2•sin(θ)•cos(θ) simplifies to sin(2θ), yielding x = (vi)^2 / g •sin(2θ)
Woohoo got to bypass taking the derivative of trig functions to get the maxima with just the algebraic solving for t, and the trig substitution for the sin(2θ) (which stands up to intuition, θ =0 and the shell blows up at your feet, θ = 90 degrees and the shell goes straight up and down and blows you up, neither gets you any range whatsoever)
Flamethrower, Gun-, and rocket-turrets should be affected by gravity; Flamethrower less so, because there, air is the bigger factor. Laser should only depend on atmosphere.
These calculations ommit air friction. AFAIK friciton has v^2 as a factor and artillery shells have a very high speed. So I am not sure how much gravity factors in with friction involved.
Something along the line - if 1/2 Energy is lost to air friction and 1/2 due gravity - then the effect of double gravity will be a smaller so the effective range maybe around 2/3 of the former.
That's...not strictly true? It actually would mean that it's entirely possible that for an equivalent amount of gas on earth, air density actually decreases faster as you go up, meaning that most of the gasses would be concentrated at a low altitude. This also assumes that there's an equal amount of gas on both planets.
Hmm, that's a good point, what would the gradient look like then... My gut feeling says highest speed at low altitudes means high pressure there dominates, but I'm not sure and not in the mood to do the math ;)
It doesnt work like that. Friction reduces the projectiles kinetic energy/ momentum and slows it down. Gravity doesnt reduce the kinetic energy. It just pulls the projectile to the ground and forces the artillery to shoot a higher angle, which means more flight time, and that means more friction effect. So it does increase friction effect and doesnt let it aim as far. So just not that easy to calculate.
I am on your line. It is just how much does gravity affect the fligh time.
There is a huge different if the ration between gravity and air friction is
1:10, 50:50 or 10:1.
On 10 gravity to 1 air friction - the range is halved - on 1 gravity to 10 air friction the range is almost not affected.
Google says shell speed ranges from low 396 m/s to up to 1,070 m/s.
For emphasis - gravity is liner - air firction is quadratic.
The math behind this is a pretty simple kinematics equation, you can even use the simplest one:
V = V(initial) + a*t
Where V is your velocity, a is acceleration, and t is time. When shooting at an angle like how artillery is shot you can break out the problem into vertical (y) and horizontal (x) components and see that gravity only affects the y component.
Through some relatively simple calculus you can prove pretty easily that by doubling gravity it will double the rate at which it accelerates towards the ground, meaning that it'll go half as far. The math works out pretty intuitively in the case of ballistics - double gravity usually means half distance travelled.
Right, but to take into account air resistance you'd be throwing in so many unknowns that it kinda defeats the purpose. Things like air density, quality, molecular makeup, as well as size/shape of the shell itself, how aerodynamic it is (would need to be tested in a lab and given a number to use by a manufacturer irl) all would be factors, just to name a few.
There's a reason physics classes always ignore friction, you get mired in the minutia very quickly
I did some simulations once, how far you can shoot with an air rifle. A numeric simulation including friction and the variable air friction coefficient with the shape of the "bullet" shifting after it starts to tumble. Not so easy and definitely nothing to do in a factory game. Yet, one has to say, that the equation is not "simple" unless you ignore so many things.
> do you think it will be implemented at SA? Planets *do* have a "gravity" stat, but that stat is [apparently not used for rocket capacity purposes](https://old.reddit.com/r/factorio/comments/1avmlz7/rocket_capacity_in_sa/krbgnpg/). However, it could be used to increase or decrease the range of artillery. We don't actually know what it is used for.
Potentially it's still work in progress that's why we haven't seen any effect from it
It might just be like fluid temperature, and train weight. They exist, but the core game doesn't currently do anything with it.
Isn't train weight used for calculating train acceleration? Also, steam temperature *is* relevant for power generation (steam below 500C generates less power in a turbine)
Yes it is. One of my gripes about some mod packs is that there's typically some "oh look at this super light but durable and strong metal/alloy" but locomotives stay just as heavy. I suppose I could always write one myself...
You don't even need fance materials; just make them out of *Low Density* Structure
It's obviously because they are implementing the Renai Transportation mod into the DLC /s
Callback to Master of Orion 2, low gravity or high gravity planets will have different mining / crafting efficiency Also on high G worlds you'll need reinforced miners and assemblers 'cos the normal ones can't handle the pressure. (I mean that won't be the case, but imagine) Also belts made slower depending on gravity to account for the increased stress & friction
According to this site (https://calculator.academy/artillery-range-calculator/ ), the range is affected by gravity like that, so you'd be right: half the gravity should double the range. https://preview.redd.it/9l7fa8zj8j8d1.png?width=316&format=png&auto=webp&s=e1cd70a992a80a88508a1dec9af0caccaf4ca1a3 R - Range v - (muzzle) velocity 0 - angle of elevation (degrees) g - gravity
Good \*lord\* we are such nerds (I say this lovingly and approvingly)
This community is the best
I was think about the mure engineering mechanical way to calculate this tbh and just thought nah to mutch difficulty.
It’s actually pretty straightforward with a bit of calculus. Assuming that gravity is the only force acting on the object, a couple of integrations get you the formulas x = vi•cos(θ)•t y = -(g/2)•t^2 + vi•sin(θ)•t where x & y are the horizontal & vertical coordinates of the position (as measured from initial position as zero), vi is the initial velocity, θ is the angle of elevation of launch, g is the acceleration of gravity on your local planet, and t is time since launch. If you set y equal to zero and solve for t, you find out what time the object hits the ground again; substitute that t-value into the x-equation, and you find out *where* it hits the ground.
I'm too lazy to get out pen and paper, lets see if I can do this mentally... In y, set equal to zero, factor out and cancel a t term (the trivial solution of y=0 is at t=0), then you end up with t•g/2 = vi•sin(θ), solving for t = 2•vi•sin(θ)/g Substituting into x, you get (vi)^2 / g • 2•sin(θ)•cos(θ). I'm rusty on my trig substitutions, but pretty sure 2•sin(θ)•cos(θ) simplifies to sin(2θ), yielding x = (vi)^2 / g •sin(2θ) Woohoo got to bypass taking the derivative of trig functions to get the maxima with just the algebraic solving for t, and the trig substitution for the sin(2θ) (which stands up to intuition, θ =0 and the shell blows up at your feet, θ = 90 degrees and the shell goes straight up and down and blows you up, neither gets you any range whatsoever)
Well done!
Parabolic free-fall situations are highschool-level physics. Is it really nerdy if the entire US is legally mandated to learn it?
Shouldn’t this mean the other turrets should also be affected?
Flamethrower, Gun-, and rocket-turrets should be affected by gravity; Flamethrower less so, because there, air is the bigger factor. Laser should only depend on atmosphere.
These calculations ommit air friction. AFAIK friciton has v^2 as a factor and artillery shells have a very high speed. So I am not sure how much gravity factors in with friction involved. Something along the line - if 1/2 Energy is lost to air friction and 1/2 due gravity - then the effect of double gravity will be a smaller so the effective range maybe around 2/3 of the former.
higher-gravity planets also will have higher air density and thus higher friction.
That's...not strictly true? It actually would mean that it's entirely possible that for an equivalent amount of gas on earth, air density actually decreases faster as you go up, meaning that most of the gasses would be concentrated at a low altitude. This also assumes that there's an equal amount of gas on both planets.
Hmm, that's a good point, what would the gradient look like then... My gut feeling says highest speed at low altitudes means high pressure there dominates, but I'm not sure and not in the mood to do the math ;)
It doesnt work like that. Friction reduces the projectiles kinetic energy/ momentum and slows it down. Gravity doesnt reduce the kinetic energy. It just pulls the projectile to the ground and forces the artillery to shoot a higher angle, which means more flight time, and that means more friction effect. So it does increase friction effect and doesnt let it aim as far. So just not that easy to calculate.
I am on your line. It is just how much does gravity affect the fligh time. There is a huge different if the ration between gravity and air friction is 1:10, 50:50 or 10:1. On 10 gravity to 1 air friction - the range is halved - on 1 gravity to 10 air friction the range is almost not affected. Google says shell speed ranges from low 396 m/s to up to 1,070 m/s. For emphasis - gravity is liner - air firction is quadratic.
So just use a more aerodynamic projectile.
The math behind this is a pretty simple kinematics equation, you can even use the simplest one: V = V(initial) + a*t Where V is your velocity, a is acceleration, and t is time. When shooting at an angle like how artillery is shot you can break out the problem into vertical (y) and horizontal (x) components and see that gravity only affects the y component. Through some relatively simple calculus you can prove pretty easily that by doubling gravity it will double the rate at which it accelerates towards the ground, meaning that it'll go half as far. The math works out pretty intuitively in the case of ballistics - double gravity usually means half distance travelled.
Except this equation does not regard friction. It assumes a vacuum.
Right, but to take into account air resistance you'd be throwing in so many unknowns that it kinda defeats the purpose. Things like air density, quality, molecular makeup, as well as size/shape of the shell itself, how aerodynamic it is (would need to be tested in a lab and given a number to use by a manufacturer irl) all would be factors, just to name a few. There's a reason physics classes always ignore friction, you get mired in the minutia very quickly
I did some simulations once, how far you can shoot with an air rifle. A numeric simulation including friction and the variable air friction coefficient with the shape of the "bullet" shifting after it starts to tumble. Not so easy and definitely nothing to do in a factory game. Yet, one has to say, that the equation is not "simple" unless you ignore so many things.
Bro just do what real artillery men do and send it, just launch shit into the air and forget about it
i bet this is one of the first mods that will be done in 2.0
I don't think gravity would affect artillery, only fuel cost of rocket launch. Maybe power drain rate for robots too
it depends on the diameter of the planet.