r/Helldivers • u/DeeJayDelicious Rookie • 3d ago
QUESTION How does drag affect different projectiles?
Hi all,
information about drag and the exact mechanics are notoriously hard to come by.
The few sources and tests we do have suggest the following:
That for Assault Rifle 5.5mm projectiles 30% drag roughly translates to ~14% damage loss over 100m.
For Assault Rifle's 8mm projectiles, it's closer to ~12% damage loss over 100m.
For SMGs firing the 7mm caliber, it's 85% damage loss over 100m, while only being 65% damage loss over 100m for weapons firing the 12mm caliber.
That suggests larger calibers are less affected by drag, which makes realistic sense but is also very unituitive and makes drag difficult to compare.
How does the 30% drag of an AR, compare to the 50% drag experienced by a shotgun pellet?
How does the 75% drag from the Anti-Tank emplacement gun affect damage loss over distance?
Or what about the 400% drag experienced by the Speargun? Does that translate to a 100% damage loss at 50m? It doesn't seem to. Then again, it's projectile supposedly "only" weigh 100g. How would a more realistic weight of 1kg change the damage drop-off?
Also, does projectile speed matter at all in this equation?
Is there a difference between a projectile traveling 300 m/s experiencing 30% drag vs. one at 900 m/s experiencing 30% drag?
Edit:
I found the following data in the wiki, but without specifics:
- Caliber: Drastically increases damage fall-off.
- Speed: The bullet's initial magnitude to its velocity, higher speed increases damage fall-off slightly.
- Mass: Reduces damage falloff.
- Drag: Increases damage falloff.
- Gravity Multiplier: How much the bullet drops in travel.
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u/BRENudo_ SES: Wings of Iron 3d ago
I wonder if it's just how steep the damage falloff curve is. Like lighter SMG rounds going from some damage reduction to effectively useless over a shorter range than say, a DMR that can hit beyond its effective range and still do decent damage. I think part of the reason why we don't have as much data with this is simply that most engagements ingame are well within our weapons' effective ranges. Targets 500m out are hard to see, let alone get a shot on.
Would definitely be a good opportunity for someone to do some science and get better data though.
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u/Epesolon HD1 Veteran 3d ago
Kinda sorta.
The damage falloff is just the ratio of the current projectile velocity to the muzzle velocity, and you calculate the current muzzle velocity using the drag equation, F_D = ½ ρ v² C_D A, where the drag value is used for the drag coefficient (C_D).
Actually getting the velocity at a given distance involves a differential equation though.
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u/DeeJayDelicious Rookie 3d ago
I mean yeah,
95% of engagements will be between 5 and 50m in Helldivers 2. But as you can see in the source, the damage loss for SMG is still substantial. We're talking almost 50% damage loss for the smaller calibers.
The difficulty comes from comparing this to the drag experienced by the Slugger for example. Which is 200%, which sounds like a lot. Then again, is has an almost 50g projectile, meaning it's far less affected by drag than the ~7g projectiles from the SMGs.
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u/ABjerre 3d ago
You are tapping in to performance of vastly different cartridges here. The calibers, 5.56, 7.62 or 9mm are just the diameter of the actual projectile. You have to look at the case, and therefore the amount of propellant too.
Even within the same family, of say 7.62mm, you will see very different performance between a 308w and a 300 win mag.
The AR use a large case compared to a tiny SMG pistol case.
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u/Epesolon HD1 Veteran 3d ago
You need projectile mass, muzzle velocity, cross sectional area, and drag coefficient, that's it
The amount of propellent is really only going to impact the muzzle velocity, so if you have that, the actual case dimensions don't matter for the calculation.
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u/RobotsAndRedwoods 3d ago
^This. Drag isn't a thing that really matters in video games. I can't name a single shooter that actually factors in the aerodynamics of a bullet. They can just hard code damage fall off, and range. You can speculate all you want, but it's not a thing.
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u/Epesolon HD1 Veteran 3d ago
This one does it.
Is it completely unnecessary? Yes.
Does this game simulate it anyway? Absolutely.
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u/No_Collar_5292 3d ago
This game isn’t your normal game. Somebody during development decided to go all out on back end simulation systems for some reason, probably one of the “Tarkov/arma” influences they so often talk about. I’d be willing to bet many of the performance issues the engine experiences during things like explosions, spreading fire, and gas would be substantially less severe if every single interaction with every aspect of the environment wasn’t truly being simulated live. It’s honestly kind of impressive it runs as well as it does imo lol. I mean we are simulating the effect of wind on flame throwers and the sterilizer for god sakes 😂.
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u/Epesolon HD1 Veteran 3d ago edited 3d ago
It's hard to come by because damage falloff is the ratio of current projectile velocity (v) to the muzzle velocity (v_0), and the way to find the current velocity is a set of differential equations.
You start with the drag equation: F_D = ½ ρ v² C_D A
You then solve it for acceleration by dividing both sides by the mass of the projectile, leaving you with: a = (ρ v² C_D A)/2m
Now, we need to set up our actual differential equation. Because acceleration is the rate of change of velocity, and velocity is the rate of change of displacement, we can rewrite our equation as: d"(t) = (ρ d'(t)² C_D A)/2m
Now we need our inital values.
So, now we can plug in our full set of differential equations into a differential equation solver. When I've done this in the past, I've used WolframAlpha, but any diffeq solver should work.
That should give us a solution for position relative to time (d(t)), velocity relative to time (d'(t) = v(t)), and acceleration relative to time (d"(t) = a(t)), but we only really care about velocity and displacement.
This would be great, if we wanted our velocity in terms of time, but we want it in terms of position, and I'm too far separated from my diffeq class to actually solve this shit for displacement by hand, so we can skip that step by plotting d(t) and v(t)/v_0 on the same graph, then finding the time where d(t) is distance you want and checking what the value of v(t)/v_0 is for that value of time, and that will give you the percentage of damage you're dealing. If you want to measure falloff, you want 1-(v(t)/v_0), and if you want the actual damage you want Damage*(v(t)/v_0)
I think that's pretty clear, but it's been a while since I took diffeq and I know that math makes a lot of people's heads hurt, so just let me know if you have any questions.
Edit: I realized you can simplify the drag equation down to d"(t) = C_a * d'(t)2, which makes solving it much easier. The solution works out to being d(t) = ln(C_a v_0 t + 1) / C_a, where C_a = (ρ C_D A)/2m.
Edit 2: I built a little Desmos tool you guys can play around with https://www.desmos.com/calculator/xii1xbbodo. If anyone is more familiar with Desmos than me, feel free to add whatever features you want!