Incorrect Physics in Call of Cthulhu!

Castellan

First Post
*gasp*

Horror of horrors! I finally began reading my copy of CoC today, and I stumbled across this myth of physics!

Page 29, Sidebar -- Cars and Other Vehicles:
If two objects of roughly the same weight or mass run into each other head-on, the speeds should be added together.

:eek:

No, no, no, no, no!

In any kind of interaction between two forces, energy must be conserved. If two cars (same make and model, for easy math) strike each other head-on while both are going exactly 60 miles-per-hour, and you add the speeds together, then you've got twice as much energy coming out of the collision as you have going into it. This can't happen. Why, oh why did WotC perpetuate this myth?!

If both cars are the same mass and traveling the same speed when they strike head-on, under ideal conditions, it's the exact same thing as each car hitting an immobile object (say, a really thick brick wall).

For a good explanation, go here.

Now, if the cars are traveling at different speeds, or are different masses, that's a completely different story.

However, to keep things simple in-game (I don't want to be doing kinematic equations and figuring out the elasticity of a collision at the table) I'd probably take the total speed and divide it by two (even though this is technically wrong) unless one car's speed or mass was orders of magnitude greater than the other (i.e. an 18-wheeler strikes a Yugo... I don't think the truck is gonna notice that much!).

:rolleyes: Good grief... I always knew I was a geek, but I never thought I'd be ranting about the physics in a frickin' RPG source-book! Is there a clinic I can check myself into?!


:D
 

log in or register to remove this ad

Darkness

Hand and Eye of Piratecat [Moderator]
Are you referring to Chaosium's CoC or CoC d20?

'cause if it's the latter, I'll probably have to move this thread to the d20 system games forum... :eek:
 



Ealli

First Post
Probably one of those side affects of having Great Old Ones running around. Roll your sanity checking now that you realize that physics aren't working the way you thought they were supposed to.

On a serious note, if you add velocities together in a head-on collision it will be a good enough estimate for the game (remember that one velocity will be in the negative direction of the other).
 


CrazyMage

4th Level Lawful Good Cleric
SableWyvern, I thought the same thing--what about Car Wars? "Add speeds to find amount of damage." D'oh!

Drive offensively!
 

CRGreathouse

Community Supporter
Do the normal rules assume hitting a fixed object (brick wall) or not (parked car)? If the former, you're right; if the latter, I'm not so sure.
 

LightPhoenix

First Post
Castellan said:
If both cars are the same mass and traveling the same speed when they strike head-on, under ideal conditions, it's the exact same thing as each car hitting an immobile object (say, a really thick brick wall).

However, to keep things simple in-game (I don't want to be doing kinematic equations and figuring out the elasticity of a collision at the table) I'd probably take the total speed and divide it by two (even though this is technically wrong) unless one car's speed or mass was orders of magnitude greater than the other (i.e. an 18-wheeler strikes a Yugo... I don't think the truck is gonna notice that much!).


I assumed it was because of conservation of momentum (which is part of conservation of energy), all things being equal they would crunch against each other in some sort of non-elastic collision, which for all intents and purposes would stop both at the point of impact.
 

Zerovoid

First Post
You're driving at 60 mph? Well I hit you from behind with my car going at 61 mph. Now the cars are both going at 121 mph. Sweet.

Luckily, the math is pretty easy. If we assume that the cars stick together after the collision, then all we need is conservation of momentum.

m1*v1 + m2*v2 = m3*v3
but since the masses stick together, and the cars are the same mass
m1 = m2
m3 = 2*m1
m1*v1 + m1*v2 = 2*m1*v3
so, v3 = (v1 + v2)/2

Of course, you have to do this for both the x and y directions, but otherwise it should work. Obviously if the cars are going in opposite directions, one of their speeds will have to be negative.
 

Remove ads

Top