Questions related to racing and physics

[Brien Martin]

I have tried finding these answers on the internet, but I believe I simply don't know enough about physics (I got a D in high school) to really know what it is I'm supposed to be looking for.

I'm working on a game that will simulate drag racing. As I want the game to be as realistic as possible, I need to know if there are formulae that I can use to calculate the following items. And, if so, what those formulae are:

1) In certain situations, a driver will shutdown the power to his car to keep from having a crash or a serious engine malfunction. What I need to know is the formula, if it exists, that will tell me how far a moving object, traveling at a certain speed, will travel if the object will do nothing but decelerate from the point of shutdown until it comes to rest.

2) In that same situation, can the time be calculated for the car from the shutdown point to the finish line?

3) In that same situaion, can a speed be calculated for the car as it crosses the finish line?

Let me give some sample numbers in the event that someone can post an example ...

Let's say that a car is traveling at 300 mph, and has traveled 960 feet of a 1,320-foot, quarter-mile drag strip. At the 960th foot, the driver must shutdown the car and let it coast to the finish line.

How far will the car travel before it actually comes to rest?
How long will it take for the car to travel the final 360 feet to cross the finish line?
What speed would the car be traveling when it crosses the finish line?

I hope that there are some formulae that apply here.

Thanks in advance for any help you can offer.

Brien

 

[russ_watters]

Did you take physics in high school? Calculus? Most of this is derived from Newton's f=ma and the relations between (definitions of?) "speed" and "acceleration".

Ie, if f=ma, then a=f/m. If your mass is 1000kg (metric is easier to use...) and your force is 1000n, your acceleration is 1000/1000= 1m/s/s. If your initial speed is 100m/s, your time is 1000/100=10s. Distance is tougher because it requires calculus, but if you use an iterative approach, you can just take the average speed and multiply by time: 100/2*10=500m.

All of this is easy to do in an excel spreadsheet and though I'm not a programmer, I'd think an iterative approach is the way to go (apply the equations at, say, intervals of .1s using the known forces on the car.

That all works fine if you are using brakes and have a constant deceleration force (over your .1s interval, at least), but if you are talking about coasting and using a combination of friction and air resistance, it gets more complicated. You have a constant force associated with friction and a force that varies with the square of speed for air resistance. Ie:

F_f=K
F_d=C_d*V²

The constants K and C_d are physical properties of the car and can't really be calculated from scratch, but only by experimentation (or, in your case, perhaps trial and error).

 

[Brien Martin]

Did you take physics in high school? Calculus? Most of this is derived from Newton's f=ma and the relations between (definitions of?) "speed" and "acceleration".

Ie, if f=ma, then a=f/m. If your mass is 1000kg (metric is easier to use...) and your force is 1000n, your acceleration is 1000/1000= 1m/s/s. If your initial speed is 100m/s, your time is 1000/100=10s. Distance is tougher because it requires calculus, but if you use an iterative approach, you can just take the average speed and multiply by time: 100/2*10=500m.

All of this is easy to do in an excel spreadsheet and though I'm not a programmer, I'd think an iterative approach is the way to go (apply the equations at, say, intervals of .1s using the known forces on the car.

That all works fine if you are using brakes and have a constant deceleration force (over your .1s interval, at least), but if you are talking about coasting and using a combination of friction and air resistance, it gets more complicated. You have a constant force associated with friction and a force that varies with the square of speed for air resistance. Ie:

F_f=K
F_d=C_d*V²

The constants K and C_d are physical properties of the car and can't really be calculated from scratch, but only by experimentation (or, in your case, perhaps trial and error).

Russ,

Thanks for your reply. I had physics in high school, yes, but it was not only 27 years ago, but I got a "D" ... which means the little bit I did learn wasn't obviously enough to tackle *this* problem

And, as an accountancy major, I took business calculus, which ignored all those cool formulas and stuff that the scientists needed. So, I stand where I stood this morning ... ignorant of the formulas necessary to make this work for me. Your assistance, however, has given me a starting point from which to continue.

Thanks,
Brien

 

[rcgldr]

Unless there's a safety hazard, drag cars are supposed to exit the drag strip at the normal exit point, which is well past the end of the 1/4 mile. Typically drivers will coast to the finish line before braking or deploying chutes.

Aerodynamic drag force is relative to speed^2. The chutes are needed, because the brakes along won't usually stop a drag car in the typical 1/2 mile of runoff past the 1/4 mile line. Maximum braking force would be somewhat proportional to downforce at the rear tires.