A writer's (not a physicist's) question

1. Dec 11, 2008

davewe

I'm hoping someone can indulge me here. I'm a writer and wannabe novelist working on a satirical thriller.

I have a character who's a young physicist and a VERY smart one. At the beginning of the story he's driving with his grandfather, a wild old coot. The grandfather is driving his Porsche at 90mph and takes a dramatic curve on a guardrailed road overlooking a precipitous drop. He does not use his brakes. It's a game for the old guy.

Our young physicist freaks out. Using a physics formula (momentum?) I want him to be able to tell the grandfather that with the weight of the car, wheelbase length, at a certain speed, the car would have gone off the cliff. Could this be done? IOW, I want him to say "Considering that momentum is mass times velocity, and considering this Porsche is 3120 lbs. with a width of 71" and a wheelbase of 92.5", at 92 mph we would have inevitably gone off the cliff." Is it possible to determine this and what might a formula be?

Thanks for any info you can provide.

Dave

2. Dec 11, 2008

LowlyPion

Welcome to PF.

On a curve you would want to be aware of the radius of the curve as well as the speed, because the force that pulls the car to the edge is mv²/r and what's holding the car to the road is the coefficient of friction that is expressed as a percentage of the weight of the car. Loose gravel might be much lower coefficient than dry black asphalt.

3. Dec 12, 2008

davewe

So here's what I have in my story. We start with just after the grandfather has taken the sharp curve and slowed down. Does what I've written make sense? Remember, as a fiction writer I'm going for the illusion of reality.

"What the f**k were you doing?" I spat and foamed.

He shrugged. "Don't know what you're complaining about. I held back for your benefit. Took it at ninety just last week."

Ninety? I closed my eyes and took a deep, cleansing breath. Clearly the strain of her illness had precipitated a death wish in the old man. My shrink says that suicidal thoughts are common with old folks after the death of a partner.

I finally was able to speak. “Ninety-one.”

“Ninety-one? Ninety-one what?”

“That’s the speed at which we would have crashed through the rail no matter how good your diving skills. Mass - approximately thirty five hundred pounds for the car and two occupants, times velocity squared, divided by our distance to the center of the curve’s radius. Now if we consider the friction coefficient…it’s ninety-one miles per hour and we're still flying.”

“Very clever.”

4. Dec 12, 2008

franznietzsche

The equations you want is

$$\frac{v^2}{R} = \mu g$$
where $$\mu$$ is your coefficient of friction. Mass cancels out because it's on both sides of the equation, all you need are the accelerations. This gives
$$v = \sqrt{\mu g R}$$
Anything faster than $$v$$ and you're still flying.

Although, note that turns are often banked for this reason, which changes things obviously. With a bank of angle $$\theta$$ the critical velocity becomes
$$v = \sqrt{\frac{\mu g R}{cos \theta}$$
All of this assumes that the curve of the turn is the same as a circle. But otherwise, makes sense yes.

5. Dec 12, 2008

LowlyPion

Your numbers are a bit high.

The coefficient of friction is going to need to be roughly equal to the V2 (in your case more than 25² or about 640) divided by g (9.8) and the radius in meters.

Since the coefficient must be less than 1 that means your radius of the curve must be bigger than 65 meters (2/3 of a football field - hardly hairpin) or you're kissing the rocks at the bottom of the Pacific Coast Highway.

6. Dec 12, 2008

davewe

When you say my numbers are too high, do you mean 91 is too high? BTW, one other element to consider is that while the grandfather doesn't use his breaks, he downshifts when he hits the curve.

BTW, in my misspent youth a buddy of mine use to drive me this way down the Pacific Coast Hwy. Lots of fun though we weren't doing 90.

Dave

7. Dec 13, 2008

franznietzsche

You must be an engineer if you actually use numbers.

90 miles per hour is roughly 40 meters per sec (why did you do with 25?), squared is 1600, over 9.8 is 163 meters for the minimum radius. But since you bring it up, for verisimilitude reasons, it is better to start with the size of the turn and determine the speed they drive around it at from that.

Also, the coefficient of static friction for rubber on dry concrete is 1.0.

8. Dec 13, 2008

LowlyPion

Decelerating in a curve is not the best recipe for control, so definitely downshift before. But that aside it's the velocity consideration in the curve.

Since the ratio is radius to speed², you can halve the required radius by dropping the speed by the √2 (1.4) which means that at say 65 you could be more believable at a 30 meter radius curve, which for your purpose still may not be the dramatic effect you are looking for.

Edit: See numbers in the next message. These are not right. They are bigger even.

Last edited: Dec 13, 2008
9. Dec 13, 2008

LowlyPion

Yep. I used the wrong conversion. It should be 40 m/s. Thanks for pointing that out.

25m/s is only 56 mph. Oh well. I was thinking of another problem I guess. So now the curve is a football field and a half at 91. Complicate that with loose gravel and a coefficient maybe closer to .5 than to 1, that's double the radius again. Almost not a curve.

10. Dec 13, 2008

davewe

Thanks for this cool discussion. Gotta wonder how James Bond or Jason Bourne do those high speed chases. I have a sporty car and know that I've taken a curve or two pretty fast. You should have seen how the salesman drove it when I bought it.

Here's what might be throwing off the numbers. I approach a turn at high speed but by the time I hit the brakes and/or downshift I have actually taken the turn at a much slower speed. So, our old grandfather approaches the turn at ninety but by the time he actually takes the turn he's going let's say 55 or 60. It's still pretty damn fast but we think it's actually faster.

What do you think?

11. Dec 13, 2008

franznietzsche

This is part of it, the other is that sharp turns are banked, the road under the turn is tilted slightly, so
$$R = \frac{v^2(cos \theta - \mu sin \theta)}{ g(\mu cos \theta + sin \theta)}$$
(ignore my above equation for a banked turn, it's wrong.) So a ten degree bank lowers the radius thirty percent (note a 20 degree turn is not a 60 percent decrease).

So if we say he's really going sixty when he hits the turn we're down to a 50 meter radius turn. Which incidentally, trying to sit up in that car will be fun, since it will be like trying to sit on the side of a wall, meaning they will be slammed to the outside of the turn. What we perceive when going through a turn is the centripetal force on our bodies more than anything.

Also at 50 meters into reasonably sharp turn territory (not hairpin, which in an outside lane would be on the order of 10-15 meter radius, bring us down to 30 mph for the critical velocity, with the bank.). Certainly the kind that would have warning signs and lowered speed limits.

Last edited: Dec 13, 2008