: How to calculate HP to weight ratio?



weister42
09-05-06, 12:20 AM
My Deville is right around 4,000 lbs.(I think) and 0-60 is about 7 seconds, obviously the more weight I add the slower it gets. So, how much speed am I losing for every 100 pound added load? Things like subwoofers, extra batteries, junk in the trunk, passengers...etc.

danbuc
09-05-06, 06:37 AM
Well, I don't know how much this will help, but they say that for every 100lbs of weight added, you loose loose 1/10 in the 1/4miles and vice versa. I would say that it would probably apply in the same fassion to your 0-60 time. Maybe not exactly, but you will definitely get slower.

Elmer Fudd
09-05-06, 10:18 AM
Physics 101:

F=MA therefore A=F/M Which says increase mass by 10%, reduces acceleration by 10%. So 100 more pounds is 1/40, therefore you reduce your acceleration by 1/40. Now then we also know:

D=1/2AT^2

Rearranging:

Time = square root (2D/A)

Just for fun, lets say you car accelerates at 32 ft/sec^2 at 4000 pounds. At 4100 it would be 31.2 ft/sec. So your two 1/4 miles times are:

@32ft/sec^2 = 9.08 sec
@31.2ft/sec^s = 9.20 sec

But it is not that simple. That assumes a frictionless situation, which is not so in the real world. Extra weight means extra friction as well, so the actual increase would be somewhat greater, though not a lot, (due to rolling resistance on the tires, bearings etc.) Air resistance would be unchanged.

So I cannot disagre with Danbucs estimate of 1/10 sec in the quarter mile. Perhaps a bit more.

eldorado1
09-05-06, 11:09 AM
Just for fun, lets say you car accelerates at 32 ft/sec^2 at 4000 pounds. At 4100 it would be 31.2 ft/sec. So your two 1/4 miles times are:

@32ft/sec^2 = 9.08 sec
@31.2ft/sec^s = 9.20 sec

But it is not that simple. That assumes a frictionless situation, which is not so in the real world.

Well not only that, it's assuming a linear acceleration, which is not the case. At best you'd probably see 0.5 G's at launch, and as your speed increases, your acceleration will decrease. This is mostly due to gear changes, partly due to air (and other) friction.

The 10th/100 rule is pretty standard though, and that does carry over to the 60mph pretty consistantly. If your 1/4 drops by 0.1, your 60mph will drop by about 0.1. This won't apply to really fast or really slow cars. ;)

Elmer Fudd
09-05-06, 09:03 PM
Actually I assumed constant torque, I guess. That would translate into constant acceleration. Torque also varies of course, but if you look at it in terms of % change, it should still apply within reason.

Looked good anyway............

Elmer...professional engineer...but clearly NOT automotive...:)

weister42
09-06-06, 02:16 PM
So say if my stereo system is about 300 pounds, that means I would loose about .3 seconds on 0-60?

codewize
09-06-06, 04:08 PM
And remember that 1 pound of sprung weight is equal to 4 pounds of un-sprung weight, or so I'm told.

Why the hell would you put a 300 pound stereo in a car then worry about performance. Heck I've decided not to put a top on my car because I don't want to add the weight or the drag. The spoiler is a must and doesn't weight that much.

eldorado1
09-06-06, 06:13 PM
So say if my stereo system is about 300 pounds, that means I would loose about .3 seconds on 0-60?

Yes. Similarly, if you have a cow in the front seat.... you'll lose a second. :alchi:

codewize
09-07-06, 11:41 PM
ROFL
Yes. Similarly, if you have a cow in the front seat.... you'll lose a second. :alchi:

Mark Bunds
09-15-06, 07:57 PM
So, as the weight approaches zero, my acceleration approaches infinity?

eldorado1
09-15-06, 08:39 PM
So, as the weight approaches zero, my acceleration approaches infinity?

Yes, but...

Once you start approaching C (the speed of light), relativistic theory starts to come into play and you start gaining mass (weight), so your acceleration will come back down.

:thumbsup: