Does anyone know the equation for converting lost lbs.(by the 100) in your C5 to lost tenths of seconds at the track? Is there such a thing?

 

Rule of thumb is for every 100 pounds is about 1/10 off your ET.

 

Damn it, Shawn, you beat me to it ;)

 

That's for wieght within the body or chassis which only has to be accelerated in a straight line (forward). Rotating weight (wheels, tires, flywheel) costs you even more, four times as much, so 25 pounds (approx) lighter wheels and tires would gain 1/10 also. :cheers:

 

I have to loose some weight :lol: ... is the reverse true also ???

 

Just curious.... But how many car lenghts is 1/10th in the 1/4mi? :)

 

.1 sec * (trap speed) * (5280/3600) = distance
For 102.1 MPH this works out to 14' 11.7" or precisely the length of a C5 :D
105MPH would be 15.4'
110MPH would be 16.13'
and so on...
So you for the speeds that C5s typically trap at, a 1/10th is a little over a car length.


[Modified by bdavies, 10:56 AM 7/18/2002]

 

What if your car only weighs 1000 pounds? Does a 100 pound reduction still result in a 1/10s better ET? What if you have 500HP instead of 350HP? Does the relationship still hold?

 

Egads, the law of diminishing returns......

 

Removing weight will give you near linear results at lower speeds, but will ultimately lead to diminishing returns, so removing 1,600lbs out of a 3,200lb car will not cut your ET in half. Similarly doubling your HP won't cut your ET in half. One reason for this is that as speeds increase, wind drag becomes increasingly important and removing weight doesn't reduce wind drag.

Another reason is that ET is heavily dependant on your average speed through the quarter and average speed doesn't vary linearly with power/weight ratio.

For simplicity's sake, let's assume your acceleration is roughly linear. Since you always start at 0MPH, increasing your trap speed by X% increases your average speed by X/2%. Consequently your ET is improved by X/2%. So a 10% improvement in ET (i.e. 10.0 to 9.0) would require approximately a 20% higher trap speed (i.e. ~130 to 155). Unfortunately, wind drag rises geometrically with speed, and frictional losses increase as well so you'll need alot more than 20% more power or 20% less weight to achieve that increase. I have simplified things greatly but I hope it helps.


EDIT:
Sorry, I guess I rambled on a little there. In answer to the question, it depends on many other factors. In a 1000lb car, a 100lb reduction is huge - that would improve the power/weight by 10% - which would help the most at lower speeds. So if it were a relatively low HP car that was not trapping very fast you would see greater than 1/10th improvement. If it were a big HP car that is mostly drag limited then you won't see much of an improvement at all.


[Modified by bdavies, 12:01 PM 7/18/2002]

 

Good point here. Weight has virutally no effect on Top Speed since a vehicle is overcome by drag. your could add 2000 pounds to the weight of a C5, and its top speed would not drop by much (only a little due to some rolling resistance increases). It would just take much longer to achive that top speed because accelleration to that point would be slow.

 

Thanks! :cheers:

 

How could you manipulate this equation to compare two different cars in a race. Say you smoked a 5.0 'Stang that trapped 108 and you won trapping 119. Is there a way you could estimate that? Thanks.

 

If you let:
Tf = the time when the faster car crosses the traps
Ts = the time when the slower car crosses the traps
Then the distance between them would be the distance the slower car travels between time Tf and Ts. Sice velocity = distance / time:
distance = (Ts - Tf) * (average speed of the slower vehicle from Tf to Ts) * (5280/3600)

The problem is you need to know how much the slower car accelerated between time Tf and Ts. If you assume the slower car doesn't accelerate much between time Tf and Ts then you can approximate using the slower car's trap speed. If it accelerated appreciably then the distance will be even greater.

So you can say:
distance >= (Ts - Tf) * (trap speed of slower car) * (5280/3600)




[Modified by bdavies, 3:01 PM 7/18/2002]

 

Just to put some real numbers, let's say the faster car ran a 12.0 and the slower car ran a 12.6 (just made up numbers). We'll use the trap speeds of 119 and 108.

The distance would then be:
distance >= (0.6s) * (108MPH) * (5280/3600)
distance >= 95.04 feet
distance >= 6.36 C5 lengths :D


[Modified by bdavies, 3:22 PM 7/18/2002]


[Modified by bdavies, 3:22 PM 7/18/2002]

 

Hmm, maybe I shouldn't have added 50lb of stereo equipment. Nah, I like my tunes... :)