The 370Z Weight Reduction Thread

[Supergoji]

found this formula on the lotus forums. seeing as they hate weight this equation most likely has some truth to it.

save ~10 lbs gain 1 hp ** EQ: Y=(190*X) / (1984-X) where Y is (HP) and X is (lbs)

[wstar]

Back on the horse jockey angle, I've been working on that too. From all the work I've been doing on my 370Z in the Houston heat sweating like crazy all day, I've lost about 16 lbs since I bought my Z (down to avg 174 now from avg 190, big change for me). Kill two birds with one stone: install your own upgrades and automatically lose driver weight

[wstar]

(I had a post here about the whole converting weight into horsepower thing, but I posted it at 2:30AM and the math was totally wrong, so I'm killing it before anyone comments on it. I'll put up a fixed one later)

[wstar]

Oh yeah, I never came back and fixed this.

So the easy formula for calculating how much horsepower a given weight reduction is "worth" (in terms of HP:Weight ratio for acceleration, obviously this doesn't take into account effects on handling and balance):

E = (W/R)*H

Where E is your new "effective" horsepower, W is the starting weight of the car, R is the reduced weight of the car, and H is your actual horsepower.

For example, take a set of numbers not unlike a stock 370Z on a DynoJet (3320 lbs, 275 rwhp), and drop 100 pounds:

E = (3320/3220)*275 = 283.54

So 100 pounds off of the stock car is going to make the same acceleration difference as adding 8.54 horsepower.

How much the car weighs and how much horsepower you actually have can have a big effect on the outcome of this formula, which is why there isn't a universal "X lbs is worth Y horsepower" number for all cars, or even for one car given all the mod variations.

Let's look at a hypothetical TT 370Z which is otherwise stock (with some added weight for the TT system, let's say 550rwhp and 3400 lbs), and see what happens when we drop 100 lbs there:

E = (3400/3300)*550 = 566.67

So 100 lbs off of that car is effectively worth 16.67 horsepower.

Either way, in both cases you're gaining about 3% effective horsepower numbers for acceleration purposes, because you've dropped roughly 3% of the car's body weight.

So a good rule of thumb to go by from all of this, is that it's all relative to your starting weight. Think of your reductions as percentages, and that's how much they'll help your acceleration. If removing 20 lbs is worth 0.66% weight reduction on your car, then it's going to make a 0.66% difference in acceleration (or effective horsepower if you prefer to think of it that way for comparing to bolt-on power adders).

Keep in mind, as I said at the top, that this says nothing about the handling effects of dropping weight, and it especially doesn't cover the special case of rotating masses in the front or rear wheels (or the driveline or either end of the crankshaft, etc), which have a more pronounced effect that I don't know how to calculate

[1slow370]

Originally Posted by Supergoji

found this formula on the lotus forums. seeing as they hate weight this equation most likely has some truth to it.

save ~10 lbs gain 1 hp ** EQ: Y=(190*X) / (1984-X) where Y is (HP) and X is (lbs)

That formula assumes a 1984 lb car also known as a lotus. It's just a power to weight formula. It seems a little off too should be something like

equivalent horsepower= (hp*original weight)/(original weight - amount removed)

or say Ehp=(332*3232)/(3232-100)

Ehp=1073024/3132 Ehp=342.6 so a 100lb savings = 10.6 horsepower

But you aren't making more horsepower it just means you now have the same power to weight as the guy with the stillen exhaust

[fuct]

^^awesome...

[dad]

It's a massive motor, in a tiny lightweight car.
Carroll Shelby

[StLRedrider]

[Endgame]

O.K. Two questions:

1. DD - you responed to one of my questions on this earlier regarding HFCs. While the HFCs will not make that car THAT much louder, you noted they would change the tone. If I was running an Amuse R1 Titanium catback and Y pipe, would HFCs make the Z louder? I am not sure if that can truly be answered, but I appreiciate any thoughts.
2. The Nismo flywheel and clutch that shaves 20 pounds is very cool, but how much would that affect the already noticiable shifter vibration??

Any thoughts/experiences/comments are greatly apprecieated guys!!

[M.Bonanni]

Originally Posted by Endgame

O.K. Two questions:

1. DD - you responed to one of my questions on this earlier regarding HFCs. While the HFCs will not make that car THAT much louder, you noted they would change the tone. If I was running an Amuse R1 Titanium catback and Y pipe, would HFCs make the Z louder? I am not sure if that can truly be answered, but I appreiciate any thoughts.

Yes they would make it louder. The stock cats restrict more noise than high flow cats do no matter what setup you have for the rest of your exhaust.

[shabarivas]

Just use moment of inertia calculations to determine how much you will save off the rotational mass change. I can do it later - too many ppl watching me @ work lol

[wstar]

Originally Posted by shabarivas

Just use moment of inertia calculations to determine how much you will save off the rotational mass change. I can do it later - too many ppl watching me @ work lol

Please do, I'd love more information on how to calculate the effects of changing the rotational masses (as in lighter wheels and brake rotors), but I'm not really familiar with the math/physics involved.

[shabarivas]

Oh my - looks like he did my calculations for me!!

[shabarivas]

ok if we assume the wheel is a uniform disc... of weight M then the moment of inertia of that wheel is 1/2 * M * R^2 ... so... with a 19'' wheel - lets say it weighs 25lbs the moment of inertia is about - 1/2 * 25lbs * 19 * 19 = 4512 lb*inch^2

now lets say the new wheel weighs about - 24 lbs so a 1lb saving... moment of inertia = 1/2 * 19 * 19 * 24 = 4332 lb*inch^2

ratio of difference = 4% less moment of I w/ 1lb diff in wheel

so ... what that tells us is... roughly - shaving 1 lb off per wheel - makes it 4% easier to spin that wheel ... hope that makes sense

EDIT: for more info: http://hyperphysics.phy-astr.gsu.edu/Hbase/tdisc.html

[wstar]

Originally Posted by shabarivas

ok if we assume the wheel is a uniform disc... of weight M then the moment of inertia of that wheel is 1/2 * M * R^2 ... so... with a 19'' wheel - lets say it weighs 25lbs the moment of inertia is about - 1/2 * 25lbs * 19 * 19 = 4512 lb*inch^2

now lets say the new wheel weighs about - 24 lbs so a 1lb saving... moment of inertia = 1/2 * 19 * 19 * 24 = 4332 lb*inch^2

ratio of difference = 4% less moment of I w/ 1lb diff in wheel

so ... what that tells us is... roughly - shaving 1 lb off per wheel - makes it 4% easier to spin that wheel ... hope that makes sense

EDIT: for more info: Moment of Inertia, Thin Disc

Nice. I have some followup questions though:

1) Does it make any difference whether the weight is being dropped from the driven (rear) or rolling (front) wheels? I'm inclined to think it doesn't matter, as the engine is pushing both rotationally, assuming no appreciable tire slip being factored in for a simple case.

2) How do we apply this to the car as a whole in order to determine how much effective % difference in acceleration a weight drop at the wheels makes? Let's say I run the numbers for the whole wheel/tire package at all 4 corners, using the combined weights of the tire+wheel at each corner and the radius of the tire+wheel. Let's say I shaved off 4% per corner. That doesn't really mean the same in terms of overall acceleration numbers as, say, a 4% drop in car body weight does it?

I would imagine one would have to calculate moment of inertia stuff for all rotating masses (so the hubs, rotors, wheels, tires, etc, plus the driveshaft, the rear axle, the rear diff, the transmission, the flywheel, the engine itself, etc??? it's a long list), and then somehow combine that and the static weight of the car into some final number for how much force has to be applied to move the car a foot, and then see what % that number drops from removing weight at just the wheel as a percentage of the whole, right?