Let's say you added 7 lbs. per wheel (I'm guessing that it's actually more than 7 lbs). Multiply by 4 because you are accelerating all wheels, not just the drive wheels. That's 28 lbs. of rotating weight to accelerate. Why wouldn't that make a noticeable difference? There was less of a weight difference when I went from a steel driveshaft to an aluminum one and it was noticeable.

 

I noticed the very same thing when i went from 16s to 17x9.5s. My car got hella slower and it didnt spin the tires as well(reason explained before). I did one simple thing that put things to even better than before. I put in 3.73s and immediately went from a 14.3 to a 13.7. that was the only mod between the times. track temps were roughly the same. The 3.73s brought the car back to life. It still gets 21mpg on the highway too. we'll not talk about city mileage. good luck. -Michael

 

Just to clarify some errors above....

A 245/50-16 is exactly the same outside diameter as a 275/35-18 (actually the 275/35-18 is a tiny bit smaller in some manufacturer's sizing)... so he did NOT lose any torque as a result of tire diameter.

When you add weight to a ROTATING (not "reciprocating") component, you must overcome the inertia of the part to accelerate it. This consumes energy. Inertia increases with weight, and also when you move the centroid (center of mass) of that weight further from the axis of rotation. An 18" aftermarket wheel/tire will typically weigh more than the factory 16" wheel/tire, particularly when the tire and wheel is wider. Further, when you go from a 16" wheel to an 18" wheel, the center of mass moves further out from the axis, because the rim weighs more than the center disc. Ditto with the tires. Takes more HP to bring it up to speed.

For a real-world measurement of this.... look at the dyno test on ws6.com, where he compares the rear wheel HP of his stock 17x9 275/40-17 setup to the same wheels widened to 17x11, and sporting 315/35 tires.... he LOST 3.5rwHP just from this simple change. Going to 18's will be even more significant.

On the other hand, I question whether anyone could actually feel these small HP losses, using SOTP

 

My method might be wrong, but this is how I calculated the outside diameters

diameter of 245/50/16 = 16 inches + 2 x 50 / 25.4" = 19.9inches
diameter of 275/35/18 = 18inches + 2 x 35/25.4" = 20.8 inches

oops.... so its 4.5%

BTW, am i forgetting something really obvious in my calculations?

I still say the few HP required to accelerate the bigger wheels would not explain the SOTP effect being described here. Im biased cuz my own *** is not very reliable.

 

The dyno does not measure the extra energy to accelerate the front wheels. This should double the hp loss, right?

On my daily driver that probably dynos at around 260 hp I would think I could detect a 7 hp (3%) difference. It's not like I was expecting it, actually it caught me by surprise.

 

from my experience my car was definitly not the same with the 17s. it was slower, i know this cuz me and my best friend would race all the time. I dropped in the 3.73s and it fixed the problem. I dont have dyno numbers to prove, but it was a significant difference.

 

do they make 14" rims for the lt1?

 

its not only the rotating mass thats taking away his power to the ground but he said the the overall (tire) height is .5" taller. think of turning the wheel the same as turning a gear.

the gear circumference is larger=faster on top but not as Quick.

Cir. is smaller=quicker but not as fast on top end.....

in any event....you need more power to make the car as quick as it was with the smaller wheel.

 

Yes sir... the section width.

tire diameter = 2 x ((section width ÷ 25.4) x (aspect ratio ÷ 100)) + wheel diameter

or

tire diameter = 2 x ((section width x aspect ratio) ÷ 2540) + wheel diameter

Run your calcs again and you'll see that the two wheel/tire combos yield a difference of only .070". Of course we haven't considered the rim width and how it effects the tire carcass. But that's good enough for this conversation.

The problem is the weight. Or do we need to run the weight calculations to prove that point?

-Mindgame

 

if the guy measured the total diameter of the two wheels and says that there is .5" difference. then there is probably about .5" difference. you can sit and do all the math you want but there are always variables that you cant equate.....so your end result will be different than what was calculated on paper.

if he measured the wheel diameters the same way on each wheel and says there is a half inch difference.........he feels less power.........then that means that the half an inch increase in tire size has taken away some torque to the ground.

period.

 

If the 16" tire had 10/32's of tread new, that's 20/32's total (top and bottom of tire) and if the tires are in need of replacing, lets just make it easy and say that there was 20% of the tread left or 2/32's, well that means that 16/32's has been used... 16/32's = 1/2 inch.

The tread depth will come into play when comparing two tires, if one or the other is used.

 

Wheels do not reciprocate. They rotate. Pistons reciprocate. Also you can't look at the weight of the wheel alone. Larder wheels add rotational inertia to the system. Even if you have 2 wheels that are the same weight, if one has more of its weight on the outer portion of the wheel, it will have more intertia. Does this make sense? Guess I actually learned something in my engineering classes.

 

Sure, there's probably close to 1/4" of tread depth on a new tire, so a diameter loss of close to 1/2" is understandable, given our example. The loss of tq due to a 1/2" increase in overall height is ~1.9% The answer lies in the formula for Work. I'll spare those of you who don't like understanding things mathematically the equation.

I don't believe anyone was trying to make a case against a "loss of power felt from 1/2" difference in tire height". You could also include rolling resistance into the equation, because the wider tire has a much larger footprint. These things aren't near as significant as the difference in weight. But that would require more math and people would be offended.....

-Mindgame

 

l = 1/2M(R1²+R2²)

R = radius

You're correct.

-Mindgame

 

My car feels noticably slower when I go from my 15" skinnies/16" DR's to my 17/18 Y2k C5's. Not a huge difference but it is noticable. Just throwing my $.02 in here.