I need a physics refresher...

[drledford93]

Swampy reminded me of something I learned in one of my physics classes in college--that is, if you have two cylinders of equal diameter, the one with the most mass closest to the center will win when rolled down an incline. This can easily be applied to a car's performance relative to its wheels.

That said, I wonder exactly how much of an advantage are lightweight rims that don't have near the mass close to the hub as our beloved OEM wheels do? Is there a formula for determining this? I wonder how much of the 24lbs of wheel lies between the center and before the outer part where the tire mounts? Anyone want to take a shot?

Daniel

 

[Sacrilicious]

something tells me that doing out the math for THAT is going to be a lot of calculus, but here's an alternative way of thinking about things that is much more straight forward:

our cars put out 263hp at the crank and ~220hp at the wheels. that means we have lost about 263-220 = 43hp throughout the drivetrain. now, we know that those 43hp cover transmission loss, wheel rotational loss, and assorted friction points (bearings, etc.). what this means is that we're probably looking at something like 20hp actually lost in the act of turning the wheels of our car. in addition to this, if you consider that our rotational mass is not just the wheels, but the wheels + tires (24lb+24lb=48lb), dropping rotational mass by losing 5-10lb off the rims is not really all that interesting. basically, for all practical purposes, going to crazy forged wheels will likely get you in the neighborhood of lower single digit whp gain...^_^;

EDIT: this is why i, personally, believe that the main reason to go to lighter rims is merely the improved handling characteristics and reduced stress on your suspension.

 

[drledford93]

Thanks, Sac.

Isn't the formula for unsprung weight this: For every 1 lb lost, that translates to 7 lbs of sprung weight reduction? For instance, if you buy 14lb rims, then your car will behave like a car that weighs 280lbs less? (10lbs*7*4 wheels=280lbs)

To me, that would both increase power:weight ratio and handling...

 

[Derek88]

Because wheels are powered, not rolled.

Push a 500lb wagon and push a 50lb wagon of the same size. It's easier to push the lighter one, but the 500lb one would be faster down (a sizable) incline.

 

[Sacrilicious]

Thanks, Sac.

Isn't the formula for unsprung weight this: For every 1 lb lost, that translates to 7 lbs of sprung weight reduction? For instance, if you buy 14lb rims, then your car will behave like a car that weighs 280lbs less? (10lbs*7*4 wheels=280lbs)

To me, that would both increase power:weight ratio and handling...

this is where the actually location of that 1lb would make a difference. if you added 1lb at the edge of your tires versus adding 1lb at the center of your hub, you will see differences. contrary to common belief, there's no easy magical formula for 1lb lost = 7lb of sprung weight, since weight distribution starts cutting into your perceived gains. all things equal, you will always lose the 40lb of total vehicle weight, but that added 6lb per lb of lost unsprung weight is not necessarily going to be the full 6lb.

 

[drledford93]

Derek, it's not that simple. I think I missed something in my original post--that is that the one with the most mass centrally is easier to accelerate than the hollow one. Thus, even though the hollow one is lighter, the solid cylinder will win because it's easier accelerated by gravity.

 

[happy and angry]

Inertia is a measure of a give bodies resistance to rotation. I'm going to have to refresh myself on a bit of calculus to figure stuff out for something of a non-uniform density (IE: more mass near the center of the body), and I can probably post on that later, but let's take a look at a few basic formulas.

Rotation of a cylinder about it's central axis has the following formula:
I = 1/2 mr^2

Rotation of a thin cylindrical shell (hollow center) with negligible thickness has the following formula:
I = mr^2

Rotation of a cylindrical (hollow center) shell with a non-negligible thickness has the following formula:
I = 1/2m(R1^2 + R2^2) where R1 is the inner radius and R2 is the outer radius.

m, mass, is the least significant term in these equations. Every time mass of the rotating body, the value of I increases linearly. IE: m = 1, r = 1 for a cylinder:
I = 1/2mr^2
I = 1/2(1)(1^2)
I = 1/2
IE: m = 2, r = 1
I = 1/2mr^2
I = 1/2(2)(1^2)
I = 1

r, radius, is the most significant term. Every increase of R will increase the value of I at a squared rate. IE: m = 1, r = 2 for a cylinder:
I = 1/2mr^2
I = 1/2(1)(2^2)
I = 2

What does this mean? More mass, more inertia, more resistance to twisting force (acceleration, braking). More radius, a LOT more inertia, etc etc. But... we don't really change our total tire + wheel radius by very much, do we? 1 or 2% maybe, depending on tire size. What changes the most is total weight, and usually not by more than maybe 15 - 20%. It'll change the moment of inertia, but not by a hugely significant amount. As it stands, the third formula I've posted is the nearest to the actual problem, because a wheel and tire still have most of the weight near the outside of the tire. It gets more complicated to calculate inertia of a cylinder with an effectively non-uniform weight distribution, and if I can brush up on my university physics enough to make a decent explanation of it, I will later.

Long and the short of it, you'll see positive performance gains in all areas of the vehicles performance, and it is a lightening of unsprung mass, but they may not be as significant as you think. A 20% less heavy wheel will have approximately a 20% lower moment of inertia. How much effective HP this works out to, I have no idea.

 

[drledford93]

this is where the actually location of that 1lb would make a difference. if you added 1lb at the edge of your tires versus adding 1lb at the center of your hub, you will see differences.

This is my point. Take, for instance, the 5Zigen FN01RC. It's a relatively "open" wheel design with plenty of space between the spokes. Thus, one could assume that a large proportion of the mass is in the rim of the wheel compared to the OEM MS3 rim. This is where the dilemma comes in--there is 9lbs difference in these rims, but actual performace difference? Hmm...

 

[happy and angry]

Oh, and just to more directly answer your question here:

Let's compare a cylinder where the weight is uniformly distributed throughout it's composition, and that of a wheel of a non-negligible thickness, because it's the closest to a wheel.

We will keep the mass the same and the radius the same. m = 1kg, r = 1m. For the wheel, we will use a thickness of 20 cm, or 0.2 meters. Meaning, R1 = 0.8m, R2 = 1m

Cylinder:
I = 1/2mr^2
I = 1/2

Wheel:
I = 1/2m(r1^2 + r2^2)
I = 1/2(0.64 + 1)
I = 0.82, or about 4/5

In general, the further from the center of mass the weight is distributed, the more resistant to rotating the body is.

 

[jaydubz]

these are all great ideas but lets not forget to factor in the cost of these incredibly light weight wheels. Sorry to be "buzz-killington" but is the cost of these light weight wheels worth the minimal horse power increase? Would it not be more cost effective to increase the already usable horse power to overcome any loss? just a question

 

[drledford93]

Happy, you gave me reaaaalllllyy scary flashbacks of college physics! Yikes! (shocked)(I'm a math guy, not an engineer...) Thanks for the solid explanation!

 

[happy and angry]

Meh, physics is just applied math, not very difficult stuff. Besides, this is what you get when you ask for a physics refresher!

 

[Sacrilicious]

these are all great ideas but lets not forget to factor in the cost of these incredibly light weight wheels. Sorry to be "buzz-killington" but is the cost of these light weight wheels worth the minimal horse power increase? Would it not be more cost effective to increase the already usable horse power to overcome any loss? just a question

lolz...at $400-500 a wheel, it's likely a lot cheaper to just get yourself more power at the crank...^_^;

 

[drledford93]

Meh, physics is just applied math, not very difficult stuff.

Yes, but when you live in the theoretical mathematical world and not what most call the "real world," the real world tends to scare you. (silly)

 

[Sacrilicious]

haha...actually, engineers are all about simplifying problems down to solvable pieces...it's the theory people that break out all the insanely complex equations that make you wonder what's up and what's down...^_~ they give the more complete answers, but engineers give you the bottom line: lighter wheels won't really net you a whole lot in power gains...;o

 

[drledford93]

Sorry to be "buzz-killington" but is the cost of these light weight wheels worth the minimal horse power increase?

I couldn't agree with you more...(yes)

 

[Sierra117]

I think my brain just stopped...

Sorry, Poli Sci major, I don't do math.

 

[drledford93]

Engineers give you the bottom line: lighter wheels won't really net you a whole lot in power gains...;o

So there is a reason for engineers to exist after all! (bowdown)

 

[Sacrilicious]

So there is a reason for engineers to exist after all! (bowdown)

yes, other than arguing about who can build the bigger turbocharger, they do serve a purpose in the world...;o

 

[SwampAss]

I'm a nerd muse!