Wonder how many IS250 drivers are old enough to remember when Mattel first introduced the Hot Wheels range of toy cars in the 1970s. These were eventually sold as "Mag Wheels."

These were revolutionary toy cars, no pun intended. With plastic wheels that were much lighter than the metal/rubber wheels of preceding toy cars, Hot Wheels cars went much, much further and faster than their predecessors when rolled on the floor.

Switching from the stock 16" alloys to super-light 17" x 8 Forged Pro-drive GC-05 rims on the IS250, I was vividly reminded me of my 1st Hot Wheels experience. Weighing only 15 pounds each and shod with light Goodyear Revspec tyres, the IS250 accelerated more responsively and launched ahead more freely. When cruising with my foot off the accelerator pedal, the IS250 rolled along for an amazingly long distance with reduced resistance and decelerated more gradually. The upside of this is improved fuel economy. The downside to this is that the brakes to work harder and wear out more when coming to a stop.

This shows that wheel weight counts big time in vehicle response, acceleration and fuel economy. I notice that the majority of Is250 drivers upgrade to 19" or 20" wheels which look impressive. However, has anyone else downsized to 17" and gone the super-light route to enjoy the Hot Wheels benefits?

 

Good info. Although I would think the brakes would work better since there is less weight? I read somewher for every pound of unsprung weight you add (wheels tires) is equal to 7-8 pounds of weight.

 

I'll be getting some forged 18" wheels that weigh about ~6 lbs less per wheel than stock. I could have gone 19" but I wanted a decent amount of sidewall protection against potholes/road hazards without straying from the OEM overall diameters. That and all of the stories about rubbing sealed the deal for me on the 18"s.

Unsprung rotational mass is the worst type of weight and will hamper performance. Decreasing it will help in pretty much all respects including braking performance. Spin an 17", 20 lb wheel on it's axis and then stop it. It will take more effort to stop the rotation than it would a 17" 15 lb wheel because of it's reduced mass.

 

You don't realise how important sidewall protection is until they save your expensive rims from being scratched or dented. The Goodyear Revspec also has protective "gills." The tyres have saved my expensive forged rims at least twice already.

 

Switching from the stock 16" alloys to super-light 17" x 8 Forged Pro-drive GC-05 rims on the IS250, I was vividly reminded me of my 1st Hot Wheels experience. Weighing only 15 pounds each and shod with light Goodyear Revspec tyres,


Hi, I thought the stock wheels were 17" instead of 16"?

How much did the rims + tires cost?

 

He is probably from Europe where 16" is stock.

I put on BBS RGRs on mine (17") with Goodyear F1 tires. I love the way they ride. I think they are 17 lbs each for the wheels, forged one piece. When I took them into Lexus they were wrapped up with the tires mounted and the service rep asked me where the wheels were. He did not realize they were already mounted because they felt so light.

 

Certainly, the brakes have much less momentum to bring to a stop due to the lighter wheels.

On the other hand, the decreased rolling resistance keeps the caring going and going, so more brakes need to be applied.

So perhaps it pretty much balances up as far as required braking force is concerned?

 

The weight of the wheels really have nothing to do with 'rolling resistance.' Rolling resistance is determined by the tire pressure, tread design, etc.

 

I'm just thinking outloud here.. and I may be totally wrong, but I'm going to attempt to analyze this with some basic physics... To change the speed of a wheel you need to apply torque, and T = I*alpha, where I is the rotational inertia and alpha is the angular acceleration. If you go to a heavier wheel, you defintely increase the rotational inertia. With a heavier wheel there is also more frictional force. But this becomes a lot more difficult to model. With traditional simplifications we like to think of two different types of frictional forces: a static frictional force, and a kinetic force. A static force is when the object is not moving and you attempt to push it, what is the frictional force that you have to overcome before the object will start moving? This is just some coefficient (dependant on the the surfaces) multiplied by the normal force (weight) of the object. You can model a wheel using static friction: at every point it touches the ground, the wheel is not moving (relative to the ground). This is a static frictional force. ( When you overcome this static force, your wheel starts skidding, and you get kinectic friction). So with heavier wheels you add more weight into the equation and should get better grip on the ground, assuming you still have the same coefficient of friction. Granted, this is an oversimplification.

When you apply the brakes you're causing torque on the rotors becuase you create a frictional force between your pads and rotors (this is a kinetic frictional force) and this slows the wheel down. If you have a lighter wheel, the rotational inertia of the wheel is less (assuming same diameter and a similar distribution of mass) , so for the same torque applied, you should get a higher angular acceleration = improved braking.

I'm not really sure how you can put "rolling resistance" into this equation. The only thing I can think of to quantify that statement is rotational inertia. e.g. if you take a bike tire in your hand, spin it, and then try to rotate it sideways, it takes a lot of force and you won't be able to simply tilt it right over. If you've never seen/done this, try it! But if you have less rotational inertia this quantity should decrease.

Just my simple analysis of how the scenario SHOULD play out...