Some Basic Tire Physics

[Flanders]

TLDR. For the purposes of acceleration and braking, the effective weight of a spinning wheel is up to twice the actual weight.

The kinetic energy of a spinning wheel on a moving vehicle is up to twice that of the same mass at the same speed without the rotation (e.g., the spare). The ratio is less than 2 - it would be exactly 2 if all of the rotating mass were on the tread - and is the same at all vehicle speeds. This energy must be supplied by the engine to accelerate the vehicle and dissipated by the brakes to slow it.

For example, replacing factory tires with 35s at +25 lbs apiece effectively adds up to 225 lbs, that is, up to 50 lbs on each corner and 25 for the spare. One NFL running back, perhaps an undersized one.

The factor of <2 only applies to acceleration and deceleration. At constant speed only the actual mass that matters.

And this says nothing about the forces on the suspension and steering due to bigger heavier tires.

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[PuddleJumper]

yeah i learned that lesson years ago when i couldn't figure out why my tundra was going through front wheel bearings every 4-6 months. E rated 35s on 0 offset wheels is way outside what the truck was designed to handle, 31s, d rated, +60 offset.

 

[Wheelin98TJ]

Diameter is a bigger factor than weight.

Here is a nice explanation from @ALVagabond :

What you're thinking of is angular momentum. A wheel takes some amount of effort to start it spinning and some amount to stop it. Conceptually, the "bigger" and "heavier" the tire, the harder these processes will be.

Breaking it down mathematically, angular momentum (L) is the moment of inertia (I) times rotational speed (w), so L = I*w. But what's I (the moment of inertia)? You can think of a wheel as a disc spinning about it's axis. For that geometry, I is 1/2 of the mass (m) times the radius (r) squared, or I = 1/2m*r^2 (note that in this case, m is the mass of the entire wheel; rim plus tire).

Basically, an increase is radius is going to much more impactful to momentum than an increase in mass because radius is squared and mass is not. If you have a 35 and a 37 that are basically the same weight, the 37 will be harder to start/stop spinning and that increases the load on the starting/stopping spinny parts of your Jeep.

But 37s will make you look cooler.

 

[Flanders]

Diameter is a bigger factor than weight.

Here is a nice explanation from @ALVagabond :

This is spectacularly wrong but it's not worth explaining in detail why on a forum.

The engine adds kinetic energy to the vehicle. The rate at which it does so is power. If, due to heavier tires, the vehicle has say 5% more kinetic energy at a given speed than it did with the lighter tires, then it takes 5% more power to get to that speed in the same time, or 5% longer to get there at the same power output. This depends on weight but not tire diameter, except insofar as changing tire diameter amounts to a change in gearing.

Angular momentum is not entirely irrelevant but the important thing is the radius of gyration, the RMS distance of mass of the rotating object from its axis of rotation.

The kinetic energy of a rolling tire is

E = 1 /2 m * v^2 * (1 + (r/R)^2)

where R is the radius of the tire and r is the radius of gyration. For a fixed vehicle speed v, the mass m and the ratio r/R are all that matters.

You might recognize 1/2 m * v^2 as the kinetic energy of a non-rotating moving mass. The kinetic energy of the rolling version is more than this amount, and less than twice it. If we assume most of the added mass of a larger tire is in the tread, then it's closer to twice it.

I only mentioned it because I've seen some utter nonsense on the question of how much more added rolling mass affects acceleration vs non-rolling. The correct answer is 2x, at most. Not 7. Not 3.

Adding 25 lbs of rotating mass to each corner affects acceleration in the same way that adding 50 lbs of non-rotating mass would, at worst.

Edit: The r/R in the energy equation is squared.

 

[mx5red]

Are

This is spectacularly wrong but it's not worth explaining in detail why on a forum.

The engine adds kinetic energy to the vehicle. The rate at which it does so is power. If, due to heavier tires, the vehicle has say 5% more kinetic energy at a given speed than it did with the lighter tires, then it takes 5% more power to get to that speed in the same time, or 5% longer to get there at the same power output. This depends on weight but not tire diameter, except insofar as changing tire diameter amounts to a change in gearing.

Angular momentum is not entirely irrelevant but the important thing is the radius of gyration, the RMS distance of mass of the rotating object from its axis of rotation.

The kinetic energy of a rolling tire is

E = 1 /2 m * v^2 * (1 + (r/R)^2)

where R is the radius of the tire and r is the radius of gyration. For a fixed vehicle speed v, the mass m and the ratio r/R are all that matters.

You might recognize 1/2 m * v^2 as the kinetic energy of a non-rotating moving mass. The kinetic energy of the rolling version is more than this amount, and less than twice it. If we assume most of the added mass of a larger tire is in the tread, then it's closer to twice it.

I only mentioned it because I've seen some utter nonsense on the question of how much more added rolling mass affects acceleration vs non-rolling. The correct answer is 2x, at most. Not 7. Not 3.

Adding 25 lbs of rotating mass to each corner affects acceleration in the same way that adding 50 lbs of non-rotating mass would, at worst.

Edit: The r/R in the energy equation is squared.

The way I read your posts, you’re saying what matters is the overall weight of the car moving.
I don’t think looking at a car’s kinetic energy is what people “feel” as much as the accel/deceleration, which is more to what people explain about tire diameters and weight and gearing.
My example would be, if I add 500 pounds in cargo and passengers for a trip and keep my 33”s, the trucks drives way better than if I’m empty with 500 pounds of extra “weight” in 37” tires and wheels. The tire size/weight and gearing, the rotational mass, matters a lot.

 

[Flanders]

Are

The way I read your posts, you’re saying what matters is the overall weight of the car moving.
I don’t think looking at a car’s kinetic energy is what people “feel” as much as the accel/deceleration, which is more to what people explain about tire diameters and weight and gearing.
My example would be, if I add 500 pounds in cargo and passengers for a trip and keep my 33”s, the trucks drives way better than if I’m empty with 500 pounds of extra “weight” in 37” tires and wheels. The tire size/weight and gearing, the rotational mass, matters a lot.

Of course gearing matters. Let's take it out of the equation and see how you feel.

Truck 1 is stock with 33s and a 4.10:1 rear diff, plus some added non-rotating mass on each corner to level the playing field with Truck 2.

Truck 2 has 37s and a proportionally regeared rear diff at 4.60:1, and is otherwise identical to stock.

The point is they're the same, except for gearing and tire size and weight, and they have the same crankshaft-to-pavement gearing.

What I'm claiming is

(1) There is an amount of non-rotating mass that when added to the corners of Truck 1 makes acceleration and response to brake and throttle on level ground identical* to those of Truck 2; and
(2) The amount of non-rotating mass to achieve (1) is less the than twice the difference in mass between the 37s and the 33s.

*identical minus some small stuff

 

[Escape.idiocracy]

This is spectacularly wrong but it's not worth explaining in detail why on a forum.

The engine adds kinetic energy to the vehicle. The rate at which it does so is power. If, due to heavier tires, the vehicle has say 5% more kinetic energy at a given speed than it did with the lighter tires, then it takes 5% more power to get to that speed in the same time, or 5% longer to get there at the same power output. This depends on weight but not tire diameter, except insofar as changing tire diameter amounts to a change in gearing.

Angular momentum is not entirely irrelevant but the important thing is the radius of gyration, the RMS distance of mass of the rotating object from its axis of rotation.

The kinetic energy of a rolling tire is

E = 1 /2 m * v^2 * (1 + (r/R)^2)

where R is the radius of the tire and r is the radius of gyration. For a fixed vehicle speed v, the mass m and the ratio r/R are all that matters.

You might recognize 1/2 m * v^2 as the kinetic energy of a non-rotating moving mass. The kinetic energy of the rolling version is more than this amount, and less than twice it. If we assume most of the added mass of a larger tire is in the tread, then it's closer to twice it.

I only mentioned it because I've seen some utter nonsense on the question of how much more added rolling mass affects acceleration vs non-rolling. The correct answer is 2x, at most. Not 7. Not 3.

Adding 25 lbs of rotating mass to each corner affects acceleration in the same way that adding 50 lbs of non-rotating mass would, at worst.

Edit: The r/R in the energy equation is squared.

Just tag me next time I corrected mine and included your explication with gratitude.

 

[HankB]

This is spectacularly wrong but it's not worth explaining in detail why on a forum.

The engine adds kinetic energy to the vehicle. The rate at which it does so is power. If, due to heavier tires, the vehicle has say 5% more kinetic energy at a given speed than it did with the lighter tires, then it takes 5% more power to get to that speed in the same time, or 5% longer to get there at the same power output. This depends on weight but not tire diameter, except insofar as changing tire diameter amounts to a change in gearing.

Angular momentum is not entirely irrelevant but the important thing is the radius of gyration, the RMS distance of mass of the rotating object from its axis of rotation.

The kinetic energy of a rolling tire is

E = 1 /2 m * v^2 * (1 + (r/R)^2)

where R is the radius of the tire and r is the radius of gyration. For a fixed vehicle speed v, the mass m and the ratio r/R are all that matters.

You might recognize 1/2 m * v^2 as the kinetic energy of a non-rotating moving mass. The kinetic energy of the rolling version is more than this amount, and less than twice it. If we assume most of the added mass of a larger tire is in the tread, then it's closer to twice it.

I only mentioned it because I've seen some utter nonsense on the question of how much more added rolling mass affects acceleration vs non-rolling. The correct answer is 2x, at most. Not 7. Not 3.

Adding 25 lbs of rotating mass to each corner affects acceleration in the same way that adding 50 lbs of non-rotating mass would, at worst.

Edit: The r/R in the energy equation is squared.

I‘m with you except for the point about additional mass being only in the tread. If you were talking about performance tires, i.e. 21” wheels then yes, probably a good assumption. However, when considering off road tires for a Glad, there will be significant meat in the sidewalls and casing, and not just in the tread cap, so the multiplier is probably going to be less than 2.

 

[Flanders]

Just tag me next time I corrected mine and included your explication with gratitude.

That was rude of me and I apologize. Could you (or anyone) enlighten me as to what "tag me" means?

I‘m with you except for the point about additional mass being only in the tread. If you were talking about performance tires, i.e. 21” wheels then yes, probably a good assumption. However, when considering off road tires for a Glad, there will be significant meat in the sidewalls and casing, and not just in the tread cap, so the multiplier is probably going to be less than 2.

You are absolutely right, it's unrealistic. Any sane choice of rim for a JT is 20+ lbs nowhere near the tread.

The case I had in mind was putting a bigger tire on the same rims. Most of the difference is near the tread. I applied the maximum multiple of 2 to the incremental mass to get a worse case estimate. It's not mathematically rigorous.

 

[Escape.idiocracy]

That was rude of me and I apologize. Could you (or anyone) enlighten me as to what "tag me" means?



You are absolutely right, it's unrealistic. Any sane choice of rim for a JT is 20+ lbs nowhere near the tread.

The case I had in mind was putting a bigger tire on the same rims. Most of the difference is near the tread. I applied the maximum multiple of 2 to the incremental mass to get a worse case estimate. It's not mathematically rigorous.

@ and their screen name.

No hard feelings, clearly your math defeats my argument.

I am curious and not in a challenging way, to your statement about the placement of weight…. My reasoning for the question, is the effects on fuel economy/energy required. My JKU is a manual 4.88’s two sets of wheels however they run the same tires. MT patagonias. A set with beadlocks wheel weight 47# and the other with factory rubicon wheels- 28#’s. There is almost a 2mpg 17% difference on the highway, and from a “feels” perspective is slower and for sure stops slower. The math indicates that there shouldn’t be a significant difference, but the energy required is leaning towards a larger difference?

Or am I just that bad at following along. ???

 

[Flanders]

@ and their screen name.

No hard feelings, clearly your math defeats my argument.

I am curious and not in a challenging way, to your statement about the placement of weight…. My reasoning for the question, is the effects on fuel economy/energy required. My JKU is a manual 4.88’s two sets of wheels however they run the same tires. MT patagonias. A set with beadlocks wheel weight 47# and the other with factory rubicon wheels- 28#’s. There is almost a 2mpg 17% difference on the highway, and from a “feels” perspective is slower and for sure stops slower. The math indicates that there shouldn’t be a significant difference, but the energy required is leaning towards a larger difference?

Or am I just that bad at following along. ???

Interesting. Just adding 19 lbs to each wheel should not affect acceleration and fuel economy that much. It would take around 3% more energy to accelerate to a given speed with the heavier set vs the lighter set, all else being equal. An impatient driver could lose more than 3% fuel economy, but 17% seems unlikely.

My guess is that wheel weight is not the only difference in play here. Do the beadlocks have the same width as the factory wheels? Are the tires running at the same pressures? Is it the same driver, and the same GVW except for wheels, on both sets?

 

[Escape.idiocracy]

Interesting. Just adding 19 lbs to each wheel should not affect acceleration and fuel economy that much. It would take around 3% more energy to accelerate to a given speed with the heavier set vs the lighter set, all else being equal. An impatient driver could lose more than 3% fuel economy, but 17% seems unlikely.

My guess is that wheel weight is not the only difference in play here. Do the beadlocks have the same width as the factory wheels? Are the tires running at the same pressures? Is it the same driver, and the same GVW except for wheels, on both sets?

The beadlocks are 1” wider, the tire pressure is the same. ??‍ the economy difference is basically 98% highway running down hwy 395. Vehicle weight/on vehicle weight is only changing with the spare on the back.

Chalked this up to increase in wheel weight before this discussion.

Driving patterns and patience are pretty well set. Fill up in Inyokern and running north to Lone pine/Bishop and refilling. Or running south to Big Bear.
Both directions have minimal stops, and fairly consistent traffic…. Lots of cruise control action.

 

[Volt0]

Alright, I’ll bite. Mr @Flanders, your formulas don’t account for the changes in the contact patch, correct? Is there a website/reference that would include that, and/or something to give the common driver a way to tell how much their mileage may change from one tire to the next? ( these silly 3.5 star ratings for AT tires is less than informative )

 

[Geoarch]

TLDR. For the purposes of acceleration and braking, the effective weight of a spinning wheel is up to twice the actual weight.

The kinetic energy of a spinning wheel on a moving vehicle is up to twice that of the same mass at the same speed without the rotation (e.g., the spare). The ratio is less than 2 - it would be exactly 2 if all of the rotating mass were on the tread - and is the same at all vehicle speeds. This energy must be supplied by the engine to accelerate the vehicle and dissipated by the brakes to slow it.

For example, replacing factory tires with 35s at +25 lbs apiece effectively adds up to 225 lbs, that is, up to 50 lbs on each corner and 25 for the spare. One NFL running back, perhaps an undersized one.

The factor of <2 only applies to acceleration and deceleration. At constant speed only the actual mass that matters.

And this says nothing about the forces on the suspension and steering due to bigger heavier tires.

Yeah, I really noticed the weight of the 35s (Goodyear Ultraterrains) when removing one. I ain't 20 anymore. A flat bladed shovel did the trick.

 

[Flanders]

Alright, I’ll bite. Mr @Flanders, your formulas don’t account for the changes in the contact patch, correct?

Ouch. Indeed, I would have mentioned it.

Is there a website/reference that would include that, and/or something to give the common driver a way to tell how much their mileage may change from one tire to the next? ( these silly 3.5 star ratings for AT tires is less than informative )

It’s outside my wheelhouse but I doubt that contact patch is useful in predicting fuel economy.

It boils down to estimating energy losses due to deformation of the rolling tire and friction with the pavement. Contact patch doesn't tell you anything about the unloaded shape of the tire, stiffness, sidewall deformation under load, distribution of mass or distribution of contact pressure.

Contact pressure varies over the contact patch. It can be more than the absolute tire pressure where the carcass provides support. It can be less than atmospheric pressure in places (i.e., a tread block pulling away from the pavement creating a bit of vacuum). All of this depends on the tire structure and varies non-linearly with speed, load and tire pressure.

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