Larger tires
2nd Gear
Joined: Nov 2025
Posts: 2
Likes: 0
Haha! Open the tailgate, HAL... Yes, I just bought a 2025 F-250 and was browsing the forum and this caught my eye.
Powerstroke is right: "The major diameter of the TIRE/WHEEL ASSEMBLY is what drives that calculation, NOT the diameter of the wheel." If you have 35 inches from the axle to the ground, it doesn't matter which part of that is wheel and which is tire. You can change either one and as long as the distance stays at 35 inches the thrust stays the same. But if you increase the distance from axle to ground, say by making either the tire OR the wheel 2 inches taller, then you've decreased the thrust. That's the physics part: force = torque / radius.
I'm not sure how that's different from what the AI said.
Powerstroke is right: "The major diameter of the TIRE/WHEEL ASSEMBLY is what drives that calculation, NOT the diameter of the wheel." If you have 35 inches from the axle to the ground, it doesn't matter which part of that is wheel and which is tire. You can change either one and as long as the distance stays at 35 inches the thrust stays the same. But if you increase the distance from axle to ground, say by making either the tire OR the wheel 2 inches taller, then you've decreased the thrust. That's the physics part: force = torque / radius.
I'm not sure how that's different from what the AI said.
Last edited by ernie.kovak; Nov 21, 2025 at 01:38 PM.
Moderator
Joined: Jul 2010
Posts: 12,050
Likes: 1,549
From: Media PA
My AI tells me that the force applied by the torque of the engine, through the wheel and tire, to the road surface is a function of wheel/tire radius. And increasing that radius by 10% (roughly the amount of increase when going from 18 to 20 inch wheel) will reduce the applied force by about 10% as well. So, increasing wheel radius will decrease towing performance.
Moderator
Joined: Jul 2010
Posts: 12,050
Likes: 1,549
From: Media PA
Haha! Open the tailgate, HAL... Yes, I just bought a 2025 F-250 and was browsing the forum and this caught my eye.
Powerstroke is right: "The major diameter of the TIRE/WHEEL ASSEMBLY is what drives that calculation, NOT the diameter of the wheel." If you have 35 inches from the axle to the ground, it doesn't matter which part of that is wheel and which is tire. You can change either one and as long as the distance stays at 35 inches the thrust stays the same. But if you increase the distance from axle to ground, say by making either the tire OR the wheel 2 inches taller, then you've decreased the thrust. That's the physics part: force = torque / radius.
I'm not sure how that's different from what the AI said.
Powerstroke is right: "The major diameter of the TIRE/WHEEL ASSEMBLY is what drives that calculation, NOT the diameter of the wheel." If you have 35 inches from the axle to the ground, it doesn't matter which part of that is wheel and which is tire. You can change either one and as long as the distance stays at 35 inches the thrust stays the same. But if you increase the distance from axle to ground, say by making either the tire OR the wheel 2 inches taller, then you've decreased the thrust. That's the physics part: force = torque / radius.
I'm not sure how that's different from what the AI said.
More Turbo
Joined: Mar 2024
Posts: 739
Likes: 304
From: SE Vermont, South Jersey
The AI answer is WRONG. It addressed the OUTSIDE DIAMETER of the TIRE, which has VERY LITTLE and probably NO relationship to the WHEEL size, which is what was questioned.
PSA: Just because you read it on the internet DOES NOT MEAN IT's TRUE!!!
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