Load, Inflation, and a Mismatched Dual Pair
Class 7 straight truck, 12,000 lb front axle, 21,000 lb rear axle with duals. Tires are 11R22.5 load range G, currently all at 100 psi cold. Scale weights with the usual load: front axle 11,600 lb; rear axle 21,600 lb, split 11,900 lb left and 9,700 lb right. Chart excerpt, lb per tire - dual application: 80 psi 4,820; 90 psi 5,350; 100 psi 5,860; 110 psi 6,350. Single application: 80 psi 5,180; 90 psi 5,750; 100 psi 6,300; 110 psi 6,820. The right outer dual measures 40-3/16 in in diameter; its mate measures 39-13/16 in. The owner wants to know why the left rear tires wear faster than the right and whether rotating them will help.
1. Using the scale weights, determine the minimum cold inflation for each position.
Read the worked answer
Work per tire, off the heavier side of each axle - capacity is set by the most heavily loaded tire, not the axle average.
- Front: 11,600 ÷ 2 = 5,800 lb per tire, single application. 90 psi rates 5,750 (not enough); 100 psi rates 6,300. Set 100 psi.
- Left rear: 11,900 ÷ 2 = 5,950 lb per tire, dual. 100 psi rates 5,860 (not enough); 110 psi rates 6,350. Set 110 psi.
- Right rear: 9,700 ÷ 2 = 4,850 lb per tire, dual. 80 psi rates 4,820 (short by 30 lb); 90 psi rates 5,350. Set 90 psi minimum.
Practical call: run both rear sides at 110 psi. Setting the right side to 90 and then rotating side to side leaves right-side tires underinflated in the left position, which is the side that already carries the extra load. Verify 110 psi does not exceed the sidewall max cold pressure or the wheel rating.
Also flag the 2,200 lb left/right split as a load distribution issue - fifth wheel position or load placement - not a tire issue.
2. The dual pair is 3/8 in apart in diameter. What is the consequence and what do you do?
Read the worked answer
The difference is 40-3/16 − 39-13/16 = 3/8 in, over the usual 1/4 in maximum for a dual pair.
Consequence: circumference differs by π × 0.375 = 1.18 in per revolution. At roughly 500 revolutions per mile, the pair fights itself about 590 in - nearly 50 ft of scrub per mile - converted to heat in the tread and sidewall, plus continuous spider-gear action in the differential.
Decision: the smaller tire comes off the dual pair.
- Replace it and match the pair by measured diameter, not by the size molded on the sidewall.
- Or move the odd tire to a steer position where it is not paired.
Never pair a new tire with a nearly worn one on a dual, and never run mismatched duals 'until the next service' - that is how you get a tread separation or a roadside fire.
3. Why do the left rears wear faster, and does rotation solve it?
Read the worked answer
Load, mostly. The left rear carries 11,900 lb vs. 9,700 lb on the right - about 23% more per tire. More load means more heat, more slip, and faster wear, and it is made worse by running 100 psi when the left needs 110.
Rotation alone will not fix it. Fix inflation and the load split first, then check for causes rotation cannot address:
- Dragging brake or a hot hub on the left rear - feel or use an infrared thermometer on the hubs after a run.
- Wheel bearing end play, a bent or shifted axle housing, or a thrust angle problem.
- A heat source on that side (exhaust routing, restricted airflow).
Rotation rules: drive-axle duals can be rotated side to side and front to back only if the measured diameters match within 1/4 in. Never move a worn drive tire to the steer axle. Then re-check wear at the next PM - if the pattern has not changed, the cause was not inflation or rotation.