effect of wheel diameter

[nomuken]

I was reading an article about car physics in games, and came across this:

Here's an equation to get from engine torque to drive force: the longitudinal force that the two rear wheels exert on the road surface.

Fdrive = u * Tengine * xg * xd * n / Rw where u is a unit vector which reflects the car's orientation, Tengine is the torque of the engine at a given rpm, xg is the gear ratio, xd is the differential ratio, n is transmission efficiency and Rw is wheel radius.

According to this the force pushing your car forward is inversly proportional to the radius of the wheel. So my question is: why do people upgrade to larger diameter wheels? Is it just vanity?

[f8sjester]

seems like most people that actually enjoy driving upsize their wheels/tires but they still keep the rolling diameter the same as stock so that the gearing is unaffected. . . with a larger wheel and smaller sidewall tire, there is less sidewall flex and the car may be more responsive . . . not to mention that many aftermarket rims will weigh quite a bit less than stock rims. . .

as for upsizing to 19" or higher. . . yeah. . . that's just "bling" factor. . .

[Q45tech]

RW is defined as the total wheel [the wheel and tire] radius. So the wheel size means nothing it is the total wheel tire size.

Since tires deform when weight is placed on them [roughly 1500 pounds per inch for a 15"] and rear might have 900lbs /1500=0.6" compressed.So on a 26" [15"] tire maybe around 12.4" [insted of 13.0] loaded radius or 0.9677419 correction factor............3.2% reduction in gearing ratio.A 27" tire might be 13.5 minus 0.6=12.9" =0.9302325 or 7% reduction in gear ratio..................on a Q in 4th gear the engine rpm change would be 78 rpm slower with 27" tires at 60 mph vs 26" tires.

The foot [12"] in the foot pound equation is the key.

!!!!WARNING tires with low aspect ratios [30,40, 50 vs 60 and 65] have much stiffer and shorter side walls so they compress less with weight load..........they may decrease the loaded radius by only 0.3" vs 0.6"...........which means the sidewall stiffness may approach 3,000 pounds per inch.

A 337 lb/ft with 9.85 gear multiplication = 3319.45 less 22.3% friction and TC losses=2612.4 x 0.9677419 [from our 26" tire]=2,528 pound feet divided to [by 2 tires] the road assumming 100% friction [which is never the case or exactly equal between tires.

2528/4300 pounds= 0.5879 G which is the theoretical maximum acceleration at 4,000 rpm in a Q.In reality it is the rare 90-93 Q that gets to 0.47-0.50G due to tire friction and lower engine output.

See why you can't smoke the tires in 1st gear. [because the tire road friction is somewhere around [0.85-0.87 u] except in wet weather when it obviously drops to half or less.

You plot the rear wheel torque vs engine rpm [from a dyno] and multiply by gear ratios equation, then you have a graph of the possible acceleration at each rpm.

As the acceleration RATE increases [as the rpm build towards 4,000] the weight shifts more towards the rear wheels and their friction [coupling] increases due to extra weight pushing tires. against the road.

The rear tires are lighter loaded than the front [sitting still] so around 400 pounds [progressively ramps up] shifts at 4,000 rpm resulting in 20-22% more instantaneous friction.

The acceleration is directly proportional to body weight [Force/Mass].

[nomuken]

wow, that warranted multiple re-readings, and i still don't get most of it

you're trying to show what would happen when switching to a different size wheel by giving an example of what happens when the (wheel+tire) diameter changes under a load, correct?

please clarify some things for me:

1) Quote »roughly 1500 pounds per inch for a 15"[/quote]

do you mean a 15" tire will lose an inch of diameter under a 1500lb load? if so, then wouldn't a 26" rear tire (that decreased 0.6" in diameter under a load of 900lb) have a loaded radius of 12.7" instead of 12.4"?

2) am i correct to assume that since you give examples of 15", 26" and 27" tires that this 1"/1500lb figure is pretty much independent of tire size?

2) what is "2528/4300 pounds= 0.5879 G" are you getting acceleration by dividing torque by weight (is 4300lb the car's weight)?

[Q45tech]

You are missing the point the wheel [in your supplied equation] is the tire and the wheel.........the total diameter.

The wheel diameter itself in immaterial as most try to maintain the same rolling diameter/circumference when going to blink blink wheel sizes.............there is only so much wheel well clearance.

I wrote 15" wheel when I assumed a 26" tire mounted on a 15" wheel.

Only the portion [radius] of the tire at the bottom [road] is decreased by the load, the upper radius doesn't change.......so the tire is flexing at every revolution.

Tire side wall stiffness varies for every tire design 1200-2000 pounds per inch........the taller the sidewall the less stiffness but tires in the exact same size may vary by half that amount. Single or dual ply sidewalls different materials, S rated or W, Y, Z speed rated.................the faster the speed the more times per minute the sidewall flexes generating heating which must be radiated to air [why tires get hot].

I was using a 4300 pound loaded Q45 for example.

A 26" tire times pi=81.68" divided into 5280' =775 revs per mile yet the tire actually does 800 revs due to the change in loaded diameter [79.2"]= 25.2" AVERAGE diameter.

http://www.geckermotorsports.c...r.htmhttp://www.informs-cs.org/wsc00papers/135.PDF

[nomuken]

i did a back of the envelope comparison to see what the change in wheel+tire radius (Rw) would be if i switched from 205/60R15 to 245/45R17, (compression under a load is ignored, because i assume both tires would undergo approximately the same comrpression), here goes:

Tire 1 - 205/60R15wheel diameter = 381mm (15")sidewall height = 123mmRw1 = (381+2*123)/2 = 313mm

Tire 2 - 245/45R17wheel diameter = 432mm (17")sidewall height = 110mmRw2 = (432+2*110)/2 = 326mm

Rw2 - Rw1 = 13mm (~4% difference)

...so assuming the above is correct, does that mean the resulting torque loss at the wheel will be = EngineTorque * GearMultiplier * 0.04?

[Q45tech]

"(compression under a load is ignored, because i assume both tires would undergo approximately the same comrpression), here goes:"

WRONG a 45 sidewall will compress less than a 60 sidewall not 75% less but maybe half that 30-40% less so the tire sidewall stiffness on a 45 is 30-40% higher per inch than a 60 series.

Tirerack charts show the loaded radius for each size tire on the ideal rim width but they never give you the weight on the tires.I would guess something like 85% of max weight shown on tire.

The tires are only round when there is not weight on them!

The 24" diameter solid steel steam roller is the only tire that needs no correction...........most tires are around 24.5-25.5" loaded radius .............2.0-6.25% correction decrease for rear torque from gear method alone.