2015 Yukon wheels

LaserBlueZ71 · October 17, 2014

Wow.

GM spends all this money hyping up fuel economy, and better MPG numbers all around, only to use a wheel/tire combo of 88 pounds!! That is more than most 35 inch off road tires in 10 ply form, on a wheel! This just seem stupid to me. How can they expect to gain MPG with that much unsprung weight on each corner.

http://blog.caranddriver.com/the-wheels-on-our-long-term-2015-gmc-yukon-xl-denali-are-heavy-really-heavy/

SouthernSilveradoGuy85 · October 17, 2014

Well I figure the weight of the rims and tires don't effect fuel economy, especially once you get them rolling. With bigger and heavier rims you mainly need to worry about braking, especially emergency braking. Plus for me it would make me feel better to have heavy rims on a heavy full size truck or SUV. Plus people are wanting bigger and bigger rims from the factory. I'm glad my '14 came with 18s since that's the perfect size rim for me.

kstruckcountry · October 17, 2014

It doesn't have as much of an effect as you'd think... long story short, an object in motion stays in motion. But even braking, I guarantee you it will still lock up fine with 88lbs on there. My wheels and stock size tires are not much lighter than these.

That said for a daily driver I definitely like smaller rims instead of those... Especially on an SUV that will more than likely be curb checked every single turn on the way to soccer practice...

white1500 · October 17, 2014

How do you guys think that heavy wheel/tire combos don't hurt economy or performance? The more something weights the more effort (power) is needed to get it moving, and ultimately stopped. This also ends up affecting the chassis/suspension

http://en.m.wikipedia.org/wiki/Unsprung_mass

Although everything is a compromise between cost, weight, strength, aesthetics, etc.

kstruckcountry · October 17, 2014

Wheels(22in+tires) > 88lb > 39.92kg

Yukon(Denali XL 2wd) > 5750lb > 2611.33kg

Assuming a travel speed of 80 mph > 128.75 kmh

Assuming a perfect disc for a wheel (too lazy to make one in SolidWorks)

KErot=12,704 Joules

x4 wheels = 50,816 Joules < (rims + tires) x 4

KE_Yukon=1,673,392 Joules

KE_total=1,724,208 Joules (Yukon energy and rotational and forward energy of wheels)

50816/1724208= 2.9% <<< 2.9% of the total kinetic energy is the four wheels. Hardly even relevant when stopping. 4 average american passengers will have about 200,000 Joules of kinetic energy at 80mph.

00Silverado4x4 · October 17, 2014

Wheels(22in+tires) > 88lb > 39.92kg

Yukon(Denali XL 2wd) > 5750lb > 2611.33kg

Assuming a travel speed of 80 mph > 128.75 kmh

Assuming a perfect disc for a wheel (too lazy to make one in SolidWorks)

KErot=12,704 Joules

x4 wheels = 50,816 Joules < (rims + tires) x 4

KE_Yukon=1,673,392 Joules

KE_total=1,724,208 Joules (Yukon energy and rotational and forward energy of wheels)

50816/1724208= 2.9% <<< 2.9% of the total kinetic energy is the four wheels. Hardly even relevant when stopping. 4 average american passengers will have about 200,000 Joules of kinetic energy at 80mph.

Ok, now let's see this in a language I can understand...lol.

kstruckcountry · October 17, 2014

All it means is that speeding up or braking, about 3% of the power goes to spinning the wheels up to speed. Which is less than goes towards even moving/stopping the passengers.

Jon A · October 17, 2014

Assuming a perfect disc for a wheel (too lazy to make one in SolidWorks)

A disc is not remotely appropriate for an approximation. Remember that whole Parallel Axis Theorem? The vast majority of the wheel's mass is quite a distance from the axis (depending upon wheel design of course) and it gets worse for bigger wheels--exponentially. And of course all the tire’s mass is beyond the wheel radius from the axis.

If looking for a shortcut approximation, use the equations for a ring instead of a disc. Guesstimate the diameter as somewhere between wheel diameter and tire tread diameter.

blackout07 · October 17, 2014

Things just got awfully scientific around here. Lord knows I don't understand it. But I'll go out on a limb and say that the engineers at GM know a little something about what they're doing.

'01LS1 · October 17, 2014

Wheels(22in+tires) > 88lb > 39.92kg

Yukon(Denali XL 2wd) > 5750lb > 2611.33kg

Assuming a travel speed of 80 mph > 128.75 kmh

Assuming a perfect disc for a wheel (too lazy to make one in SolidWorks)

KErot=12,704 Joules

x4 wheels = 50,816 Joules < (rims + tires) x 4

KE_Yukon=1,673,392 Joules

KE_total=1,724,208 Joules (Yukon energy and rotational and forward energy of wheels)

50816/1724208= 2.9% <<< 2.9% of the total kinetic energy is the four wheels. Hardly even relevant when stopping. 4 average american passengers will have about 200,000 Joules of kinetic energy at 80mph.

So that is the class I failed several steps down. That is cool stuff!

Am guessing the avg. 20" wheel with tire that is on a majority of the trucks produced in the last several years is pushing the upper 70 lb weight as is.

Looked up my stock Goodyears & they were 43 lbs each per GY's site.

SnakeEyeSS · October 25, 2014

I'll bring it back down to earth.

Wheels(22in+tires) > 88lb > 39.92kg

Yukon(Denali XL 2wd) > 5750lb > 2611.33kg

Assuming a travel speed of 80 mph > 128.75 kmh

Assuming a perfect disc for a wheel (too lazy to make one in SolidWorks)

KErot=12,704 Joules

x4 wheels = 50,816 Joules < (rims + tires) x 4

KE_Yukon=1,673,392 Joules

KE_total=1,724,208 Joules (Yukon energy and rotational and forward energy of wheels)

50816/1724208= 2.9% <<< 2.9% of the total kinetic energy is the four wheels. Hardly even relevant when stopping. 4 average american passengers will have about 200,000 Joules of kinetic energy at 80mph.