anniversar

I read somewhere in another post that the rpm's showing at 70mph should be 1600 for the AWD model. I will have to recheck mine but I think it is showing 2000 rpm. Can anyone verify what it should be?

Vvman

Hello anniversary

Let me verify the theoretical rpm at 70mph mathematically.

The 8 speed tranny has an overdrive 8th gear ratio of 0.685 and a final drive ratio of 2.937 according to the LS460 product info (link below). What this mean is the overall drive ratio is 2.011845 (0.685 multiply by 2.937, assuming you at at 8th gear) and that the engine rotates 2.011845 times per revolution of the driving wheel/s.

The stock wheel is 235/50-18 inch (theoretical height 27.252 inches fully inflated) and this translate to an overall tire rolling circumference of 85.62578 inches (27.252 multiply 3.142) or 0.001351417 mile. This is also the distance your tire covers in 1 revolution.

In order to cover 70mph or 1.16666667 mile per minute, your driving wheels would need to rotate 863.2914 times every minute.

At all overall drive ratio of 2.011845, your engine will turn 863.2914 X 2.011845 = 1736.8 revolutions per minute when you are travelling at 70mph - assuming you are on stock size tires.

As you probably know by now, several factors can ‘throw off’ these calculations, including different tire sizes, underinflate tires and even the accuracy of the speedometer.

http://www.google.com/url?sa=t&rct=j...F1JUpx_vebbyFw

robert1408

Nice calculation, Vvman! I just have two items to add here.

The final drive on an AWD LS 460 is a little lower than a RWD at 3.1331 for 2010 models.

Tire squat under load will decrease the rolling radius and therefore raise the tire revs/mile by about 2%.

Using your workable and valid method of calculation and adding the new final drive ratio and tire squat factor, I
get 1889 rpm at 70 mph. Of course, this squat factor will vary depending on tire size, load and air pressure.

anniversar

Wow

Thank you both for these detailed calculations. As mine is an AWD, the 1900 rpm is just about right with my gauge is showing.

Devh

This is one of the other reasons why I liked the RWD. Having the higher final gear ratio higher keeps the RPMs lower for a dead silent drive at 65-70mph without any lugging.

Vvman

Thanks guys.

Spot on! The final drive on an AWD is indeed 3.1331! Thanks for pointing that out.

As a matter of fact, tire diameter actually changes at various speed.

The low engine rpm even at relatively high vehicle speed (engine barely at 4,000rpm when vehicle speed at 150mph) is one reason why the LS enjoy such good FC for a V8 at its time.

And I'd just discovered a Speed versus RPM calculator while searching online. Cheers

http://wahiduddin.net/calc/calc_speed_rpm.htm

Fhobbs

Down here a cricket cricking at 100 times per minute has a great bearing on how fast an ant runs. That variation can account for at least 37 rpms at 68 mph or more.


Just had to throw that in due to my amazement that the calculations above even existed. I am truly amazed at the depth of information that can be obtained from this site. I assume the rpm calculations are exactly true in a vacuum. Now I wonder how much head wind affects the rpms. I am going to really notice my rpms now. I have noticed that my LS has lower rpms at 75 than my Tahoe (also a v-8)

robert1408

Headwinds won't affect your cruise rpm on an LS460 unless they're extreme enough for the trans to actually down-shift. Automatic transmissions these days use a clutching torque converter lock up in the higher gears. Like a manual transmission with a clutch, the rpm's are dependent on road speed alone. Torque converter slippage is eliminated when this clutch engages.

Your Tahoe may allow the torque converter clutch to unlock without downshifting under load which would give you maybe a 200-500 rpm increase if it has GM's 4 speed automatic. I don't know about GM's 6 speed automatics.

Vvman

Engine RPM and vehicle speed is a pure function of gearing – tire sizing is part of gearing, so to speak.

At a fixed rolling circumference of the driven wheel, the engine RPM corresponds to the road speed at a predefined set of gearing ratio regardless of whether the car is travelling upslope, downslope or into headwind – assuming no transmission or wheel slippage.

So unless your Tahoe has exactly the same overall drive ratio as your LS460 and runs exactly the same tire profile, you will not likely run the same engine RPM at the same travelling speed.

Nospinzone

I believe you should take "tire squat" out of your calculation. While the radius of the tire will decrease on the road side, the circumference of the tire does not change. One revolution of the tire will travel the same distance whether the tire is on the car or freestanding.

robert1408

Oh, now you've gone and done it Nospinzone! I had to fetch my arguin' hat for this one…heh heh.. You are correct that the circumference of a tire dictates how many times the tread rolls around over any given distance. But here we are interested in the revs/mile of the axle driving that tire. And that is a product of the rolling radius, not the circumference. Seems counterintuitive doesn't it? Consider the following examples:

1) Tire pressure monitoring systems fall into two groups. Direct systems use pressure sensors
and the now less common indirect system which relies on the ABS wheel speed sensors to
detect a low tire. They do this by sensing the increase in rotational speed of the ABS sensor
as the tire gets lower. It's rotational speed, as measured at the axle or hub, definitely increases
due to the shorter rolling radius.

2) Look at the road wheels of a tracked vehicle like an army tank. Those tracks have a huge
circumference but those little road wheels driving the track are spinning many more revs/mile
than the tracks are. An extreme example here but the geometry is valid.

I have to confess here that I was about 45 yrs. old before a Master Tech schooled me on this. I had considered myself pretty knowledgable about general automotive things. I sure ate some humble pie on this one!

Nospinzone

Robert, I can't agree with your first point. In #1 the old method was only an approximate measure and seems irrelevant to this discussion. In #2 the geometry is correct, but imagine that one little wheel had its own track wrapped around it. Then one revolution of the wheel would equal one revolution of the track.

This same principle holds true for the tire. You can easily test this yourself. You have a full size spare in your trunk. Take it out, be sure it is inflated to the same psi as a tire on your car. In your driveway make a chalk mark on the 6 o'clock point on the spare and a chalk mark on the driveway. Roll the tire one revolution until the mark is back at 6 o'clock, and mark that spot. Now put a tire of your car at the first mark and chalk it. Drive to the next mark and you will find that the tire chalk mark lines up perfectly. The car travels the exact same distance because the circumference remains the same.

A good example of this principle is when people put snow tires on their car. Most people will put narrower snow tires on for better traction. However, when you do that you should also get a higher profile size in order to eliminate or minimize the distortion to the speedometer. For example, on your LS with summer tires of 235/50/18 your tire's diameter is 27.3 inches. If you went to a snow tire of 215/50/18, your diameter would now be 26.5 inches. This because the profile number (the middle number) is a function of the relation to the tread width. This of course would distort your odometer/speedometer readings. So the proper snow tire size would be 215/55/18 as this tire would have the same 27.3 inch diameter.

The flaw in the thinking of that master tech is while the radius of the tire on a car from the center to the ground is shorter than a freestanding tire, he misses that the tire on a car is no longer a perfect circle.

robert1408

Well, I did say my argument was counterintuitive. I can only try to clarify my point in example 1. For purposes of gearing, road speed, speedometer and odometer readings the car only senses the speed of the hubs and driving axles which is defined by the rolling radius of the wheel/tire assemblies, not the tire circumference. This is how indirect TPMS systems work. As you say, it is only an approximate measure but it does alert by sensing an rpm increase in a deflating tire caused by a reduction in its rolling radius.

You are also correct that loaded or unloaded, the tread will will roll the same number of times over any given distance. However, if an axle were bolted to that spare tire you would measure a change in its revolutions over distance loaded vs. unloaded. This effect is not a lot, maybe 2% or so on a fully inflated tire. But it is there.

Nospinzone, I intend no disrespect in my replies but I do love a good argument. Especially with someone like you who is even interested enough in this stuff to take the time to dig into it and express your thoughts.

Nospinzone

First off Robert, no disrespect taken. This is purely a gentlemanly argument/discussion. And hopefully my responses are also accepted in the respectful way they are intended.

Now to say to you what a dear uncle used to say to me with a wink, "Paul you are usually not right, but this time you're wrong".

Logically, how can this be? The circumference of the tire is the circumference of the tire. If the axle turns one full turn, the tire has to make one full revolution, and hence travel the same distance whether loaded or unloaded. If it didn't it would mean that the bead of the tire would have to slip on the wheel.

I'm willing to be convinced!

robert1408

Nospinzone, I spent a few minutes looking into this and what I found out was interesting. Tire Racks technical FAQ had a very short paragraph on this and concluded that actual road rpm of a loaded tire is impossible to accurately calculate. They cited factors of tread and sidewall flexing at the contact patch as the problem but didn't elaborate. Goodyear acknowledged the problem and uses a correction factor of .967on medium truck tires. Their formula had the overall unladen diameter multiplied by by .967 to yield a then slightly reduced overall circumference to use in calculating true revolutions/mile.

It's what happens at the contact patch that resolves the seemingly impossible difference in unladen revs/mile and loaded revs/mile. Measure the length of the contact patch and mark its leading and trailing edges. Call these points A and B. On a perfect circle A and B are always the same distance apart. As the contact patch forms, A and B are now separated by a straight line. This is now a shorter path between them compared to the longer curved path of the unladen tread surface. This causes the tread in the contact patch to compress lengthwise to conform to the now shorter path. Since the tire is "seeing" this shorter distance, it has to roll more times per mile.

As you can imagine, all this flexing and compressing at the contact patch is by far the main cause of wear and heat buildup. On a seriously under inflated tire this A to B path discrepancy becomes quite large and serious. The tire is conflicted between trying to roll at a rate dictated by its circumference and the much greater rate dictated by the reduced rolling radius. This generates an amazing amount of heat.

So, Nospinzone, in a way we were both right. A tire will indeed try to roll at both a rate dictated by its circumference and the loaded rolling radius at the same time. Heat and wear are the products of these conflicting geometries. And in a way, we were both wrong. Identical size tires at the same load and air pressure can have very small differences in their revs/mile depending on their construction alone. How they respond to the A B problem is the difference.