coltsneckbob

I have a friend that just bought a 1960 MGA. Now I know that is not an Olds, but he has an interesting problem that might apply to any classic car owner.

He was trying to determine his tach and speedo accuracy. He also knows his differential ratio and the factor specs for tire revolutions per mile (RPM) which is 809.3. He also knew his engine was 1:1 with trans in 4th gear and should therefore register 3500 RPM at 60 mph. However, his tach shows 3750 at 60.

So, he jacked up car and measured tire circumference at 80 3/16 inch (6.67 feet). When he divided this number into 5,280 he got 791 RPM.

Now he asked my if I know about "rolling circumference" (a measure front the road surface to the center of the rear axle) which is less then the radius of the tire when car is jacked up because the weight of car flattens the tire. I didn't. So, he made a test by rolling his car down the road until the tire made 10 revolutions. He them measured the distance covered. It turned out to be 64 ft and 11.75inches. Which was puzzling because that would mean the tire had a circumference of 6.5ft.

So, what is going on here? Why does the distance covered after 10 revolutions of the tire come out to less (by about 2-3%) then one would calculated based on the unweighted tired circumference. After all even though the tired is flatten a bit when weighted....the rubber that must go around is still the same length. I can understand how the so called rolling circumference might affect the speedo, but how does it affect the distance per revolution?

What are we missing? Is there some phenomenon about tread compression that accounts for the 2-3% difference? Is there some sort of slippage somehow....even though the test was made at 2 mph?

Are there any automotive engineers out there that can explain??

compedgemarine

if I am right it is because the diameter of the tire when "flattened" is less than when jacked up. that essentially makes the tire smaller.

coltsneckbob

Well that is the first reaction one would have. However, that does not change the lenght of the circumference. True the distance from center of axle to the edge of tire on road is less then the distance from center of axle to top of tired, but that still does not make the amout/length of rubber that must revolve less. Though this compression might certainly affect the speedo's registered speed.

compedgemarine

the length of the circumference makes no difference. drop the air pressure to 5 lbs and the tire will flatten much more and the diameter that the rotation of the axle 'see's' will be even shorter and the circumference will just push out front and rear of the flat area. the axle does not care the circumference, only the diameter from axle center to road.

coltsneckbob

I am not sure if I understand that. The tire flat or inflated still has the same lenth of rubber. That rubber must go around as the axle spins. Therefore, if you mark the rubber then ten revolutions with it flat should cover the same distance as if it were inflated. This analogous to a tank tread.

Rocket Richard

Don't forget that the rubber of a tire is flexible.

Imagine looking at the tire when you're standing in front or behind the car. Think of a cutaway view of the bottom of the tire when it is jacked up, and then think about how it would look when it is on the ground supporting some of the car's weight. The same amount of rubber is there in each case, but the shape of the tire's cross section is different.

When there is a load on the tire the effective diameter decreases, and so, the effective circumference decreases. The result you had makes sense to me.

-Rich

svnt442

Lets look at it from a slightly different perspective.

Have you ever watched a top fuel car do a burnout before a race? Ever notice what happens to the tires when it does that? They expand a great deal and thus increase the diameter and circumference giving the car the ability to achieve the 300+ mph speeds they are able to hit.

The same theory applies to a tire sitting on the ground. It shrinks slightly under the weight of the vehicle and thus the circumference and radius are reduced slightly.

svnt442

The difference is a tank tread is made of steel, not rubber and is not pneumatic.

MDchanic

You've just answered your own question.

The only distance that matters is the distance between the road and the centerline of the axle, and that is a constant (on a smooth road at a constant speed) regardless of tire position, so as far as the mathematics is concerned, that is the actual entire diameter of the theoretical "circle" of the tire (which is not actually a circle).

- Eric

GTI_Guru

What I have not seen mentioned yet is "slippage". When the weight of the car is on the tire, yes, it decreases the rolling radius. Moving at a slow speed (i.e. the 10 revolution driveway test he performed) the "squish factor" is going to give you a number that does not correspond to the true unweighted diameter because there is going to be a slight bit of inherent slippage in the friction between the tread of the tire & the road surface.

If you want to test this theory yourself, here's how.
Over-inflate the tire to around 45-50lb so that it removes most of the squish.
Mark the tire & pavement with chalk.
Roll it forward 10 revolutions.
Deflate it to proper inflation (30-32 I suspect)
Roll it BACK to your original spot and see if the marks line up.
If I'm right on this, they will be slightly off.

I remember studying this exact problem back in auto/diesel school, but it was 20 years ago. Performing this test at slow speeds Vs. driving speeds is also going to make a difference due to the centrifugal forces that act on the tire itself (the dragster tire principle mentioned above)

Now... I'm going to throw another caveat into the mix, and this is one that I have never truly gotten an answer to that satisfied me, but it's along the same lines.

If you keep your tires properly inflated (or slightly over-inflated), are you actually increasing your mileage (MPG) or are you offsetting your odometer to make it APPEAR that you are changing the mileage.

The way I have always figured it for both questions, is that there's a mechanical connection to every piece of the equation from the crankshaft to the tire itself that can be measured and calculated, but once you get to the tread of the tire meeting the road, then the mechanical connection ceases and a friction connection starts, rendering everything from that point a "best guess" or "approximation".

rustyroger

A good quality timing light should give you a pretty accurate base for checking the tachometers accuracy.
BTW is the tacho driven from the camshaft like early Jaguuars or MGBs?.

Roger.

coltsneckbob

I get that the rolling radius gets smaller. However, I cannot fathom how the circumference gets smaller. If you wrap a tape measure around the car tire while jacked up and then lower the car, keeping the tape in place, the overall circumference cannot shrink. If it is 80 inches of rubber while up, it is 80 inches of rubber under load. Again think of a tank tread.
Therefore, if I lay out 10 tires flat (if I could cut them) I should have a total length of 800 inches, but apparently this does not happen - at least as my friend experimented by pushing his car while counting 10 tire rotations.

There are other interesting aspects to this. Since the rolling radius (from road to center of axle) is less then the radius of the parts of the tire not touching ground then that means the parts not touching the ground are moving faster then the part that is. So, now that I think about that perhaps that is the answer....the part that touches the ground suddenly decelerates and consequently must slip (ever so slightly) on the pavement.

coltsneckbob

OK, read this after my previous post. I think this makes sense and I think agrees with what I just wrote about slippage.

Your point about MPG is also very interesting.

compedgemarine

and just for the record, mechanical speedos are only really close at their midpoint (if it is a 120 speedo they calibrate it at 55 to 60) electronic ones are closer all around because it is just a math equation based on what the tire diameter is "supposed to be" and the rpm.

coltsneckbob

The tach is driven from the cam in the MGA.

How is the tach driven in a 1970 442? There is no cable so it must be electric......where does it pick up an electric pulse from?

Rob delta

with each rotation the entire 360 degrees of the tire is compressed. Imagine the whole tire being the compressed size. So what I'm thinking is if you did a calculation of the circumference using the compressed radius, maybe that wouldbe closer to the actual length covered by each rotation

aliensatemybuick

Assuming the speedo in Russ' car is driven off the trans, isn't all this discussion of tire dimensions irrelevant?

coltsneckbob

No cause the tire is what touches the road.

MDchanic

Sorry, gotta point out the error here.

The tire / pavement interface is NOT the only point of "slippage."

You also have one between the engine and the transmission.
It is obvious in an A/T, where we talk all the time about the slippage or loss of a torque converter.
It also is inherent in the clutch, where under relaxed conditions, it might be microscopic, but it is still there, and under aggressive conditions, clutches slip all the time.

Slippage between the tire and the road is also minimal, as it is within the clutch, under low-load conditions, such as a smooth, level road at a steady, low speed (such as pushing the car on a driveway). As with the clutch, the more power transferred from the tires to the road, the more slippage.

Okay, look, you're missing the point that the tire is rubber, and that, even when constrained by the tire's belts, the rubber stretches and deforms.
The tire stretches not only in its circumference, but also side-to side, as has been alluded to above; the bottom of a radial bulges out, and some of that is stretch from the tread surface.

Also, if you must see a countervailing movement to the axle centerline moving closer to the ground, understand that the axle centerline also moves away from the top of the tire.
It's not just the bottom of the tire squashing down - the axle moves downward toward the ground within the tire, making the distance from the axle to the ground less, and that from the axle to the top of the tire more. Remember those old videos of the moon buggy rolling along the surface of the moon? Those tires had something like 1psi in them, and you could see the axles bouncing up and down within the centers of the tires as the car bounced along.
Your car does that, too, just not as much.

Does this make sense?

- Eric

Rocket Richard

An under-inflated tire has more rolling resistance so your MPG will go down. Also, an under-inflated tire would have to turn more times to reach the same distance as a properly inflated tire... the odometer would display more distance than you actually travelled, and the speedo would tell you that you are going faster than you actually are.

So I guess if you're using the odometer to figure out your MPG, an under-inflated tire might seem to be improving your milage because the odometer would tell you that you travelled further than you actually did. And if the tire is over-inflated the odometer might indicate a lower MPG because it will display fewer miles than the car actually travelled.

GTI_Guru

*facepalm*

Yeah, and one other slip point that I forgot. In the speedo head itself, there's no mechanical connection either. The magnet/speedcup assembly doesn't have a physical connection either.

Taking all the variables into account, it's a wonder they even worked as well as they did new, let alone ones that are still working darn near perfectly decades later.

-Jeff

rustyroger

Normally from the ignition, often a lead from the coil. it counts the pulses as the coil fires and translates them into a signal driving the tachometer.

Roger.

aliensatemybuick

Sure, but will its size impact on the relationship between displayed speed on and displayed RPM?