Ok, How About An XC-70?

The pressure you are referring to is the internal pressure of the tire. The external pressure is very different since the surface is not smooth and the tire contains the internal pressure. For example, when the car is up on jacks, the internal pressure is the same but the ground pressure is zero.

I think you assumption about very near the melting point is reasonable.

Thanks again for the advice. I feel like I am in the United Nations and engaging in Peace Talks with my dealer. They won't budge a whole lot and I can't get screwed on my trade in. So, I talked to my friendly Honda dealer tonight who will deal on a loaded 2005 Pilot. I know, it's not a Volvo. But, don't you have to shop for a dealer as well as a car? The ratings on the Honda are tops. Maybe I do need to sit up high after all. Maybe it's my mid-life crisis??? I'll keep you posted, but my Volvo dealer told me that "we don't really have to haggle on Volvos, they sell themselves at the sticker price." So, other than the '04's they have, the others are sold at sticker! Whew....thanks for listening!

--Fred

My experience is that if I want a fair price for my used car, I must sell it privately. I always ask half-way between wholesale and retail and calculated by edmunds.com and kelley.com. This makes the playing field flat and I know exactly what discounts dealers are giving me. I actually choose my dealer by their service department rather than their sales department. Even though my Volvos have been very reliable, when I go in for service, I don't want to have to worry about it. My previous dealer had service writers that listened to your description of a problem and then guessed what they would have the mechanics do. It seldom worked. My present dealer has mechanics that have a good knowledge of the car and allows them to use their knowledge. It is a big difference.

From a physicist, I find that a puzzling answer.

The pressure inside the tyre is constrained by tension in the carcass of the tyre. If the tyre is on the ground with a load on it (the car), the tyre deforms until a state of equilibrium is reached. This point is when the force acting from the ground upon the tyre equals the force required the deform the tyre.

Clearly, if there was no pressure in the tyre, the only force resisting collapse is that due to the strength of the sidewall. As the tyre is flexible, this is not enough. The additional force due to the air pressure inside the tyre stops the rim touching the ground.

Thus force exerted upon ground = pressure within tyre + force required to deform sidewall.

If the tyre was rigid, then the strength in the sidewall alone would be able to withstand loads without deforming. But it ain't, so it doesn't.

As a side point, it is interesting to find (well it interested me, anyway) that the pressure inside the tyre stays pretty much the same regardless of whether there is a single person in the car, or 5 people plus luggage. This is because the contained volume within the tyre stays pretty much the same. What does change is the size of the contact patch.

I once carried out the following experiment: Put a piece of graph paper on a set of bathroom scales. Jack the car up, and coat part of the under side of the tyre tread with some printing ink. Lower the tyre onto the scales and record the reading. Remove graph paper and count the squares within the contact patch. Repeat for various loads and pressures.

Within the limits of the experiment (I didn't test to the point of destruction), the results will show that the size of the contact patch changes with the load, and in inversely with the pressure. The pressure inside the tyre hardly changes at all in relation to the load. Conclusion: the overall load on the ground increases as the load increases, but as the tyre deforms, the contact patch increases in size, and the force per unit area stays the same.

Don't know much about skates, but if the blade has a contact area of, say, 1/8" x 12", then a pair of skates supporting someone of 120 lbs will be exerting a force of 40 psi on the ice. From the above, this would seem similar to the unit force of a typical tyre upon the ground. But I imagine there would be other factors invloved in liquifying the ice, like its temperature, for example.

hehehe....looks like you've hit a nerve amongst the Volvo hard core nerds !

Force does not equal pressure. Force equals pressure times area.

The area is the area of the tire >The pressure you are referring to is the internal pressure of the tire.

To visualize this. Consider a piece of a tire that is 1 inch square. Place a 20 lb piece of lead on it. The pressure that it transfers to the ground could be almost anything over 20 psi. If the tread area is

We purchased a 2004 XC-70 in July. The mileage computer has not been reset since purchase and shows about 21.4 mpg average since new. This is with a lot of small town driving with occasional (once-a-month) 150-160 mile drives to go shopping (Montana...). Averaging about 1,000 miles per month at the moment. Reliability has been good, but then it's new, it better be...

Indeed, there is the contact patch to include in that calculation. I wasn't intending to present an specific equation, rather than a summay of factors involved.

Well, I sort of see you point, but now find myself confused.

You see it was explained to me something like this: The tyre carcass is under tension as a result of the air pressure inside. When you lower the tyre onto the ground, at the point of contact, the tension is opposed by the upward force of the ground. As things are in equilibrium, the force of the ground upon the tyre must equal the tension in the tyre carcass - the force pushing the tyre in equals the force pushing it out. Plus a bit for the mechanical strength of the tyre.

It all made sense to me then.

Further, it was put to me, and demonstrated rather convicingly, that if you put more load on the tyre in the form of extra weight in the car, the contact patch will expand, meaning the force per square inch will remain the same.

Now I figure from your comments, that if the tyre was to perch on the top of a vertical rod of, say, 1 foot long and 1 sq inch section, then the contact patch could not possible expand beyond 1 sq inch.

So now I'm confounded. I intuitively grasp the first explanation, but now it seems lacking.

Damn.