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Metalworking (rec.crafts.metalworking) Discuss various aspects of working with metal, such as machining, welding, metal joining, screwing, casting, hardening/tempering, blacksmithing/forging, spinning and hammer work, sheet metal work. |
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![]() "Dave Baker" wrote in message ... "Rich Grise" wrote in message news ![]() Saturday morning TV is mostly cartoons, yes, but there's one show that's not a cartoon, but it's still fun: http://www.beakmansworldtv.com/ It's like Stealth Educational TV - like Mr. Wizard for the 21st century kid. And they do Real Science. But today, they said that you can weigh your car by using your tire pressure. What you do is measure the footprint of each tire, take its pressure, and then the footprint of the tire times the PSI equals the number of pounds that the tire is supporting, and their sum is the weight of the car. (except for the tires themselves, I presume). But they did the experiment - they measured the footprints of the four tires on Some very well-preserved Nash Rambler, and their PSI, and did the arithmetic, and came up with a number that was within 10% of the car's "official" weight. Well, I'm a little uncomfortable with that. What about the pressure that's in a tire when it's not on the car? Where does that pressure go? Is that that 10% fudge factor that they admitted to in the show? How much does the pressure change when you take a standalone tire, mount it on the car, and let the car down on it? Or do they get around that by saying, "Well, you can ignore that, because the tire's not supporting any weight". Or is it entirely ( or mostly) due to the flattening of the bottom of the tire? Does the "bias pressure" (a term I just made up now, for the pressure that's there when it's not on the car) get lost below the noise floor? There are two sources of error in that method. The first, which is the one I think you are referring to is the increase in pressure as you put weight on the tyre. This will be the inverse of the reduction in volume of the tyre as the bottom of the tread flattens and will be very small. The second, which is the significant one, is the amount of the load which is being supported by the stiffness of the tyre tread and sidewalls. If the tyre were a balloon, i.e. perfectly elastic, the measurement as above would be exactly correct. So the result they should have obtained is a calculated weight somewhat less than the actual weight of the car. -- Dave Baker - Puma Race Engines There's at least one more error, Dave! When you bring the rubber close enough to the tarmac Van der Waals forces kick in and give you some extra attraction, so some extra "contact area". |
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