You can do this for (almost) free if you go to the local dump. They usually charge you based on weight of what you dump by weighing your car on the way in and on the way out.
Just ask them what the reading is on the way out. If you bring a couple of pounds of garbage, it will cost a few dollars at most.
From there, you could then simply average out the results - or better still, the data would help you reason about the pressure<->area relationship. Just going by the P=F/A formula, pressure should be inversely proportional to area, but people have pointed out things like the tyre sidewall stiffness will affect this. Taking several pressure<->area measurements would help you reason about this - perhaps the sidewall means P = F/A + x, and with multiple datapoints, 'x' can be calculated.
The ensuing discussion was wide-ranging, and covered various limitations of this technique. It was more true of older tires, but tires now have more structure, so some of the car's weight is supported by the tire structure, and not just by the air pressure.
In turn, I remember reading about this in a Martin Gardner book, probably nearly 40 years ago.
(Sorry people are downvoting you for that, I appreciated it.)
Such a false sense of confidence for a product whose actual performance specs aren't properly published. And it isn't just Longacre playing this game.
If you're in the recreational motorsports scene, you've probably noticed what I'm referring to. If you're an engineer in the same, it probably has driven you nuts. I'm not even talking about full NIST traceability (although if a serious customer wants that and is willing to pay, it shouldn't be an issue...good luck with that) but just a really basic warm and fuzzy.
Intercomp pulls the "Certified Accuracy" bullshit marketing wank, but doesn't even specify the overarching standard that their products are certified against.
Joes Racing labels their analog "Pro" model as "ANSI grade B40.1", but if you've actually read ASME B40.1 proper (currently published in ASME B40.100-2005, which contains 5 related standards including B40.7 for digital gauges), you'd know that's bullshit marketing wank...for all anyone knows, they're selling pseudo-Grade D chinesium trash at a premium price point.
Pick any consumer-grade product and you'll discover the same pattern.
The only markets I've seen any semblance of meaningful published specifications with respect to pressure gauges are industrial and scientific/metrology. Of course, the typical players in those markets aren't nearly as naive.
Analog gauge ~100 USD -https://www.mcmaster.com/pressure-gauges/pressure-and-vacuum...
Both of the above gauges come with Calibration Certificate Traceable to NIST
And of course the contribution from the sidewalls (as a percent) will change as tire pressure increases.
On the other hand, if you had hollow steel "tires", the stiffness would be everything and the contact patch would change only imperceptibly with a change in tire pressure.
Car tires are obviously somewhere in this range.
A car? Not much. At any reasonable driving air pressure the air is doing most of the work.
A pickup/van with E or better rated tires? Eh, some but not much in proportion to the weight said tire is expected to carry, especially if the tires are anywhere near the top of their pressure range.
A skid steer or some other off highway machine with "bajillion ply" tires expressly designed to be resilient to sidewall punctures that runs low pressure because all of the suspension is in the tire? Yeah, sidewall stiffness accounts for a lot in that case.
Consider a section of tire wall, with the same pressure on either side. It is curved outwards, and if you put compression on it, it will buckle outwards. If there is higher pressure on the inside of the curve, that would help with the buckling, not prevent it.
In an inflated tire, as in a bubble, it is the tension along the curved surface that resists the internal pressure. What keeps the sidewall of an inflated tire from buckling is that the buckling would decrease the radius of the curve locally, increasing the net inwards force of the tension over the area that is buckling.
I have some data on this for a few different tyres but I don't think I can legally quote it here :(
Here are a few:
There are many.
Most of the trucks I've driven have either an analog "rear suspension pressure" gauge, or a digital calculation of approximately how much weight is there for the given pressure. With an analog gauge, you as a driver start to get a better idea of how much weight is on your drives, possibly saving yourself from shelling out $20 to scale up on a regular basis. As I sit in my truck without a trailer, I currently have a reading of ~10psi; this truck for instance, every additional ~5k lbs over the drives is an additional ~10 psi in the bags.
As an added bonus, you can basically ignore sidewall stiffness for bicycle tires.
Also: hookless? Like they got rid of in, like, the '70s or '80s? My beater road bike (that's older than I am) has 630mm hookless rims, and I got a discount because of it.
Are the new ones hell to change tires on? Changing tires on that bike is the most miserable bike maintenance task I've done. And I've done most anything short of a headset or suspension fork rebuild.
Zipp and ENVE seem to be in some kind of internal width arms race with the release of the Zipp 303-S, and the ENVE SES AR series rims. I was exaggerating a little bit :) You'll still find many, many road bikes with disc brakes maxing out with clearance for 32mm width, but that's an improvement over their rim brake cousins. But look around in the growing "all-road" segment and you'll find wider tire clearances on road geometry frames showing up.
I am unsure if this hookless is the same hookless that is from way back in the day. With the way things go in the bike industry (cyclically, no pun intended), I wouldn't be surprised if it were.
> I am unsure if this hookless is the same hookless that is from way back in the day. With the way things go in the bike industry (cyclically, no pun intended), I wouldn't be surprised if it were.
I’m sure they’re holding way tighter tolerances in manufacturing now than they were back then at the very least. Probably more on the tire side than the rim side.
You used to have to get matched sets of V-belts for multi-belt applications, table saws being a common one.
Nowadays, people mostly accept that the tolerances are tight enough that you don’t need to do that, at least for the lighter-duty uses.
The new tubeless rims are different than the old ones, and There’s a lot that went into reducing the variance in the rim the rim / tire interface.
Then you can weigh arbitrary objects using your phone! Be careful, don't break your phone (don't try to stand on it to weigh yourself!)
I have a Melitta funnel for making single servings of coffee, but when I'm using it I can't see how full my mug is without lifting up, and they are mostly opaque, and the funnel doesn't hold a full mug's worth of water. So I started putting my mug, funnel, filter, and coffee on a digital scale with tare, and remembering how much each size holds in dry ounces. It avoids the situation in which I'm waiting for the water to run out to make sure it won't overflow; I just pour until I hit the right weight and then go do something else until it's done.
The compass is still terrible though!
A V4, yes V in a car
A compact 4 speed transaxle with 4 on the tree
So much fun. Torquey. Enjoyable. Gets good looks and sparks conversation
Easy to work on!
I do love my 900s as my daily drivers but the 96 is a masterpiece
That's actually a harder question than getting the weight of one... which makes the entire exercise ridiculous.
You should be able to estimate this contribution by fitting to the curve of area vs pressure at lower pressures. For example, the linear F=PA relationship clearly breaks down at 0 pressure, as F>0 and A>0 when P=0.
You'd get better accuracy from a bigger added weight, but the bigger the added weight the more you risk changing what you're measuring by forcing more tire rubber into contact with the ground.
Yes you can. It's called "chalking" tires and is commonly used to figure out what pressure achieves optimal contact patch.
My guess is that the tire's resistance to being bent out of shape is also a significant factor in the usual case of the tire's resting on flat ground and that he got as close as he did through a combination of coincidence and cherry picking (deciding after he knew the right answer how much the added gas weighs, changing his estimate of how much of tire's surface consists of grooves after he knew the right answer).
As a bit of cool technology trickery, the Air-Weigh system encodes the weight data onto the existing tractor-trailer electrical connection so that the weight can be displayed in-cab without any extra wiring to connect. They don't say much about how but I'd guess RF modulation on the electrical for the lights or something.
Additionally, the grooves are rather stiff and will help hold the treads - the air pressure on the inside of the grooves will be helping push the treads onto the road. Likewise, tire sidewalls have some stiffness and will also be helping hold the car up beyond just the air pressure, as you've noted in your addendum.
This is a good heuristic for approximating the weight of a car but you won't be getting particularly accurate weight measurements out of it.
If course that would only give you the sprung weight, but you could jack up a corner and then weigh just the unsprung components with a single bathroom scale. (Subtracting any contribution from the spring, if it's still compressed at all.) Either that or remove and weigh a wheel and then estimate the rest.
And that they'd be reporting that back to the central database along with their gps etc.
Reading the comments here, that probably is more accurate than this hack. It isn’t as educational or entertaining, though.
Can someone explain why the air pressure inside the tire is the same as the pressure that the tire applies to the ground? Ignoring sidewall stiffness as pointed out to others, why must this be true?
The author also did not ensure the ground was level, which of course would have an effect on the weight distribution.
I don't think this works. To see why, suppose we make the tires out of a very stiff material. We control the pressure P, but changing the pressure will not change the surface area A. Therefore, by changing P we can set F to whatever we want, which shows that we are not actually measuring the weight of the car. The basic issue here is that the air pressure is not the only thing that is supporting the car.
I don't know how important the stiffness effect is for real tires, but I suspect they are sufficiently stiff that it matters a lot.
For the case you suggested above, this is essentially how run-flat tires work. Of course, the sidewall of the tire isn't as stiff as the support ring, but they do provide some support.
Let P_m be the pressure recommended by the manufacturer. A better sanity test would be to fill the tires to .75P_m, then to 1.5P_m, and see if the area in contact with the ground has shrunk in half.
It's odd, because as I was reading your comment, I was thinking the exact opposite. I am considering the force it takes to squish a flat tire. A person can squish a (regular car's) flat tire with his/her hands.
My gut-check tells me that the stiffness of each tire can lift less than 10kg/20lbs... multiplied by 4, that still is not a huge contributor to the overall weight.
my father was erroneously pulled over by a CHP that used a similar method, taken to a weigh-scale, and was actually under-weight. Wooops!
This is a negative result.