> The first thing I wanted to do with my Saab 96 was to establish a baseline weight. The cheapest scales used for racing are over $700,
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.
Or go to the metal scrapyard instead, and leave with a few dollars more than you came with! (and get weighed) .. and if you have a pickup truck, you get to see them use this cool magnet to suck your junk out of the bed ;)
When I would bring stuff to the dump, I would see people pick up all the metal items from the pile after they dump their trash. It would add weight to their car, so they were charged less, and then they could recycle the metals for even more money. I found it to be a brilliant system. Not entirely sure if it's fraud or not...
Not as convenient as a dump, generally, but there are also regularly unattended weigh stations for trucks all over the place. The scales are always on. You can drive across those and you don’t even have to stop, just slow down. I have even driven over them while the weigh station was attended and the dude manning the booth did not care at all.
I mean, public weighbridges are a thing pretty much everywhere, whenever you have trucks bringing in any kind of goods they usually have to be weighted before and after. There's at least 3-4 public ones that you can turn up to and use without an appointment around me, it's £10-20 depending on the vehicle you're weighing.
In Santa Clara near the airport there was (and probably still is) a paper and cardboard recycler. The scale is off a public street near some train tracks. At some point in college we noticed that the scale stayed on at night and the weight was clearly visible in the kiosk across the street. So me and my car buddies used to weigh our cars there for free at night!
Both (in and out) numbers are usually printed on the receipt, as the charge is per weight dumped (with some minimal fees; for our waste management station that includes 320 pounds).
Another way to possibly get better accuracy is to pump the tyres to several different pressures, and for each pressure, re-measure the tyre contact surface.
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.
A 12v pump which plugs in to a cigarette lighter socket costs about $15 / 12 GBP. They're not the fastest things in the world but they work and there's no labour other than screwing them on and off.
It’s possible (but laborious) to use a bicycle pump to fill a car tire: they use the same kind of valve and bicycles use a higher pressure than cars do, there’s just a lot more air to move.
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.
> I have a decent pressure gauge that I got for $25, so I feel I can be reasonably accurate.
Such a false sense of confidence for a product[1] 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[2] pulls the "Certified Accuracy" bullshit marketing wank, but doesn't even specify the overarching standard that their products are certified against.
Joes Racing[3] 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.
Fluke[1] 700G06 or 60psig span Ashcroft[2] 2089 with choice of option(s). These are just gauges with standard 1/4" NPT male interface, so you'll have to retrofit your own tire pressure adapter. To be sure, these are not your typical consumer-friendly price points.
How much does tire sidewall stiffness contribute to the force holding the car up? I know that can vary a lot. Some cars have run-flats that won't sag when they lose pressure (or won't sag as much, anyway). Low profile tires probably transfer more of the load through the sidewall than doughnut tires.
And of course the contribution from the sidewalls (as a percent) will change as tire pressure increases.
Oops that's a great point. I completely forgot about sidewall stiffness. My messing around with treads was probably just me adjusting things to make myself look good then!
It's useful to consider the extremes. If your car was somehow supported only by party balloons on the rims, then their stiffness would contribute nothing and the calculation would be spot on.
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.
>How much does tire sidewall stiffness contribute to the force holding the car up?
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.
I'm sure the sidewall of a quality passenger car tire could hold 50 lbs just fine. A decent quality passenger car tire will hold my body weight if I sit on one tilted upright. It could be a significant percentage.
When the tire is inflated, the sidewalls (and the tread area) are under tension, right? (They would just fold up under any appreciable compression.) If so, does that mean there's no downward force from the sidewalls? Perhaps we should be asking if there needs to be extra pressure in the tire to keep the sidewalls under tension?
The pressure of the tires pushes horizontally outward on the sidewalls and helps them hold their shape, so they can absolutely be in compression and not fold up.
I am not convinced by this. It would seem that, if it works the way you say, a bubble's surface could be under compression.
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 don't follow F1 very closely these days, but I think I've read that they recently changed, are will change in the near future, to lower profile tires. Is that right? And that the teams have made, or will made, the suspension softer because low-profile tires contribute less?
Didn't somebody test a couple years ago with a set of wheels/tires that more closely resembled passenger car sidewall heights? You know, like 80-100mm sidewall heights instead of whatever they run currently? It seems nothing came of that; can you shed any light?
Commercial trucks with air suspension do this, there are various "on-board" scales that tap into the air lines and try and give a weight. It's obviously much more of an issue when you have regulations involved in how much you can carry, so weighing is a big part of trucking.
Bingo, came in to throw my two cents out as a driver. Only difference between the OP and trucks is that trucks typically measure the pressure in the suspension (virtually all trucks built in the last 20 years use air suspension, at least for the rear axles), which gives a surprisingly accurate reading.
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.
Relatedly: if you and your bicycle weigh 200 pounds, and your tires are inflated to 90 psi (a very reasonable pressure for a road bike; racers run much skinnier higher pressure tires), your total contact patch is a little over 2 square inches. Think about that the next time you’re watching somebody fly down a descent going 45mph or more. That’s not a lot of rubber keeping you on course.
As an added bonus, you can basically ignore sidewall stiffness for bicycle tires.
I guess I'm way out of date. I remember hearing that people who ride e.g. PBP run way wider tires than normal road racers because of the road conditions, but I didn't realize that the mainstream road racers were running that wide now.
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.
Well, a lot of this is driven by the gravel racing gang... the people partying at gravel events are already happy running 40+mm width, or MTB tires, on very wide rims. But there's also a growing segment of people who have figured out that you can have a bike that's a little bit more comfortable off pavement by going wider, without totally compromising speed on the pavement, and it also makes tubeless less likely to blow up in your face.
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.
That was informative; thank you. I’ve been pretty out of touch with the cutting edge for a while.
> 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.
I’ve gone from 700x25 to 700x35 to 650x42 and I’m never going back. The wider tires are no slower, and a hell of a lot more comfortable. That may be because I am in a land where bad chipseal is good pavement, but even on Germany’s smoooooth roads, they still fly.
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.
Wow, I just road ride for leisure on my own, don't know any racers/university types, and my tires are 5 years old 19mm on an old carbon frame. I'm way out of current trends.
My ass hurts just reading that. I bike toured about 7000 miles on 622x28ish tires, and I can't even imagine riding 19mm tires on anything other than completely immaculate pavement (or maybe a velodrome? What do they run?).
With ideal friction assumptions, the contact area doesn't matter. The friction force is determined by the normal force (weight) and the coefficient of static friction.
You can weigh small objects with most smartphones with decent accuracy these days using the barometer. Put your phone into a ziplock bag, blow full of air, zip closed. Take a small object with known weight to calibrate your scale. Carefully place the object on the ziplock bag with your phone inside. The barometer in your phone can measure the air pressure change by having the weight placed on top of the bag. Knowing the weight of your first object and the change in pressure observed, you can calibrate the scale to know how much pressure change correlates to weight placed on top.
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!)
On measuring things - I prefer going in the opposite direction and weighing even where it's unconventional.
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.
That's amazing - the accuracy of phone barometers is great, I found they could reliably detect the difference in pressure from raising or lowering the phone a meter or so.
Kudos, for My first inclination would've been to jack up the car and look at pressure differences of unloaded vs. loaded tyres... then look at loaded wheel center height above ground vs. nominal radius, compute the the reduction in volume, and assume contact patch to be the area of the flattened section of the tyre cylinder... giving me a whole lot of parameters with wild inaccuracies and probably equally wild results :)
Back when Microsoft was doing those obscure job interview questsions like "how much does a 747 weigh", this was my solution, measure the deflection of the tires and the PSI to calculate weight. Apparently the answers they were looking for were more along the lines of "put the plane on an aircraft carrier in a lake and measure the water displacement" <shrug>
This method doesn't account for the support of the tire itself through the sidewalls, as opposed to the pressure contained within it.
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.
I wonder if you could do even better by putting paint or ink on the outside of the tire, driving over a piece of paper, and seeing what fraction of the paper touched your tire (and thus what fraction of the rectangle boundary actually touches the ground).
And to measure how much of the grooves are in contact with the ground, you could add a known weight to the car (like 100kg of people), measure it with and without that weight and see what proportion of the grooves you need to account for to get 100kg difference.
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.
>I wonder if you could do even better by putting paint or ink on the outside of the tire, driving over a piece of paper, and seeing what fraction of the paper touched your tire.
Yes you can. It's called "chalking" tires and is commonly used to figure out what pressure achieves optimal contact patch.
You'd basically just be measuring the maximum width of the contact patch, not its area. The best solution would be to lift up each tire, put a piece of paper under it, let the car down, then remove it, measuring the darkened area.
If you drive the tire onto a rail half as wide as the width of the tire (specifically, the width of the area in contact with flat ground), then according to his logic, the length of rail in contact with the tire will be twice the length of flat ground in contact with the tire (assuming the contact area to be roughly rectangular in each case). I have no doubt that the length would be more, but severely doubt it would be anything close to twice as much because I think a large percentage of the "normal force" (the weight of the car) goes into keeping the tire bent out of shape in ways that do not contribute linearly to the area in contact with the rail.
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).
Well, the idea is nice, but, as others have said, we need to account for sidewall stiffness. But we can do better with law of large numbers and employing metering tape to measure difference in car height with different pressure in tires. First we need to collect baseline data - tire brand, model and size, tire pressure, and car height difference , for known car weights, for two different pressures (say, 20 and 35 psi), this way we could deduce tire sidewall stiffness for given make/model/size of the tire. After that we can do the same for all cars with same tires, by measuring car height difference between two pressure points. Measuring height changes on both axles (wheel center to lowest point of the wheel arch) will get us weight distribution ; but we'll need to account for staggered tire setup .
A related method is used commercially for self-weighing trucks. Typically it's done by measuring change in the pressure in the suspension airbags and needs to be calibrated for the specific suspension and airbags being used. One major vendor lists the accuracy as +/- 300 lbs which is pretty good for use in the 30-40 ton range but not so good for a passenger vehicle.
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.
The edges of the footprint won't be bearing full pressure because they will be in tension (in fact, load distribution profiles of tires tend to look like a rounded trapezoid).
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.
The tread should not matter, because the force gets spread under the tread groves before you reach the inner diameter. You should also consider the thickness of the tire rubber, as this could reduce the pressurized contact area.
Another option is to measure the compression of the suspension springs and then use the spring rates and the suspension motion ratios to calculate the corner weights.
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.
Can you do this purely with pressure?. Lift the car off all 4 tires then add a known pressure to all tires, just enough to seal the beads. Then lower the car and measure the pressures again. I'm sure you could get to the ratio of weight distribution for your car. And with the right coefficient for your tire you could probably accurately approximate weight.
I've been assuming that public buses would be able to detect how many passengers might be on board by detecting the weight changes from an internal sensor of the tire pressures. Count the jumps up and jumps down in weight as people step on and step off, etc.
And that they'd be reporting that back to the central database along with their gps etc.
The quickest way to realize the contribution: Imagine the tire is made of steel so it does not deform when the air is let out. This would be terrible to drive on, but it would not require inflation even though it's hollow like a rubber tire. A rubber tire will usually not "go flat" even with the rims/wheel installed even without pressure (aside from atmospheric which is present when it's put on the rim). So some load is supported by the structure of the tire itself through the side walls. How much will vary with the tire.
> We want to find the force at each tire. So by rearranging, we get F = PA. The pressure is gotten by a tire pressure gauge.
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?
Because the tire would deform, if there is a net force on it. Therefore the force from the inside is the same as the one on the outside. (However, that argument has only any chance to hold at the contact point from the tire to the ground, since obviously everywhere else the tire does not perform because of stiffness of the tire material.)
It actually isn't true at all in practice. The inflation pressure and contact pressures vary dramatically even for a static tire due to curvature, tread depth, and many other factors. See e.g. https://www.sensorprod.com/news/white-papers/tcp/wp_tcp.pdf
The force must be the same (because the net force must be zero, otherwise the tire surface would move up or down), and the area is the same on both sides so the pressure must be the same.
Assuming all tires are the same, I think this would be most applicable for weight distribution alone, but not really for the weight. For performance applications, distribution is most important
The author also did not ensure the ground was level, which of course would have an effect on the weight distribution.
In the Netherlands they use driving axle load systems usually in the right most lanes on the highway that feature trucks. There is also a service that automatically calculates the tire pressure and sends a message to the company if the tires are deflated when driving over these weigh points.
Maybe use a lever instead? If you have a decent length of steel pipe or 2x4 and a trolley jack. Lever one end of the trolley jack up on a pivot, you should easily be able to get 20:1, within range of a cheap luggage scale or convenient heavy objects. Have to do front and back of course.
There's a period, after it rains, when, as cars that have been standing in it drive off, they leave behind a wet print of their tire contact patches. With a jug of water and some patience, perhaps you could use this to get a more accurate measure of the contact area?
I got this question asked in an interview for a mechanical engineering internship. I was asked "you are given a ruler and a parked car, how can you measure the weight", took me about 10 panicked minutes and 4 hints before I figured it out.
By the way, a car with electric propulsion and regenerative braking should be able to "know" its weight quite precisely (from the amount energy that is needed to accelerate or the amount of energy that is returned when regen-braking).
After you've subtracted the effect of thousands of hard to measure variables like the slope, the tyre rolling resistance, the resistance in the wiring, the magnetic flux saturation in the motor, the accuracy of the current sensors, etc, etc, etc...
Not thousands but a few and most of them are measured anyway, like slope with a gyro or battery drain.
No need to over complicate it just measure the amount of regen the more you get the more the car weighs.
Look into WIM (Weigh in Motion) and Tire Anomaly technology. I don't know how much you'll find out on the internet about specific details, and I can't elaborate, but it's pretty interesting.
The suggestion is to use F=PA to measure the car weight, where P is the tire pressure, A is the area the tires in contact with the ground, and F is the unknown: the weight of the car (measured in terms of gravitational force).
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.
I think he nixes your case at the start, when he goes through the sanity check thinking about the flat tire. Also, I think your rigid body case use the walls of the tires to apply force to the care, not the force from the pressure to keep the car static, which is why it fails.
I agree that higher tire pressure results in smaller surface area, but it's not clear to me that the tire itself does not support the car, skewing the result. So I'm not convinced this sanity check is sufficient. For example, as another commenter mentioned, run-flat tires can support the car on stiffness alone.
I agree, the sanity check is not sufficient. It confirms the concept that as pressure decreases, area increases, but is not sufficient to confirm a linear relationship. In the limit, the area clearly does not become infinite when pressure drops to 0.
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.
He mentions the flat tire, but he ignores the fact the pressure of a flat tire is 0, leading his measuring method concludes that the weight of the car = F = PA = 0A = 0. His measuring method flunks his sanity check, but he never notices.
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.
With a completely flat tire, the rubber sandwiched under the rim will transfer weight directly from the rim to the floor, so the pressure equation won't apply.
> 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.
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.
It’s certainly possible my intuition is wrong. Run-flat tires can support the whole car, but it is possible this support only kicks in at very low pressure.
A slope and a crane scale could have been relatively cheap and I'm guessing estimating the friction would be more accurate than making assumptions about the tire wall strength.
Or just use one bathroom scale, and a lever to adjust weight of a single wheel to within the scales limit, then measure each wheel individually and add the results.
Random thought. Could you compare acceleration going uphill, downhill, and on a level street to estimate mass, somehow keeping the RPM the same on the engine?
You would have to know the rolling resistance of the tires, aero load, and the exact amount of horsepower the engine is outputting to the wheels. It is impractical, once you consider engine wear, air density, and probably a few other factors required to get a correct reading.
a similar was (maybe is) used by the California Highway Patrol to determine whether or not truck drivers are carrying too heavy a load, or too heavy a load down specific streets or highways.
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!
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.