For anyone who wants to learn more about this subject, I'd recommend Carroll Smith's book "Tune To Win", almost the entire first chapter is about tyres and how they respond to load and cornering forces, and it's explained in a fairly common sense way. It's from the 1970s, so there are a few strange anachronisms, but it explained to me pretty well why we worry about camber, weight transfer, slip angle etc.
@pgrmonteiro
9 жыл бұрын
That's a really nice book!!
@MeiGunner
8 жыл бұрын
so if i have a hatch back ,, the back is going to be lighter ,, if i put tall tires on front and short tires on back so the front of the car sits higher then the back making more weight, lean away from the heavy motor ,, ( will This make each tire have more Equal-weight ) , then having all the weight on the front two wheels
@SebastianLopez-nh1rr
7 жыл бұрын
The book looks great !!
@z1522
5 жыл бұрын
The car's center of gravity determines what percent is carried by front and rear tires; it is entirely unrelated to wheel size, nor to apparent inclination of the car. You will only raise the CG by putting mammoth tires front or rear, creating different issues like flipping or rolling, but also by screwing up the lever arms of the suspension geometry you will destroy the inherent steering and handling traits.
@stevenjackson8226
4 жыл бұрын
Have all the Carroll Smith books. Been using them as references for decades.
@labradormcgraw2409
7 жыл бұрын
I've said it once, I'll say it again - this guy embodies everything that's great about KZitem. Unrivaled knowledge coupled with a natural talent for teaching... add to this the seldom-seen ability to include only what we really need to know, and what we have is an absolutely indispensable learning resource. The fact that, through KZitem, we get all that for free necessitates the burning question: why are we paying $$$$s for inferior quality tuition from college? Huge respect to EE!
@C-A-L-M
5 ай бұрын
qualifications my friend is why college is necessary needed one from college to work even tho i was taught better from my uncle for free
@bradleylechton4964
9 жыл бұрын
Learned about all this in my physics class, its cool to see a car related application. Keep up the good work!!
@markusdammasch9108
7 жыл бұрын
Very good video - I would have liked to see more on how changing tyre width (without changing weight) affects the grip of the tyres (on dry tarmac) . As you mentioned, pressure is equal to load over contact area. So if the contract area increases then the pressure decreases... Which is why I believe that increasing tyre width doesn't proportionally increase grip - for example, a 30% wider tyre may only give a 10 or 15% increase in grip... Also wider tyres mean an increase in unsprung weight which has undesirable side effects.
@JackMott
2 жыл бұрын
A wider tire may DECREASE grip if you end up so wide that it can't reach optimum temperature or load. However in practice with road cars it is very hard to reach that point.
@d1sturb3d119
9 жыл бұрын
Thank you for this video! I've been trying to explain this to so many people and have been called nuts so far. Everyone believes more weight creates more traction stating aerodynamics as an example claiming that downforce creates the same forces that additional mass will. If you could talk about the ill effects of additional mass in the winter that would be great too or how adjusting the width of your tire can affect traction in bad conditions. For example, wider summer tires for better heat dissipation and longer tire life in the summer a positive effect to cornering. Also narrower tires in the winter to increase pressure in the contact patch creating better grip rather than increasing the mass of your vehicle.
@GregoryEvansRacing
9 жыл бұрын
Yep, more mass = more inertia, but download does not produce inertia, so it's preferable.
@d1sturb3d119
9 жыл бұрын
Gregory Evans Yeah I've explained that but many believe that if downforce creates more grip mass should have the same effect.
@GregoryEvansRacing
9 жыл бұрын
Time to find new friends then! lol
@d1sturb3d119
9 жыл бұрын
Gregory Evans Eh I'm faster than them so it balances out :P
@wannak00kie
9 жыл бұрын
Wait so, and please excuse me if this sounds ignorant I'm just trying to learn---the whole theory of if you have a rwd drive car , getting better traction in the winter can be achieved with winter tires and sandbags in the trunk ..is false? The weight is bad? And also aren't winter tires a little wider than summer tires? I know in places like india all the cars have relatively (compared to AMERICA) thin tires, which they do to increase fuel economy. Please break it down for me, just trynna learn, thanks!
@willdriverc
9 жыл бұрын
I just saw the link to the patreon website. I admire your courage for quitting your job to do this. I really hope this channel gets even bigger for you. I will be sure to pledge later this week. Keep up the brilliant work!
@EngineeringExplained
9 жыл бұрын
willdriverc Thanks, truly appreciate it! If nothing else sharing the videos with those who may be interested is just as helpful. :)
@MsKaran0000
9 жыл бұрын
you are the creator of many upcoming automobile engineers.. _/\_
@EmmaHopman
4 жыл бұрын
Including me apparently
@beatle497
9 жыл бұрын
Something I have been trying to understand for years. Thank you for that simple and straight forward explanation.
@crate2819
9 жыл бұрын
Question: Are you ever planing on going back into mechanical engineering one day? You seem to have a lot of real world knowledge and the world sure could use more mechanical engineers with hands-on experience. Great video, btw.
@pedropestana2443
6 жыл бұрын
Man, that's something i've aways wanted to know. Once, i discussed with my physics teacher about the equation F=N*u questioning him about why formula 1 cars have wide tires if it gives more weight and bad aerodynamics when they were not supposed to increase grip with the increase of the area in contact with the floor. He was clearly not able to explain me why. Thanks for this insight! haha
@ProfessorDingus
3 жыл бұрын
Had the same discussion with my physics professor! She was basically claiming tire companies were running a scam by selling more expensive wide tires since the friction coefficient is not dependent on surface area.
@Gilamang
7 жыл бұрын
Great explanation! Ever since high school physics, this question has lurked in the back of mind. Keep up the good work.
@TheGiagoskap
9 жыл бұрын
This is one of your best videos. It is on a topic that I imagine many wanted an explanation, since wider tires do result to better traction but the basic formula F=μN most people learn does not show that. Plus, your 'simple' explanation with the rubber band is remarkable for a topic this complicated. Amazing job. Keep it up.
@hotflashfoto
8 жыл бұрын
I'm not an engineer. That said I really appreciate all the videos I've watched. This is the only one that has me mixed up. I've read some of the comments already posted and only a couple allude to my question. When doubling the weight of the car one also doubles the mass, and therefore its inertia. With the brakes assumed to be more than capable of applying a 1G force to slow the vehicle then they are effectively removed from any equation, and that makes sense. So now we have 2 vehicles, 2K and 4K, attempting to apply a lateral force to the tires. Each of the 4 corners will respond differently due to things such as the setup of the suspension, CG, roll center, etc. For my example we'll take the front right tire during a left turn, which will be the tire with the greatest load. As the tire is loaded to a real-world sub-1G lateral force it will experience more downward pressure, which will deform the contact patch. In one of the other comments an equation showed that it will nearly double in size with 2x the weight, so I think (possible error here) that the increased pressure during cornering will also yield a larger contact patch, even if the pressure is not evenly distributed across that patch. Since the patch grows at a slightly slower rate as the load increases it stands that there is a higher loading per square inch. And suspension geometry can be modified so that the higher load is more evenly distributed across the contact patch, so we can effectively remove that from the equation as well for the purpose of my example and question. In your video you show an example of viscoelasticity by stretching a rubber band. Then at 5:58-6:00 you apply that same principle to a tire under compressive load. A partial transcript by myself, starting at 6:00: "as you start to smash that rubber down, and it starts to kinda press itself out and thin out these polymer chains it’ll start to break away rather than deform with the road and you don’t want that to occur. You want just an elastic deformation with the road, and that gives you your maximum frictional coefficient, and thus giving you the greatest force you can go around a corner that you can resist.” In a few other comments the question is raised about downforce and weight (mass) and why they don't provide the same traction. I didn't see any comment about downforce being dependent on speed whereas mass is constant. It makes sense to me that downforce will allow the contact patch to grow the same way that mass does, but the cornering force on a lower-mass vehicle will be less at the same speed, allowing it to corner at higher speeds. However, we're not talking about increasing speeds as a result of downforce vs mass; we've effectively thrown that out of the equation as well by specifying a lateral force applied while the tire is loaded. The load does not specify downforce or mass, just simply load. What we're trying to do is simplify the equation by removing certain variables that can change how a tire responds. This discussion is supposed to isolate specifically the applied downward load and the lateral force and then show why a tire loses traction in a corner, or during acceleration and braking. I can kinda understand that 'starting to break away' means the tire surface is beginning to act more like an eraser as particles are removed. So, as pressure increases on a contact patch how is it that more downward load, which increases pressure on the patch, finally results in traction loss instead of increased traction? This takes into account that the contact patch does not increase appreciably past a certain point, resulting in more pressure per square inch. So, after all that rambling here's my conundrum (or several): I need help understanding how viscoelasticity explains how a tire effectively loses traction as the lateral force increases. Is this once it reaches a certain threshold? Is 1G of lateral force the same as specifying 2K pounds of lateral force? If not, and I believe it is not, then using "1G" as the lateral force assumes a load that is different from downforce in that 1G on a 4K vehicle is twice that of a 2K vehicle, and therefore changes the equation, which we're trying to simplify. But it also means that I may have missed or overlooked something, which is why another discussion and/or video would be most helpful.
@2testtest2
8 жыл бұрын
+Hot Flash Foto (Bob) I'm currently studying engineering, though I do not know quite how this viscoelasticity works. Not yet at least. However I can clarify some of your question. The term "G-force" is slightly misleading. 1G is 1 times the acceleration due to gravity. So 1G of latteral force does not really make sense, because 1G is not a measure of force. What is usually meant however is the force required to accelerate something at 1G. Simply put a mass with a weight of 2k pounds will require a force of 2k pounds to accelerate at 1G, a mass with a weight of 4k pounds will require a force of 4k pounds to accelerate at 1G. Now the formula that defines the max friction force between two dry surfaces is F = N*μ, where F is the max force, N is the normal force, which in your question would be the weight put on a certain tire. From this equation one would think that the weight of the vehicle would be of no consequence to the cornering ability, since the max force would increase proportionally to the force required for turning. The point made in the video however is that μ is not a constant, but rather a function of the pressure of the contact area. That is the weight/area. So the more weight you put on the same area, the lesser the μ, making for less friction, and thus lesser cornering ability. Let's look at an example. I do not know what is realistic values here, so I will choose some arbitrary ones. Let's assume a contact patch of 2 square inches, and a weight of 2k pounds. This gives us a pressure of 1k psi. Let's again assume this has a μ of 1 for our tire. This gives us a max force of 2k pounds, and so 1G acceleration. Now let's increase the weight to 4k pounds. Assuming the contact patch is the same, this gives us 2k psi pressure. Now, since we know μ is less at bigger pressure, let's assume that μ is now 0.9. that gives us a max force of 3.6k pounds, and thus 0.9G acceleration. Hope that leaves you at least slightly less confused.
@troy7481
9 жыл бұрын
Some high level stuff explained on a very understandable level. Great video!!
@stevenjackson8226
4 жыл бұрын
Jason you are brilliant. I've been watching you for years. And I'm late to this party. 50/50? Those of us who've been driving rear mid-engined race and road cars forever, they have a weight distribution more like 60-65 rear/40-35 front.
@jess81086
9 жыл бұрын
Very good explanation of a difficult concept to explain. The information on the tire viscoelasticity is very interesting and really helps explain why larger tires are used for better grip performance. It is always difficult to explain to people that friction is independent of surface area, and then in the same sentence say the bigger tires produce more grip. I have had this debate with many people when they try to justify their big brake kits by citing "larger tires produce more grip so bigger brakes should as well." Sometimes common sense can lead one astray. Thanks for making this video. I can finally point people somewhere for a good explanation!
@EngineeringExplained
9 жыл бұрын
Jesse Crutchfield Thanks!
@jess81086
9 жыл бұрын
Engineering Explained After thinking about it more carefully, mu is a function of the cross-sectional area of the shearing plane of the tire (which has a fairly set height, but the width is set by the width of the tire) and the visocelastic properties of the rubber compound (which is behaving like a non Newtonian fluid that is "shear thinning"). For your example the width of the tire doesn't really change so it doesn't make much difference, but when you look at the concept of tire loading and weight transfer it does make an important difference. If the suspension is able to keep all four tires loaded well and is able to keep the weight from transferring too much then the cross-sectional area of the shearing plane for the tires (added together as a single unit) is four times greater than if all the weight were to be transferred mainly to one tire. In the case where the weight is evenly distributed not only is the normal force per tire lower (thus putting less stress on the polymers) but that stress is also spread out over more cross-sectional area thus increasing the system mu more as well. This concept is probably a little too complicated to go into in great detail in a video, but in reality mu for the tires is effected by both the cross-sectional area of the shearing plane and the viscoelastic nature of the polymer chains in the material.
@geoffreyanderson4719
7 жыл бұрын
Thanks for this lecture. It cleared up a lot of questions for the first time. SInce high school physics, this question has dogged me.
@shankarbalan3813
4 жыл бұрын
I like all your videos. They are objective, sensible and highly informative. Thanks!
@StereoSoundAgent
6 жыл бұрын
You should profess a course at your local university “Motorsports Physics 1” ... physics was something I loved in school but professors were bewildered when I asked them questions about “unsprung weight ” or “mechanical grip” ... they couldn’t explain to me why larger tires had increased surface area and subsequently increased grip when friction didn’t depend on surface area... anyway. I think car guys are naturally inclined to take an interest in physics and it would be a cool opportunity. Thanks for this one!
@pgrmonteiro
9 жыл бұрын
Engineering Explained It would be awesome a video detailing more about wider tires advantages, there's a lot to it!
@lichtbrechunggg
9 жыл бұрын
Great video as always! I would just say that, while it is strongly implied, you failed to openly state that for a given pair of identical tyres and a given load for that pair, 50-50 is the best distribution as it gives the largest total force BECAUSE load transfer will take more cornering force away from one tyre than it adds to the other.
@Kyle_Duffy
9 жыл бұрын
This is one of your best explanation videos ever. I'm glad you answered the "why" and that rubber band analogy was on point. Weight reduction remains king! (to an extent)
@silverslvr5185
7 жыл бұрын
Wow a lot of this went over my head, though I have a good basic understanding. I was lead to believe (high school physics teacher) all wider tires did was to stay cooler, and not offer better traction. Again, I learned something new today. Thanks!
@projectilequestion
4 жыл бұрын
Yeah, in 'classical' physics, they do go up at the same rate, but tires behave in a different way. Makes me wonder why we use rubber tires lol.
@adairhobbs2196
7 жыл бұрын
great video thanks, I'm studying Mech eng at the moment and we were talking about factors of friction and how surface area is not a factor, that confused me about why they made car tyres wider to increase traction , no one could answer me haha, but your video has. your videos are informative, relate physics to the real world and I always learn something. keep them coming please
@peterers3
7 жыл бұрын
Great explanation, a bit fast at the begin but at the end totally understnadable. Came from your Tesla Video. Very informative!
@tylerpercy8199
8 жыл бұрын
the statement "As load increases on a tire, the coefficient of friction between the tire and the road decreases, thus the amount of lateral grip the tire has decreases" is incorrect. As normal load on a tire increases, the coefficient of friction does decrease however adding normal load to a tire will always result in more "grip". When a car goes into a corner the outer tire of the observed axel does not gain grip as fast as the inside tire is loosing it. This is why lateral load transfer of an axel reduces the grip of that axel vs both tires being equally loaded.
@spencerphilippinedream3706
4 жыл бұрын
i just spent 30 minutes picking apart the video and backing up my claims on how this comment misquoted EE but then saw it is typed in the description. and deleted my comment. so any of you wondering "when" or "where" this is claimed in the video.....its in the description. this is the comment i deleted, my appologies. time stamp? i cant find where he stated what you are quoting. at 1:30 he says "coefficient of friction decreases as normal force increases." and then at 6:10 (closest statement i found to what you are quoting him as saying) he says "you want just an elastic deformation of the road, and that gives you the maximum frictional coefficient and thus giving you the greatest force...". individually, these statements are both true, you have taken 2 quotes and joined them together with a "thus". i dont know about this, which is why im watching the video in the first place. but your comment implies that he believes that a heavier vehicle has less friction (for cornering), which we know isnt true. alternatively, i think he believes that a heavy vehicle is worse at cornering, but going to wider tires to increase the coefficient of friction will help. thus, putting into context his quote "you want just an elastic deformation of the road, and that gives you the maximum frictional coefficient and thus giving you the greatest force..."
@raynic1173
4 жыл бұрын
@@spencerphilippinedream3706 it's way simple: he said to the F will increase to a point, just look at the shape of the curve.
@murraythiessen8216
7 жыл бұрын
Very helpful explanation. This is all true on a nice clean dry paved road. And the extra contact area helps dissipate heat in the rubber for extended cornering and braking as in racing. However things change when gravel sand rain snow and ice are introduced. The average car driver has more than adequate road holding friction on a good road. We typically need the extra traction when the road is slippery. Then wide tires are our enemy. Here less contact area and especially a narrower tire helps the weight of the car to drive down though the snow, .... to get good contact. I sometimes think we need traction control so much now because our oversize tires are floating on the water snow and gravel.
@rafaellastracom6411
7 жыл бұрын
Excellent video. This is why Formula cars want tires as wide and as flat as possible.
@e1337prodigy
7 жыл бұрын
Nice touch with putting what appears to be a youtube award in the video ;). Well deserved. Love the videos, keep it up.
@HusamGibreel
9 жыл бұрын
you have done a great job putting this together, very informative, keep it up.
@IgorBagayev
5 жыл бұрын
This might be the best of your videos.
@gottahavefun13
9 жыл бұрын
Just wanted to let you know that as someone just starting to learn about cars, and even more importantly physics, your vids are incredibly insightful and helpful. Keep it up! Please!!:)
@BillRize
3 жыл бұрын
Great explanation. You teach kids the basics about Coulomb and linear friction with square blocks and flat surfaces and they think they have the knowledge to rule the world. Knock, knock, viscoelasticity and even contact patch (to some extent).
@Al0011235813
Жыл бұрын
Not only is it good to have 50/50 weight distribution, but anti-roll bars will minimise lateral loading, and heave springs minimise pitch during braking and acceleration.
@JoseLikesCars-ln9qc
7 жыл бұрын
This is an older video but I think it needs to be revisited because it makes the assumption that coefficient of friction, μ, decreases as the normal force, N, increases. This is true if you follow Orowan's friction model but it is not the case with automobiles because the asperities of the road have not been flattened by the tire since the N of a typical automobile is not great enough for this to be the case. If this were the case then downforce wouldn't work and we know it does lol. The coefficient of friction of a tire does change with force but it has more to due with temperature. There is a certain temperature that a tire is going to be at is most effective coefficient of friction. A larger tire is better at staying at that temperature once warmed up because the pressure on the surface of the tire is less due to larger surface area for a given force. Therefore a larger tire is going to be less susceptible to overheating and having its μ decrease. The second advantage of larger tires has to do with the asperities on the road itself. Any given road is not perfectly flat. And because of this voids can be created were the tire is not in contact. Yes friction force should remain constant regardless of surface area of the contact point, but the tires temperature is not constant, and as discussed before, less area equals more pressure which in turn generates more heat at a given force, and more heat past optimal decreases μ for the tire. A wider tire ensures that the tire is more likely to be in contact with the road surface despite of the surface irregularities. Edit: I wanted to add that this video is not without merit, but the concept you discuss here speaks more towards why one rubber compound might be better than another, i.e low treadwear vs high treadwear, but it doesn't explain why for a given compound, a wider tire will grip more. I think that has to do with what I said above.
@BoostedFilms
9 жыл бұрын
Thanks for making this video! Good day sir
@DangerAngelous
9 жыл бұрын
Can you make a video about high-profile vs low-profile tyres? or the benefits(and bad parts) of putting bigger rims on a car
@cojo139
9 жыл бұрын
I have wondered this for years!! Thank you so much!
@mibars
7 жыл бұрын
I'm trying to understand why shifting weight around dynamically during driving (accelerating or decelerating) tends to give more traction to heavier loaded wheels? Let's take a car that goes around corner at the limit of grip on all wheels, when you suddenly lift mid-corner so that weight shifts towards front it will fishtail, despite now rear wheels have capability to support more g's with higher coeff. of friction. Works in FWD, AWD and RWD... And the opposite situation: Take a 4x4 50:50 torque split so that we can rule out traction issues and step on it when it is on the limits of grip - It will tend to understeer (unless you overpower it but then it will be sliding all 4 wheels). Why so?
@danielmay7107
6 жыл бұрын
Thank you! Wonderful explanation!
@ericb.4358
4 жыл бұрын
For my 2019 MAZDA CX 5 2.5 L. turbo, when my OEM 225/55/19 tires wear out I'm replacing them with 245/50/19 (same circumference). A decent upgrade that is the max for the OEM 19" wheels that I like. Plus I won't lose too much MPG BETTER TIRES ARE THE EASIEST WAY TO IMPROVE HANDLING.
@stephenmcphee3316
7 жыл бұрын
A little lesson in Physics for everyone (Bryce Ring hit the proverbial nail) ... the size of a tire's contact patch with the road is a function of the air pressure ALONE, and nothing else. If there's 30 PSI in a tire, then each square inch of the tire's surface, by definition, can only support 30 pounds of the car's mass. So, fitting a wider tire with the same air pressure DOES NOT increase the contact patch's area, it widens and SHORTENS it. (The only way to put more rubber on the ground is to reduce air pressure). That's why super wide tires are dangerous on wet roads as the patch does not have the same amount of time to disperse any water and is more likely to aquaplane. I've heard lots of tire shop owners get this wrong, but never an engineer working with a major organisation. Wider tires might give more grip simply because the proportionately shorter sidewalls might flex and distort less under cornering forces, leaving more of the contact patch properly in contact with the surface, and less of the sidewall possibly making contact. I remember one of the prominent Formula One guys saying the optimum tire profile for cornering has been 50 series, and has been for a long time.
@stephenmcphee3316
7 жыл бұрын
G'day mate. Regarding tire contact patches: at rest, the sole significant force acting on the inside of the tire to keep the wheel rim from smashing into the ground is certainly that compressed air. And the term "PSI" absolutely defines the amount of force that air can push against each unit of surface area it occupies, regardless of any loss of tread pattern or rubber-temperature variations (remembering the area between treads doesn't count as it's not in contact with the surface!). So, for a wider tire (and contact patch) the shape of the area must shorten fore-and-aft to keep the same mass per square inch supported constant. Regarding grip, however, I can accept that a softer compound, for example, may offer more resistance to slip from it's ability to better occupy the small diverts that exist in any road's surface. That would be referred to as the friction co-efficient in a formula that might be applied to such a situation. And on the move, there would be some centrifugal (centriPETAL for the physics nuts) forces at play helping the tire occupy a shape closer to a correct circle, but they'd surely be marginal compared to the 300-400kg each tire supports. But, yes, sidewall flex does lead to warmer tires, and a corresponding change in that friction co-efficient.
@isahaque7929
8 жыл бұрын
I have a very simple question here. I know "F = N x μ" but is there any engineering equation relating the CONTACT PATCH (Surface Area) and the maximum TRACTION (for Driving force). I raised this question because I swapped my Corolla's motor to 4A-GE translating to more torque and power. I was initially running Yokohama C-drive 185/60R14 after the engine swap but those tires slipped under hard launch, high rev shifting from 1st to 2nd gear and high speed corners so I got them swapped for the same C-drive 225/45R16. Now everything else was in factory configuration including body panels and suspension (ride height and downforce constant) but my tires won't slip under the same conditions. Being a car guy I always knew wider tires translate to greater traction but never worked out an engineering equation for that. What is your opinion ?
@strangenorms9802
5 жыл бұрын
I don't think anyone will be able to give you an all-encompassing equation, but I can certainly say that the vast majority of your improved longitudinal grip was owed to the fact that the second set of tires boasted a 5.7% larger diameter than the skinnies. These taller tires yielded you a longer contact patch, as well as a helpful moderation of torque at the wheel, reducing the likelihood of wheelspin. As for the improved lateral grip you experienced in high-speed corners -- I'd chalk it up to the wider, stabler contact patch and the shorter sidewalls.
@Thee1Muffin
9 жыл бұрын
More on the topic of grip please.
@PilotDamian
7 жыл бұрын
Once again, another excellent video!!!!!
@BavariaR
7 жыл бұрын
Not sure you already took the topic up tire friction versus tire pressure and temparature. Would be a nice add on to this one
@peterhewson4568
3 жыл бұрын
Your videos are great, but you might want to revisit this one, because it contains much which isn't true. Your starting point is correct; friction coefficient drops with load. But the contact patch area grows almost linearly with load, so trying to explain the load sensitivity of tyres via visco-elastic strain (and elastic bands!) is never going to work. The load sensitivity is actually due to a complex thermal response, so it is (sort of) to do with visco-elastic properties, but not the strain aspect.... Friction modelling is only just becoming possible, and almost nothing is published, so it's an area that is virtually impossible to research. So I totally understand why this topic is so poorly understood. Keep up the good work, because I reckon 99% of what you publish is spot on.
@glenwoodriverresidentsgrou136
3 жыл бұрын
Jason, if you have a minute please answer this question. I understand from your video that increasing rubber pressure reduces the coefficient of friction. And you explanation for this makes sense. But I’m wondering why wider tires generally have more grip than narrow tires. Ignoring carcass support, two tires of different width with the same air pressure will have the same contact patch area, the same pressure on the road surface, and the same coefficient of friction regardless of width. Yet the wider tire will generally have more grip. What is the reason for this? A given segment of contact patch will be in contract with the road for a shorter period of time on a wider tire because the contact patch will be shorter and wider relative to the direction of travel. I wouldn’t think this would be a huge influence, but it could result in less local hysteresis which might lower the local temperature of the rubber in the contact pass. What say you? Your video would also suggest that lowering tire pressure will increase the contact patch area and road surface pressure which should increase the coefficient of friction. But we also know that low tire pressure does not always increase grip. What say you about this?
@itsame1277
7 жыл бұрын
I would just like to expand my understanding of the concept of viscoelasticity with regard to tyre pressure and the coefficient of friction (cof). The viscoelasticity of a material describes the changing behaviour of the material as the polymers are subject to varying conditions. As explained, under normal pressures the polymeric tyre material can deform and behave like an elastic solid, therefore maintaining the cof. However at higher pressures, the material now behaves more like a liquid which cannot be compressed and therefore loses the ability to deform thus reducing the cof.
@Greasyspleen
9 жыл бұрын
I'd like to see you elaborate on this vid some. You could get into how the contact patch changes as a tire is over or under-inflated, and what determines the optimum tire pressure. And "square" vs "staggered" tire setups.
@bward76
7 жыл бұрын
You stated incorrectly that the contract patch increases in pressure with load increase. the tires typically are loaded with pressurised air. the contact patch area is equal to the psi of the tire times the load of tire. this is fundamental to as to why low inflated tires have high rolling friction losses, the contact patch grows proportionally to the internal air pressure.
@mauricioordaz4905
2 жыл бұрын
Jason, I once took this driving courses where they taught us that by inflating more your tires you could eliminate deformation in curves and so avoiding rolling on the shoulder of the tire, reducing slippage. How correct is this, 'cause I have been driving so the Last 20 years without any incident. Let's say 38 PSI Cold.
@florianfixman9604
6 жыл бұрын
This simple fact about tire load sensitivity is also the reason why low center of gravity height and wide wheel track and wheelbase. It also the reason about shift car balance when adjusting load transfer distribution balance between front and rear (by anti roll bars or roll centers)
@richardkoo2043
Жыл бұрын
Learning something new on friction confidence and load relation. Question on tire pressure: does higher pressure reduce braking ability and increase tire wear? Thank you.
@someassarkumar9410
9 жыл бұрын
On normal road cars, how do manufacturers manage side to side weight distribution since the driver is offset to one side.Or is it being overlooked?
@AutomationGame
9 жыл бұрын
I can't remember which one, but I swear I've heard of a car that gets close to perfect lateral weight distribution when it has an average weight driver and no one else in it.
@MeiGunner
8 жыл бұрын
so if i have a hatch back ,, the back is going to be lighter ,, if i put tall tires on front and short tires on back so the front of the car sits higher then the back making more weight, lean away from the heavy motor ,, ( will This make each tire have more Equal-weight ) , then having all the weight on the front two wheels
@Lehpurdzzz
7 жыл бұрын
On high performance cars with adjustable ride heights, the left front suspension is usually raised slightly higher than the other corners. This is called corner balancing.
@sergeyakinin997
7 жыл бұрын
Yep, the left front corner of my gen 6 Camaro sits a tiny bit higher when the car isn't loaded... I'm sure most modern high performance cars are like that from the factory.
@spankeyfish
7 жыл бұрын
+AutomationGame Bit late to the party but that's probably the McLaren F1 as the driver's seat is in the middle.
@DoctoreDoom
5 жыл бұрын
Nicely explained, very informative X)
@adrianrodrigoguitar9482
9 жыл бұрын
If I'm correct then, manufacturers of sport cars use wider tires because wider tyres can handle more load (pressure) and the coeficient of friction won't decrease as much as thin tires under heavy loads?
@Oockeshoek
7 жыл бұрын
Hey Engineering explained! I'm studying mechanical Engineering at the Technical University Delft (Netherlands). A couple of months ago my teacher had us do an experiment with different materials on a surface that we had to rotate. We had to rotate it until slip occurred and then we could find out what the friction coefficient was. Conclusion was that surface area didn't matter. This seemed strange to me as I've always been interested in cars and I always thought wider tires would be better. Then just a couple of weeks ago I talked to my project mentor about it (we have to build a mechanically powered drag car that can do 5m as fast as possible and then stop in the shortest amount of road) and he told me that the only reason cars have wider tires is because of wear and tear (wider tires wear less). I still couldn't believe this, as wear and tear at drag cars is obviously not really a (big) factor. Then he told me that if the tires get really big it's only gonna make a difference because at that point the usual rules don't apply. Now for our project, my university gave us the option to use their skating wheels (because they have a small friction coefficient), but they also have a tiny surface area. So according to my teacher they offer the same grip as wider skating wheels. After such a long story, I wonder what your opinion on this whole deal is, as my mind is still in conflict with what my teacher says. Looking forward to your reply :)
@smartcatcollarproject5699
7 жыл бұрын
The wider the tire, the larger the surface in contact with the road. A larger surface allows the use of softer compound (as a narrower tire with soft compound would wear too rapidly). Of course softer compound have better grip !
@russianrick8403
7 жыл бұрын
Oockeshoek in so many words, what is being explained here is that the properties of rubber change as the pressure on the rubber changes. In order to maintain maximum grip, you want to increase your normal force, but you don't want to cause the rubber in your tires to become overloaded. Making the tires wider adds more mass of rubber for heat distribution, but in increasing the contact patch, you are going to lessen the pressure on the rubber and reduce the effect that the loading has on hardening the rubber and you will thereby maintain a higher coefficient of friction. As is stated here rubber begins to break down as it is overstrained. The marbles that are flung aside on many racetracks, I believe, are evidence of the extreme strain on the tires.
@Oockeshoek
7 жыл бұрын
Rick Harris Aah thanks for the explanation! :D
@TunerToolsLLC
9 жыл бұрын
On thing you failed to mention is weight transfer - and actually why race engineers do not like to use 50/50 STATIC weight distribution. Under braking weight will transfer to the front of the vehicle as a function of CG height and wheel base (and certain suspension parameters.) So, under braking in a 50/50 car you will have a higher tire load on the front tires, as more of the load it transferred fore/aft. This means less grip for other functions, such as turning, and also a different requirement for braking balance (hence the reason for larger front brakes, even on cars with 50/50 or rear biased weight distribution). This is the reason LeMans cars, Formula cars and many exotics favor rear weight bias (there are others as well). That said - a 50/50 car is more neutral to drive (and typically more on the under steer side of neutral.) This makes them easier and safer to drive, as a slight push on the front end is less dangerous than the car rotating in oversteer. A neutral-to-oversteer car (in the right hands) will be more responsive at turning into a corner and inducing a yaw-moment to get the vehicle to turn. Tire loading matters and technically great and correct information - but 50/50 distribution is not always desired, specifically because of tire loading.
@florianfixman9604
6 жыл бұрын
I agree mostly with all your comments. But you tell one result too quickly about 50/50 weight distribution is the best. It is true when you have the same tire front and rear. You can achieve neutral balance and use evenly your tires grip in a 40/60 weight distribution if your tires present same contact patch distribution. You can take F1 car for exemple with rear weight distribution and wider rear tires. Or lmp1 cars close to 50 weight distribution and same front and rear tires
@thorlax
9 жыл бұрын
Great video, very informative. You got me curious, though, so I have a couple of questions: 1) How does temperature play into all this? Will the sensitivity or the weight/"coefficient of friction curve" change significantly with external temperature? I would think that lower temperatures would cause both less deformation and less propensity for the rubber to stretch, but I'm not sure which "direction" this would take the load sensitivity, if it all. 2) Is this "phenomenon" (a not-constant coefficient of friction) something that effects certain types of tires more than others? I know that different "season" tires (winter, summer, all-weather) can have different tread patterns, different types of rubber, and different rubber thicknesses. I'm wondering if these factors come into play as well.
@giorx5
7 жыл бұрын
Nice video! If you haven't done it yet, try to explain in the same scientific way, how can a very light car wearing a very thin tire (i.e. Caterham) grip much better in latera G's than a massive, heavy car, wearing very wide tires (i.e. Bentley Continental). And I am talking about the same lvl of rubber quality in the tires of both cars. IOW, how much the factors of tire width and car mass is affecting lateral grip?
@benson4u215
5 жыл бұрын
Boils down to pound per square inch, the higher the number the more itll dig. The lower the number the more likely it is to spin
@AcerbicGangrene
6 жыл бұрын
Coefficient of friction lowers for higher speeds of surface slip, not normal force. You can of course plot that it lowers for an increased normal force, but it just an effect of inertia. If you were to drive a weightless vehicle and you would have a normal force acting on your car equivalent to it's weight as a usual vehicle, there wouldn't be any corelation between coefficient of friction and normal force, as it correlates only with speed of slipping between surfaces. Also we have other formula for friction between flat and circular surfaces. Momentum of friction equals normal force times coefficient of friction for circular surface, which is equivalent to the distance between given normal force and furthest point of contact with flat surface in the direction of movement.
@johnjonlee
7 жыл бұрын
Which Physics course teaches as you increase the mass which in turn increases the Normal force results in a decrease of the coefficient of friction (which is determined by the materials in this case the rubber of the tire and asphalt or whatever you're driving on). If you increase the Normal the Friction force increases; assume temperature doesn't affect the coefficient. If you have a 2000 lb car or object and you push it until you break the friction force to move it; the friction makes it so you need to apply X amount of force to push it; now if you doubled the weight (we'll make it easy and make weight the same as mass for now) and you push it til it moves it doesn't take the same amount of force does it? (this would be your result if coefficient of friction all of a sudden reduced); what actually happens is the Friction force now increases in this case doubled... why? because you increased the Normal force. Normal force is F=ma in your diagram a=g (gravity) so F=mg; basically the object has a force going down due to gravity; and therefore there must be an equal force in the opposite direction which is the Normal force. If you increased the Mass (m) then you increased the Normal force. You rationalize your explanation by keeping the Friction force constant; but why is this the case? My example shows that you increase the Normal then Friction force increases. Try pushing a wheeled chair without someone in it and then with someone in it; the force required to push the chair increased with the increase of the mass making it harder to push. You didn't magically decrease the coefficient of friction and require the same amount of force to push the chair in either scenario.
@helixworld
7 жыл бұрын
This video explains one property of rubber that is very relevant. The temperature dependant properties and tire deformation due to loading are probably bigger factors IMO.
@LightWaIker
4 жыл бұрын
On ice and snow, how would this traction argument change? Thanks.
@roadracing3
7 жыл бұрын
Thank You! Excellent!
@klemenschen6503
9 жыл бұрын
Great video! But I am just wondering. If the molecular force inside the rubber decrease by high pressure. Then it will lead to a tear apart of the tire rubber but not to decrease the friction between the road surface and the tire , which is the key force to the μ. So I think the more pressure to the contact path will result in decreasing the molecular force (which is the main part of the Friction) between the Road surface and the Tire at the same time. Then the the F,which is in the video, will decrease and following μ decrease.
@lofro1717
8 жыл бұрын
Cool very interesting
@UgurAkdag
6 жыл бұрын
Expressing in a very simple way, with the same car, considering tire compounds are same, wider tires (means wider than recommended by manufacturer) and narrower ones (recommended by manufacturer) mainly differs at contact patch shape. While narrower tires, lets say "longitudinal contact patch") better on accelerating and braking, wider ones (lateral contact patch) good at cornering and cooling off. Though there are more factors than I mention above (i.e slip angle, surface, temperature, purpose of use etc). For safety reasons, if you are not confident in this issue, stick to the recommended tire sizes.
@SiriusBlack-qo3xi
9 жыл бұрын
But ultimately does the product of Normal force and the coeff. of friction increase or decrease with increase in load?
@yshreya4947
7 жыл бұрын
Is the friction coefficient you are talking of ... the tire-road friction/adhesion coefficient? Because as far as I know, when we accelerate a car.. the load transfers to the rear wheels and the grip on the rear wheels is better. Hence, if we use a RWD, we can accelerate faster because we assume the friction coefficient is going to be the same. In an SAE paper that I read, it was said that the effect of load on "mu" is very very little. When you said load increases, it means the inertia of the body is high which makes it difficult to stop at lesser distance.
@Sam19930420
9 жыл бұрын
it's explaining basic ideas well! but I think you assume every tire is slick lol cause different tires have various shape of grooves.
@whackamolasaurus
8 жыл бұрын
I understand that a heavier or unevenly loaded car will have a reduced friction coefficent, but how does this explain wider tires resulting in more grip? Although contact pressure would be reduced over a larger contact patch, the overall vertical load would remain constant per tire?
@dilysi156
3 жыл бұрын
So if my car is loaded with lots of stuff, I just need to lower the pressure inside to keep the coefficient of friction right? Which is actually totally the opposite of what I've always heard till tonight hehe. Love you videos! Keep going!
@projectilequestion
2 жыл бұрын
I know, tire physics are funny. However if you lower the tire pressure, you increase the contact patch and that can increase grip. But a lower pressure contact patch means the tire will deform and take a higher slip and at the same time. So yes, you should generally pump up your tires when under load.
@rpventure
9 жыл бұрын
I don't think this is the whole picture. You lump "lateral traction" all together, but there is a whole range of possible slip angles. There are cases where the maximum coefficient of friction doesn't occur at the smallest normal load like you would expect.
@motiv-8
4 жыл бұрын
Fantastic. I had to dig deep for this. I was looking for confirmation of my own understanding of how this works. Would I be correct in thinking that the increased pressure would also build heat? If the heat is sufficient it will excite the tire molecules, resulting in their bonds breaking even easier?
@404nobrakes
4 жыл бұрын
I think it’s a bit of give and take. More pressure means less deformation of the tire, and therefore less heat buildup, but it also means a smaller contact patch and therefore an increased risk of friction shearing which leads to excessive wear and reduced grip. You have to find the optimal balance.
@jdc4aub
9 жыл бұрын
Would deflating the tires help the situation at all? Of course the load would be almost the same as air doesn't weigh much and the contact patch should be the same size, but I would think that by releasing some of the air and therefore force on the tire, the coefficient of friction wouldn't decrease as much .
@thegoldenduck456
9 жыл бұрын
Well I race karts, and when we want more rear grip one way of doing it is reducing rear tyre pressures. There's more to it than that but essentially yes you get more grip with less pressure because of the larger surface area.
@alimirza7907
7 жыл бұрын
closed captions in English would be more awesome , enjoying your videos.
@ronroberts8036
Жыл бұрын
Your presentations are excellent! The red line in drawing #2 is flat to declining. Is there a point at which a tire will lose grip due to not enough weight, providing the contact patch us unchanged, (by letting some air out to compensate for N decrease)?
@carnut7521
Жыл бұрын
Tire engineer here - and I'm sorry, but this video is wrong because tires don't behave according to classical friction theory: F= mu*N - because the rubber penetrates the texture of the road surface generating more grip. And that means that wider tires give more grip because more tread surface is in contact with the road.
@gabycapoirizarry
7 жыл бұрын
Well that's pretty cool! Makes a lot of sense indeed. Now, having had finished mechanics of materials just about a month ago at school myself I feel I just need to also ask, does rubber also observe any instantaneous strain hardening at all also before the actual "shearing-off" (a.k.a. failure) that leads to the friction coefficient drop ? I feel that if it actually did, It would also help to account for the reduction, as of course, harder tire rubber compound... less grip. Imagine rubber, if you will, microscopically as porous foam. Wouldn't a compressed porous material just become harder as the tiny air packets reduce in size due to the extra loading by definition?
@benwilms3942
4 жыл бұрын
Wouldn't it have experience local strain hardening just that instant before it shears? Super high viscosity, super low elasticity, plastic behaviour, compound shears off. Anyhow, what's failed to be mentioned is that when the material is made stiffer under load, like 'meta-rigid' because of strain, then the compound can't elasically spill into the road surfaces topographical gullies, it will only lay taught over the topographical peaks, thus significantly reducing your actual contact surface area, and generating huge contact pressures locally at those peaks, exacerbating the effect of compound shear.
@W2L5G
8 жыл бұрын
The size of tire i am using now is 205/55/R16 on stock rims. But according to original spec, it should come with 195/55/R16 on that stock rims, which is 1cm wider. Do you think it is okay to use wider tires on the same rims? 1. Will it affect the handling? 2. Will it affect the safety? 3. Will it affect the tire wear? 4. Will it affect the comfort? 5. Will it affect the NHV?
@CaptainCrunch99
6 жыл бұрын
good stuff !
@RobManser77
7 жыл бұрын
This explains vehicle weight and weight distribution very well, but how does it relate to wider tyres? Wider tyres don't give a car a bigger contact patch, because contact patch area is only proportional to tyre pressure and vehicle weight. Wider tyres just change the shape of the contact patch - it becomes shorter and wider rather than longer and thinner. This means that a greater proportion of the tyre is cooling compared to heating during the tyre's rotation, allowing a softer compound to be used, which gives a higher coefficient of friction. I'm confused as to how the normal force decreases with a wider tyre though, giving a higher mu; Could you explain please?
@danpearson1911
7 жыл бұрын
As far as I understand it: The normal force does not decrease with wider tyres but the contact area does increase. The wider tyre gives a larger area which means a lower pressure can be used since pressure x area = force transmitted through the tyre. The lower pressure means that the coefficient of friction is higher because the rubber is not as stressed (pressure and stress are almost interchangeable).
@RobManser77
7 жыл бұрын
Dan Pearson I always thought that wider tyres do not increase contact patch area, they merely change its shape, as described above.
@danpearson1911
7 жыл бұрын
Rob Manser I think that the wider tyres do not necessarily increase contact patch but they do allow lower pressures to be used. If the tyre pressure is lower then the contact area must increase because the pressure x the area has to be equal to the amount of force being transmitted down through the tyre. In the static perfect weight distribution case this force would be 1/4 the weight of the car.
@RobManser77
7 жыл бұрын
Dan Pearson yes, contact patch area is proportional to vehicle weight and tyre pressure. I've not noticed a decrease in pressure with wider tyres and similar vehicle weight though, but that's just my anecdotal experience, there are probably other factors.
@danpearson1911
7 жыл бұрын
Another way to think of it would be that if the tyre pressure is constant then the contact area must also be (as long as the force transmitted though the tyre is constant too). This would mean that a narrow tyre and a wide tyre at the same pressure would have the same contact patch however the narrow tyre would have a more deflated look while the wider tyre would look more rounded because the area can be short in terms of along the car but long in terms of running parallel to the axel.
@RallyRat
9 жыл бұрын
Could you maybe do another video showing how a tire's contact patch stays roughly the same size as normal force increases?
@AJTalks
11 ай бұрын
Why did the BAC Mono need special compound tires to account for it's weight if the coefficient of grip is highest when the normal force is low? Seems like you could put on a PS4S or something that works at slightly lower temps and have amazing grip and amazing tire longevity.
@ujjawalaggarwal9985
7 жыл бұрын
I am not an expert on the subject but I have done experiments that suggest that the decrease in friction coefficient occurs much much before the fracture shear stress is reached. So I guess that your reason might not be correct. Please enlighten.
@iCeQubeTomato
3 жыл бұрын
Does this apply when the road is wet? Hydroplaning is supposed to be more effective with narrower tires.
@christianlewis7055
9 жыл бұрын
How common does a car actually reach the point where the tire's resistance to strain decreases? I was going to suggest it's occurring every time a tire slips dramatically i.e. drifting, but this coefficient of friction is proportional to the pressure on it's contact patch, so my question is - does a regular road car ever actually produce this much pressure when turning vigorously, for example?
@willk3807
4 жыл бұрын
My question... Starting with my situation... I have a 350z with 255/35/18 front tires and 325/35/18 rear tires. When driving it tends to go left and right on its own. It has like new suspension, it's properly aligned, and the steering system is also like new. Nothing is loose or needing replaced. Now my question.... Could the wide wheels be causing this swaying while driving at 55?
@baselvisions33
7 жыл бұрын
Came here from the 'How Tesla Hit 60 MPH In 2.28 Seconds!' video. I really understood the concept, 100% from this video. It makes sense.
@billtimmons7071
7 жыл бұрын
Point # 5, subsection 4 a., I don't understand why contact area does not increase with weight. 2000 lb car with 500 lbs per wheel @ 36 PSI tire pressure = 13.9 sq inch contact patch. 4000 lb car with 1000 lbs per wheel @ 36 PSI tire pressure = 27.8 sq inch contact patch area. Contact area must go up with increasing weight on tires. There is no increasing pressure on contact patch as vehicle weight increases the contact area simply has to increase. Did I not understand your video and missed something? Contact area must increase.
@mikep2421
7 жыл бұрын
I have some issues with this explanation. Maybe it is over my head? For a given tire pressure, shouldn't contact patch area be proportional to vehicle weight? Shouldn't normal force per unit area be proportional to tire pressure and therefore maximum friction be more affected by tire pressure than vehicle weight? You say adding weight to a tire doesn't much increase the contact patch, but why is this? Is it due to a corresponding increase in tire pressure? What effect does tire volume have on this? If the contact patch does not expand because of the sidewalls carrying a greater proportion of the weight, what effect does the uneven distribution of normal force over the contact patch have?
@Lexoka
7 жыл бұрын
Interesting, but hard to reconcile with the fact that adding downforce to a race car makes it turn better. Is it that they have such good tires that the coefficient of friction remains close enough to constant?
@bangyourhead99x
8 жыл бұрын
Great video! But I have a question. Are there any cons to putting wider tires on your car? Like can you go too far? I am thinking about a widebody kit for my car and some much thicker tires. Will there be any cons to this other than increased rotating mass?
@Al0011235813
Жыл бұрын
Turning circle (for front tires), and handling in the wet (depends) are the main cons.
@prdoyle
7 жыл бұрын
I'm surprised that the contact patch remains the same size. I thought the pressure on the contact patch was roughly equal to the tire pressure. So if you double the car's weight, the tire deforms, causing (1) the tire pressure to increase and (2) the contact patch to enlarge, until a new equilibrium is reached. The pressure is inversely proportional to the volume inside the tire, which I wouldn't expect to change very much as you double the weight of the car, so I would have expected the contact patch to almost double in size.
@upgrtanglancalsmith1782
9 жыл бұрын
So with the polymer chains being stretched out, is this where harder and softer tires fit in (ie. harder tires when there is high downward force (or normal force))? my thinking is that the harder the tire rubber the more pressure needed to reach the point of decreased resistance to frictional shearing.
@codygoodman7909
7 жыл бұрын
So the coefficient of friction for rubber is not exactly constant? And to clarify what you were saying, the frictional force between the tires and the road increases with the increase with normal force by F=mu*N, but at some point the decrease in the coefficient of friction decreasing due to normal force beats that out?
@t500k2l
3 жыл бұрын
Can we expand this question to: why are the racing brake calipers larger? Is it also because of the load sensitivity or something else?
@tonyblighe5696
2 жыл бұрын
Heat dissipation comes into it. Small brake pads with high pressure would get hotter - same heat generated but in a smaller space, so the temperature would be higher. At a high enough temperature the brake pad material will disintegrate or maybe before that it's coef of friction will reduce. At race tracks I've seen brake discs glowing red hot.
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