If you can get even as little as 0.4g out of your tyres (so, uhh, we're talking much worse than just wet), then you can safely go down to 20°. If you have decent tyres, 1g is pretty much a guarantee in warm dry weather (and on good tarmac), so 45° is fine. The yanks' road authorities use 0.9g as a safe assumption in wet weather, which means you can still do nearly 42°. More than you'd comfortably use in daily riding, and way more than you'd get out of most cruisers 😂 It's apparently very slippery when it first starts to rain after it's been dry for a long time (rain lifts all that debris, etc.), so for the first half hour of rain or so, the first example is a good guideline. But if you're smooth enough, you'll get the chance to feel the approaching slide slightly before it happens.
@@233kosta Just one question out of pure interest: where is it common to make these types of calculations with g instead of the CoF? You mentioned something with flight dynamics earlier... I was just wondering because in my engineering environment everybody is mostly talking about the coefficient of friction 😊
@@GiuliaonGasoline It's fairly common in aerospace as a means of nondimensionalising inertial loads. In this case, it's a simplification based on the assumption that the tyres are loaded correctly, there's no gradient change and no downforce at play (because bikes), so μ becomes the acceleration limit in g, per your derivations too. I further assumed that there's no acceleration in progress (neutral power, no braking), which means that all of that grip can go into changing direction. Yes, I did mentiom flight dynamics (if you know 'em - use 'em, right?), that was the simplest path I could think of to solve this problem. The equation I used is for the load factor (in g) an aircraft would experience while turning at bank angle β in level flight. In the aircraft, vertical acceleration remains 1g (from gravity), but as the lift vector (normal to the wings) changes direction, its magnitude must increase in order to maintain a 1g vertical component (cosβ). The horizontal component (sinβ) is the centripetal acceleration that actually turns the aircraft. On a bike it's slightly different. The vertical component comes from solid ground contact and the horizontal comes from tyre grip (our limiting factor). Crucially though, we know the resultant reaction vector *must* go from the contact centroid through the centre of mass, because at peak lean the bike is in a steady state. If the reaction vector were slightly offset, the bike would either lean further or return to upright. This means that the forces in this state resolve exactly the same as in an aircraft in a steady coordinated banking turn, so the same equation applies. Which is good, makes life much simpler 😉 Perhaps one day it'll be fun to derive the full dynamics of turning bikes, but for today I'm happy to keep it simple 😊
@@233kosta Oh that sounds pretty interesting! And i get why it makes perfect sense to use g for these types of calculations, since well you don't really have coefficients of friction I guess? (maybe somehow with the air, but that's something different then?) I honestly don't know a thing about flight dynamics, apart from maybe the things which are the same for ground vehicles. But what you explained with the calculations with g are kind of the same calculations we do, when we want to know something about a vehicle in a banked curve. Then you also have "lateral g" that doesn't come from tyre grip. 🙂 It's always great to find these similarities in different fields of engineering 🥰
You can get about 1.3g out of good tyres in good weather. Assuming you don't do anything silly, nor accelerate, nor brake, we can say all of that grip goes into lateral acceleration. Per Pythagoras, that's a resultant load factor of 1.64g. Per load factor from flight dynamics (yeh, I'm going there! 😁), n=1/cos(β). Thus β=acos(1/n). In the case of 1.3g tyres, it's about 52.45°. Highest recorded in MotoGP is 68°, which is a load factor of 2.67g, suggesting a lateral tyre load of 2.475g. Either that dude was sliding hard, or there would have been a lot of downforce at play.
Good calculation, thanks! In vehicle dynamics you can always go a level deeper - as you did right now 😊 - and then you'll get a more specific answer. For not making everybody here too much hope I assumed simply 1g for average conditions, then everybody's on the safe side 😊
@@GiuliaonGasoline Yeh, I get your point, 1.3g is never a guarantee, especially when you have a rider who isn't very smooth. I quite like this particular simplification because it translates to bikes so well, but with other ground vehicles it doesn't apply, so the bigger equations you had up are more appropriate. I ought to mention, the 68° figure is a number I plucked out of the internet (onestopracing), I've made no effort to confirm the veracity of that claim.
Good looks, now I can use these formulas when I'm out on the road about to hit a turn 🤣Good video, it reminds me of Engineering Explained. Also I like that SLS poster you have in the background 💪
There was tan in the equation and then the final result is 45 degrees with a coefficient of 1 Would you break it down a bit for those who left school long ago? (Some of them may even be high) How did you find 45 ? Thank you
That's a good question! I calculated the 45° by using arctan(x), which is the inverse tangent function. arctan(1)=45° Here the 1 is the coefficient of friction. If you'd calculate this in your calculator it's possible you get 0.7854 instead of 45. It depends on whether your calculator is set on rad or degrees, but you can easily convert it by multiplying 0.7854 with 180/π. Hope this helped:) If anything's unclear yet let me know. I'm thinking a lot about how "technical deep" my level of explanation should be because I want to make it understandable for as many people as possible. And maybe some people don't even wanna know these small details like you 😇 but I'll try not to leave out information like this in the next videos!
@@GiuliaonGasolineI got lost at some point when you said Fc because I haven’t done math in years, it takes some time to go back to talking “math”. Instead I suggest you use the full name like centrifugal force, gravitation force etc.. I know it takes longer than using math terms but it’s a way to break things down without reducing the complexity. You could talk about the most complex topic using simple explanations and reach any audience. A smile helps thinking “it’s easy and fun” , so good job on that !
@@googletropcurieux8670 that's also a good idea, took a note for the next videos 😊 And right, it's always better to explain it like this, using simpler words, and maybe some people already know some things then, but this won't hurt them :)
@@GiuliaonGasoline It was at moto gymkhana, I was going 10km/h or something, didn't even get a proper bruise out of it 😒 That day the crash bars earned their keep, though 🤣
Sorry but this is terrible. You are saying the maximum lean angle is 45 degrees? Have you ever ridden a motorcycle? Ever WATCHED riders going through corners on a racetrack or mountain road? Modern hypersport tyres are letting bikes lean 55*+ very easily. Even 60* with alot of the race specific compounds, like the diablo slick 200/65, R11's, michelin power slick, gpa-pros etc etc.
Where exactly did I say that without mentioning that the 45° are ONLY the case for basic assumptions like a coefficient of friction of 1 and a simple model of a motorcycle (for example without taking into account the width of the tires or the center of gravity)? This video is supposed to explain the basics to people who don't work in this field and just want to learn more about their motorcycle. Oh and coefficients of friction between 0,8-1 are quite normal with typical tires on a typical road. I just did measurements at work to justify this two weeks ago...
@@littlereptilian7580 yup but on a normal street with usual conditions it's usually around 0,8-1 I actually made measurements last week 😊 Then depending on the tires characteristics, the roughness and texture of the road and also the temperature...
