![]() 10/23/2015 at 11:28 • Filed to: None | ![]() | ![]() |
For cool physics stuff. Also, can you get me the best 200 ft skidpad numbers (with sources, i.e, Road and Track, etc), that you can remember for road cars? Off the top of my head, the new Z06 did 1.17 in one of the magazine tests...which is absurd. Can anyone think of a higher one?
![]() 10/23/2015 at 11:32 |
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Like 20 tons?
![]() 10/23/2015 at 11:33 |
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thanks breh
![]() 10/23/2015 at 11:36 |
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I didn’t do the math so probably a bad idea to use 20 tons.
![]() 10/23/2015 at 11:38 |
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lol, I can do the math. I need like an actual measured number as confirmation.
![]() 10/23/2015 at 11:39 |
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at 40mph it is going to be negiliable . it will make more of a difference how big the dude inside is and how much gas is in the tank.
![]() 10/23/2015 at 11:41 |
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Thiiiiiiiiiis much *holds out arms*
![]() 10/23/2015 at 11:42 |
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Actually, at 40, its not negligible. It’s actually enough to affect skidpad numbers.
![]() 10/23/2015 at 11:44 |
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Are you mathematically inclined? I can explain.
![]() 10/23/2015 at 11:44 |
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african or european corvette?
![]() 10/23/2015 at 11:45 |
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Royal Navy.
![]() 10/23/2015 at 11:49 |
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I believe at 40 mph the skidpad difference would be negligible as well and would come down to more of the driver and how skilled they are. I’m mathematically inclined if you care to type it out :)
![]() 10/23/2015 at 11:50 |
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I am not sure if a 2015 Corvette can actually reach 40mph.
I’ll see myself out.
![]() 10/23/2015 at 11:51 |
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for skid pad numbers, 40-50 pounds here or there isnt a lot.
sure, if i’m holding a 6’x6’ at 45degrees, its going to make a difference. a little CF splitter? not as much.
take the viper. from a road and track article the (base) viper makes 75 lbs at 150mph....
the viper ta makes 460 at the same speed. even if we scale it linearly (which it doent happen, because aero) (460lbs/ 150mph for a lb/mph number) then x 40) we end up with 122lbs. thats a passengers worth of weight.
this is a aero chart:
at 130, the drag is 16x what it is at 30.
lets, for the sake of this argument, say the viper makes 460 lbs at 130. at 50mpg, its making 1/16th of that. or just under 30lbs.
30lbs...that like having 2 of my cats next to my in the car! (in other words, fuck all difference)
![]() 10/23/2015 at 11:52 |
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Well, apparently the Z06 makes 350 pounds of downforce at 150 MPH. Assuming downforce is correlated to the square of the velocity, at 40mph, we’re looking at about 25 pounds...for a Z06 with a bunch of wings and shit. A base would be less.
![]() 10/23/2015 at 11:57 |
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and yes, i’m school for mech eng. lets pull out the numbers ladies and gents :D
![]() 10/23/2015 at 12:01 |
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You forgot the mic drop.
![]() 10/23/2015 at 12:04 |
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Sig figs or gtfo.
In all seriousness, I am curious to what extent you’re talking about, as I’m inclined to agree with the sv wrangler here and say yes, it matters, but less than the amount of gas and washed fluid in the car, or how big the driver’s ass is.
![]() 10/23/2015 at 12:05 |
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So basically, the only force keeping the car on the skidpad without sliding is friction, which provides the centripetal force on the car. If the Z51, as per Car and Driver’s test data, does 1.03G on a 300 ft skidpad, that makes the coefficient of friction for the tires (Ff= mu * N) 1.03, which is technically an impossible number. The only way the coefficient of friction could be greater than .99999999 is if the “weight” of the car when moving around that skidpad is actually greater than the weight of the car when standing still, and the difference thereby must be downforce. I’m using Car and Driver’s measured curb weight + 100kg for driver and fuel, both in my normal force calculation and my skidpad friction force calculation.
The point is, the difference in apparent weight between .999 g and 1.03G actually matters on the skidpad. 1.03G on a 300 ft skidpad equates to roughly 87.08 MPH, but the difference in apparent weight (if the tires are have a batshit crazy static friction coefficient of .9999, which they definitely don’t) is a minimum of 11 pounds, but probably way more, since there’s no way the tires’ friction coefficient is actually as high as .9999.
In conclusion, downforce matters on a skidpad. In fact, it’s the only physically possible way to have lateral acceleration numbers above 1.0G.
![]() 10/23/2015 at 12:05 |
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So basically, the only force keeping the car on the skidpad without sliding is friction, which provides the centripetal force on the car. If the Z51, as per Car and Driver’s test data, does 1.03G on a 300 ft skidpad, that makes the coefficient of friction for the tires (Ff= mu * N) 1.03, which is technically an impossible number. The only way the coefficient of friction could be greater than .99999999 is if the “weight” of the car when moving around that skidpad is actually greater than the weight of the car when standing still, and the difference thereby must be downforce. I’m using Car and Driver’s measured curb weight + 100kg for driver and fuel, both in my normal force calculation and my skidpad friction force calculation.
The point is, the difference in apparent weight between .999 g and 1.03G actually matters on the skidpad. 1.03G on a 300 ft skidpad equates to roughly 87.08 MPH, but the difference in apparent weight (if the tires are have a batshit crazy static friction coefficient of .9999, which they definitely don’t) is a minimum of 11 pounds, but probably way more, since there’s no way the tires’ friction coefficient is actually as high as .9999.
In conclusion, downforce matters on a skidpad. In fact, it’s the only physically possible way to have lateral acceleration numbers above 1.0G.
![]() 10/23/2015 at 12:06 |
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I don’t want to make the downforce = v^2 assumption though. Is there any more accurate data for that?
![]() 10/23/2015 at 12:07 |
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Yeah I just copied my reply to you, check it out.
![]() 10/23/2015 at 12:08 |
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So basically, the only force keeping the car on the skidpad without sliding is friction, which provides the centripetal force on the car. If the Z51, as per Car and Driver’s test data, does 1.03G on a 300 ft skidpad, that makes the coefficient of friction for the tires (Ff= mu * N) 1.03, which is technically an impossible number. The only way the coefficient of friction could be greater than .99999999 is if the “weight” of the car when moving around that skidpad is actually greater than the weight of the car when standing still, and the difference thereby must be downforce. I’m using Car and Driver’s measured curb weight + 100kg for driver and fuel, both in my normal force calculation and my skidpad friction force calculation.
The point is, the difference in apparent weight between .999 g and 1.03G actually matters on the skidpad. 1.03G on a 300 ft skidpad equates to roughly 87.08 MPH, but the difference in apparent weight (if the tires are have a batshit crazy static friction coefficient of .9999, which they definitely don’t) is a minimum of 11 pounds, but probably way more, since there’s no way the tires’ friction coefficient is actually as high as .9999.
In conclusion, downforce matters on a skidpad. In fact, it’s the only physically possible way to have lateral acceleration numbers above 1.0G.
Also fuck sig figs. Fuck them in their stupid faces.
![]() 10/23/2015 at 12:15 |
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I agree wholeheartedly with your last sentence. I think we’re leaving out rolling friction, weight transfer to the outer wheels, deformation of the tires, etc. I see where you’re coming from, but I think the discrepancy is more from an oversimplified system. For example, no where in the friction equation is surface area accounted for, but wider tires equals more grip.
![]() 10/23/2015 at 12:30 |
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okay, for some reason i was thinking that we take the skid pad at 40mph. and at 40mph, aero doesnt come into effect
also, you can have friction more then 1. silicon rubber can, for instance
check out this list here:
http://www.engineeringtoolbox.com/friction-coeff…
87 mph is a totally different story
![]() 10/23/2015 at 12:34 |
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that makes the coefficient of friction for the tires (Ff= mu * N) 1.03, which is technically an impossible number.
Pretty sure this isn’t true. It’s been a while since I completed a physics or engineering course, but as far as I’m aware there’s no reason that the coefficient of friction cannot exceed one. It can’t be less than zero, that’s for sure, but I have read about some combinations of materials can have coefficients of static friction well over one.
![]() 10/23/2015 at 12:36 |
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Well I think some testers use a 200 ft skid pad and get similar numbers, which means the velocity is right around 40 mph on those skid pads.
![]() 10/23/2015 at 12:37 |
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hmmmmm
![]() 10/23/2015 at 12:40 |
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i edited the reply, you can have friction greater then one.
with modern car tires, esp high performance rubber on a hot day, i wouldn’t surprise me if that list was slightly out of date.
![]() 10/23/2015 at 12:40 |
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Edit: on further reading, apparently it is possible to record a less-than-zero coefficient of friction, at least according to the researchers in this study: http://physicsworld.com/cws/article/ne…
![]() 10/23/2015 at 12:40 |
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Yes I just figured that out. There's the problem.
![]() 10/23/2015 at 12:41 |
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Yep, you are right. Thanks.
![]() 10/23/2015 at 12:43 |
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Whoa that's interesting
![]() 10/23/2015 at 12:45 |
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A more correct term for real analysis of tires would be adhesion coefficient. Take a look at this table:
Tires can create more lateral force than normal force.
![]() 10/23/2015 at 12:49 |
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when
![]() 10/23/2015 at 13:24 |
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The equation you are using for friction assumes rigid bodies, tires are not rigid bodies. Your coefficient of friction should be coefficient of rolling friction. A small technicality, but it makes a difference: http://www.school-for-champions.com/science/fricti…
![]() 10/23/2015 at 13:28 |
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Since when is the Drag Coefficient an assumption?
Frontal area (A) doesn’t change, fluid density () doesn’t change, “2” is a constant, Co D doesn’t change ‘cause it’s a coefficient, so the only things to change are Velocity and Drag Force.
It’s not an “assumption”, it’s Fluid Dynamics. Grab your fluids book and look it up.
![]() 10/23/2015 at 13:29 |
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Wild stuff for sure. Back when I was in undergrad I remember being taught that a less-than-zero coefficient was impossible. Then someone had to go and prove that wrong. The universe is a crazy place.
![]() 10/23/2015 at 14:04 |
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Larger surface area equates to lower probability of slippage, but has no bearing on actual force applied against anything if I recall my physics classes.
And in the case of wider tires, it also offers a more stable base by distributing the car’s weight out further in either direction.