## Max velocity in banked corners

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Ok, I know that I shouldnt use info from Wiki with big confidence in it´s corectness but that was the only place with such a formula I could find so I used it anyway

It´s a formula used to determine the maximum velocity for a car in a banked turn and the problem I encountered with it was that it showed higher velocity on smaller radius turns than at ones with bigger radius

So i wonder if someone can verify that it is correct, or if not post the one that works, here is the link btw, http://en.wikipedia.org/wiki/Banked_turn and the formula I refer to is the one under

banked turns with friction > "solving for v we get..."

Thanks a lot guys
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tomislavp4
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Joined: 16 Jun 2006
Location: Sweden & The Republic of Macedonia

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Not sure how you are reading the equation but the maximum V is a function of sqrt of r. Increase the radius r and the potential maximum velocity increases.
The only difference between the friction and non friction cases is the inclusion of the "slide down off the banking" component.
All well and good, but Nascar has long proven that the maximum speed on a banked corner is entirely (and I do mean entirely) a function of the racing rules currently in effect at the time.
Personal motto... "Were it not for the bad.... I would have no luck at all."
Ian P.
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Joined: 8 Sep 2006

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tomislavp4 wrote:Ok, I know that I shouldnt use info from Wiki with big confidence in it´s corectness but that was the only place with such a formula I could find so I used it anyway

It´s a formula used to determine the maximum velocity for a car in a banked turn and the problem I encountered with it was that it showed higher velocity on smaller radius turns than at ones with bigger radius

So i wonder if someone can verify that it is correct, or if not post the one that works, here is the link btw, http://en.wikipedia.org/wiki/Banked_turn and the formula I refer to is the one under

banked turns with friction > "solving for v we get..."

Thanks a lot guys

This one?

Shows V is proportional to sqrt(r). As radius goes up, velocity goes up. Not sure where you saw V proportional to 1/sqrt(r) which would mean small radius = high speed.
Grip is a four letter word.

2 is the new #1.
Jersey Tom
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Joined: 29 May 2006
Location: Huntersville, NC

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Thank you for the reply, I really appreciate it, here is what i did: I used 100 meter as the radius, 1 as the coefficient of friction and 50 degree as the banking angle...

V = √ 100 * 9,8 (sin50 + 1 + cos50) / cos50 – 1 * sin50
V = √ 980 * 1,409 / -0,123 = 1380 / -0,123 = -11219
V = √ 11219 = 105,92m/s

What am I doing wrong here?
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tomislavp4
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Joined: 16 Jun 2006
Location: Sweden & The Republic of Macedonia

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Well for one, that equation may be a total load of crap. Things should not be going negative. But 50 degrees is a little ridiculous. No track in the world that I know of has 50 degree banking.

Use something like 15 degrees banking instead, and try a 100m vs 250m corner radius and see how the speeds change. As radius increases, speed will as well with that equation.
Grip is a four letter word.

2 is the new #1.
Jersey Tom
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Joined: 29 May 2006
Location: Huntersville, NC

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Damn I really should´ve checked my calculations twice I did them again and this time they´re ok, 105 m/s for 100m radius and 149 m/s for 200m radius, must´ve missed something

Sory for wasting your time guys, but thanks alot for trying to help
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tomislavp4
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Joined: 16 Jun 2006
Location: Sweden & The Republic of Macedonia

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The problem lies with the denominator of the equation and the fact that it lies inside a square root:
$Cos(\theta) -\mu Sin(\theta)$
The numerator will always be positive for $0\leq \theta \leq 90 deg$ and more than 90deg is nonsensical. So if the numerator goes negative, we're trying to take the square root of a negative numbers - so unless we're on an imaginary track something is wrong.
Indeed if the numerator is zero, then we have an even bigger problem....
So $Cos(\theta) -\mu Sin(\theta) \> 0$
Skipping the algebra and trig lesson,
$\theta < arctan (\frac{1}{\mu})$

So for Tomislav's assumed $\mu$=1, as we approach an angle of 45 deg, we hit a divide by zero error. Any angle of bak above 45 deg gives a square root of a negative number problem....

Physically, it's easier to consider a static situation than a dynamic one- for $\mu$=1,45deg is the critical angle at which friction just balances the component of weight down the slope. If the angle increases beyond 45 deg the object will slide down the slope. So if you want static balance on a slope greater than 45 deg, you need $\mu$>1.

The dynamic case yields the interesting result that to corner on a banked slope steeper than 45 deg, the friction coefficient must be less than 1.... and now the math part of my brain has had enough.
ReubenG
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Joined: 21 Apr 2004

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CART drivers were doing 235 mph and 5 sustained g's at Texas Motor Speedway (24 degrees banking, I think) a number of years ago...Drivers were flirting with g-loc so the race was cancelled!
Best regards. I guess this explains why I'm not at my post!
tk421
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Joined: 12 Jan 2009

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Yeah, I read that too, the funny thing is that the example above, the one with 105m/s @ 100m radius means 11G´s!!!

But as ReubenG said the whole equation becomes useless if the banking angle is larger than 45 degrees, thanks for pointing that one out Reuben I´m not much of a maths/physics guy myself...
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tomislavp4
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Joined: 16 Jun 2006
Location: Sweden & The Republic of Macedonia

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When lateral accelleration obviously is v^2/R, I think a 200 meter radius at 100 m/s could be more realistic numbers.

Seem to remember that IMS is some 250 meter radiuses (with 9 degree banking), which at 360 km/h would mean 4g?
"Bernoulli is a nine-letter name"
xpensive
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Joined: 22 Nov 2008