Sep 272013
 

Dover is a fascinating track – twenty-four degrees of banking, but only a mile in length.carturning  A student approached me with a question:  Higher-banked tracks generate higher centripetal forces – so why doesn’t the track banking appear in the equation for centripetal force?

I’ve talked about centripetal forces in detail before, but let’s have a quick reminder.   I tie a string to a tennis ball and swing it at constant speed in a horizontal circle around my head.  Assuming I am coordinated enough not to hit myself in the head with the tennis ball (don’t laugh – it happens), the reason the ball goes in a circle is because of the string.

The string exerts a force on the ball.   At every moment, the ball tries go straight and the string forces it to turn.  The direction of the force the string exerts on the ball is along the string – which means it is always toward the center of the circle.  Centripetal literally means “center seeking”.

Just like the tennis ball, a car needs a force to make it turn.  The amount of force depends on the mass (weight) of the car, the speed of the car and the turn radius of the track.

EQ_CentripetalForceWords

As my student pointed out, the degree of banking doesn’t even show up in this equation.  All this equation tells you is that you need more force:

  • to take a tighter turn
  • to turn a heavier car
  • to turn faster

The numbers are interesting, in part because they are so big!  Every track in the NASCAR circuit has different parameters, so we have to do a different calculation for every track.  Here’s the numbers for Dover

BSPEED_TN_Dover_EnglishUnits

Teachers:  You can download metric and English unit version of this figure, along with some fun science facts about Miles the Monster.

The Numbers

Using a typical weight for a Gen-6 car (3300 lbs of car and 180 lbs of driver), we can figure out how much force it takes to make a car turn.  (Disclaimer: Parts of this table are from my previous blog.  But I did double check them and added this week’s track – Dover.)

Track Turn radius
(ft)
Speed
(mph)
Turning Force
(lbs)
G’s
Talladega 1000 180 6,848 1.97
200 8,456 2.43
Richmond 365 180 20,636 5.93
100 6,370 1.83
Bristol 242 100 9,606 2.76
Dover 500 130 7,858 2.26

We can see from this:

  • Even though Talladega is a higher speed track, the g-forces are comparable to smaller tracks – because Talladega has wide, sweeping turns.
  • Speed in the turns is limited at a small track like Richmond – the turns are very tight and you’d need a) an unrealistic amount of turning force and b) the g’s would be so high that the drivers would be likely to pass out.
  • Look at how much a 2o mph increase in speed increases the g’s experienced at the track.

“Gs”

The ‘G’ is quite possibly the most misunderstood unit in racing.  A ‘G’ measures acceleration, not force.   We use ‘G’ because the unit is equal to the acceleration of any object due to Earth’s gravity.

You are standing (or sitting) and the ground (or your chair) is exerting a force upward equal to your weight.  As a result, you do not accelerate up or down.  If the chair were to spontaneously disappear, you would accelerate toward the ground.

We use the unit ‘G’ just like a unit like ‘dozen’.  I can express anything in terms of dozens:  a dozen eggs, a dozen jellybeans or a dozen beers.  Likewise, we can use the unit ‘G’ to express the acceleration of anything.  I can measure the acceleration when you step on the gas after stopping at a red light in ‘G’s.   I can measure the acceleration you feel on a rollercoaster in Gs.

Just for reference, most amusement park rides top out at about 3G; however, some roller coasters go up to 4G (SheiKra Rollercoaster at Tampa) or 4.5G (e.g. the Titan Rollercoaster in Texas).

Although Earth’s gravity pulls down (toward the center of the Earth), I can use the unit ‘G’ to measure acceleration in any direction:  up or down, back or forth, or sideways.  Drag racers experience accelerations of about 5G backward at take off.  When you’re turning at constant speed, the acceleration is sideways (which engineers call ‘lateral’).

G-Force

Please don’t use the term “G-force”.   It’s wrong because a ‘G’ is a unit of acceleration, not force.  When you experience ’3Gs’ of acceleration, the force you experience is the number of G’s times your weight.

Compare Danica Patrick (who weighs about 100 lbs) with Ryan Newman  who (according to Yahoo! Sports) weighs 207 lbs.  They both experience 3 ‘G’ in a turn.  That means Danica experiences a force of 300 lbs, while Ryan experiences a force of 621 lbs.  (Their cars, which weigh the same, experience the same force.)

If they experience such different forces, why do we use ‘G’s?  Because the ‘G’-value doesn’t depends on who or what is accelerating. It’s the same number.  You may remember that force = mass x acceleration.  If you take the mass out of the equation above, you have the formula for centripetal acceleration.

But What About the Banking?

As I noted above, the equation for centripetal force doesn’t address banking.  Banking doesn’t directly enter into the equation;  it’s just mass, speed and turn radius.

BUT:  A banked track lets you go faster around the turns.  If Dover were flat, you would still experience 2.26 ‘G’ at 130 mph – but you wouldn’t be able to go 130 mph in the first place.  The greater the banking, the higher speeds around the corners and thus greater G-values; however, a high-banked track with very wide corners could conceivably have lower ‘G’-values than a relatively flat track with really, really tight turns.  It – like so much of racing – is a tradeoff.

Related Posts:

Why Concrete Races Differently than Asphalt

Why Turning is Hard

 

  3 Responses to “Turning, G-Forces and Banked Tracks.”

  1. For a banked turn,. the tangent of the angle of banking is equal to the square of the speed divided by the radius times g…for the case where there is no sideways loading on the vehicle. For an aircraft in a level turn, this equates to requiring a 2g pull to stay level at 60 degrees of bank. For cars, banking the turn will add the speed for a turn with no sideways g; after that, the friction of the tyres on the track comes into play as for a normal unbanked track. It’s a matter of resolving the forces…the force due to the banking acts at 90 degrees to the banking surface, and that of the tyres along the track surface. This gives a resultant higher than each individual force acting at an angle between the two. Every banking will have a speed at which you can drive around “hands off”…no sideways force required from the vehicle. Some highways, built for fast traffic but now with lower speed limits,actually require drivers to steer UP the banking to stay in their lane!

  2. To me, the banking is equivalent to the string on the tennis ball. On a flat turn, tire friction acting parallel to the track surface is the entire string. Banking requires less friction to act as a string because part of the string is made by the track pushing against the tires normal to the track. Thus, as long as the cars can withstand the track compressing it through the vertical axis it can go faster on the turns before braking loose due to static friction becoming kinetic friction

  3. You always amaze me with your logic..

Leave a Reply

%d bloggers like this: