Re: OT Back on the Track

In article <j4o7p5$674$1@xxxxxxxxxxxxxxx>,
Dave Lehnen <dclehnen@xxxxxxx> wrote:

Michael Press wrote:
In article
<5560b2f2-1954-4c54-ba4f-7c4ab62bd4c1@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Jay Beattie<jbeattie@xxxxxxxxxxxxxxx> wrote:

On Sep 12, 7:54 am, "dustoyev...@xxxxxxx"<dustoyev...@xxxxxxx> wrote:

How fast do you have to go to stay up on the banks?

I started on the rental bikes at Alkek. Yes, the banking (only 33.3)
was truly a thrill, the first couple of corners (i.e., the first lap)
was a thrill I will never forget<g>!
All Roadie Impact Alert Warning Horns blaring full blast, no doubt
about it.
--D-y- Hide quoted text -

- Show quoted text -

12mph depending on conditions (wet/dry). http://www.flickr.com/photos/bikeportland/142592414/
I was always a crappy track racer, but I raced in San Jose and at
Alpenrose, and the deal with Alpenrose is that it is much more 3D.
You look up, you look down -- not just across.

bank angle: a
turn radius: r
speed: v

We want the speed for which the tangential force at the contact patch is 0.

vv
g.sin a = ---.cos a
r

vv = g.r.tan a

For a 40 deg bank and a 15 meter radius
v = 11.106 m/sec = 40 km/hour.


Alpenrose at 43 deg, 16.6 meter is a 27 mph turn by this method.
How do you get 12 mph? By setting a maximum tangential force at
the contact patch?


This does not answer the question of minimum speed, but gives a speed
where the tires need no appreciable coefficient of friction. Minimum and
maximum speed limits may exist, depending on coefficient of friction,
bank angle, and acceleration of gravity.

for coefficient of friction = u
lateral acceleration = z = v^2/r
bank angle alpha = a
acceleration of gravity = g

minimum speed exists if a v can be found where
u(g cos a + z sin a) = (g sin a - z cos a)

maximum speed exists if a v can be found where
u(g cos a + z sin a) = (z cos a - g sin a)

for the 40 degree bank, 15m radius, g = 9.8 m/s^2, u = 0.75 turn,
minimum speed is about 2.835 m/s or 10.2 km/hr or 6.34 mph.
maximum speed is about 25.1 m/s or 90.37 km/hr or 56.16 mph.

for the Alpenrose turn, 43 degrees, 16.6 m radius, same u & g,
minimum speed is about 4.18 m/s or 15.05 km/hr or 9.35 mph.
maximum speed is about 30.17 m/s or 108.6 km/hr or 67.5 mph.

Thanks.

--
Michael Press
.

Relevant Pages

  • Re: OT Back on the Track
    ... Alpenrose, and the deal with Alpenrose is that it is much more 3D. ... First, I think you mean lateral friction force, not tangential force. ... For a 40 deg bank and a 15 meter radius ... 16.6 meter is a 27 mph turn by this method. ...
    (rec.bicycles.tech)
  • Re: OT Back on the Track
    ... Alpenrose, and the deal with Alpenrose is that it is much more 3D. ... We want the speed for which the tangential force at the contact patch is 0. ... For a 40 deg bank and a 15 meter radius ... 16.6 meter is a 27 mph turn by this method. ...
    (rec.bicycles.tech)
  • Re: A disturbing statistic
    ... they'd have had room to spare even if the crosswind ... along with the bank angle. ... the turn was such that it was a nearly direct tailwind. ... | The reports say his ground speed was 112 mph. ...
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  • Re: Cooper-Harper Handling Qualities Rating Scale
    ... it could both bank AND turn! ... Imagine that! ... Makes me wonder where a Gee Bee rates on the scale though. ... It had a top speed in level flight of 110 mph and the first ...
    (rec.aviation.homebuilt)
  • Re: OT Back on the Track
    ... was truly a thrill, the first couple of corners ... minimum speed is about 2.835 m/s or 10.2 km/hr or 6.34 mph. ... maximum speed is about 25.1 m/s or 90.37 km/hr or 56.16 mph. ...
    (rec.bicycles.tech)