# matt17

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Cornering Apprentice

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## Previous Fields

• Have you attended a California Superbike School school?
Yes
1. ## Fast Turns Vs. Slow Turns - Where To Make Time?

+1 We aren't considering that really high-speed turns may not be limited by lat acceleration, think e.g. of Turn 1 at Laguna Seca.
2. ## Fast Turns Vs. Slow Turns - Where To Make Time?

Matt, I should mention that I made a typing error in my example above, having stated the dimensions for the second turn were the same as the first. They are not and are corrected in that post. I followed your math, but don't see that it supports your quote above, as stated. Taken literally, we could simplify the example to remove the turn and say we're riding the same 1,000 feet of straight road. If we make a pair of passes, one at 50 and the other at 51 mph, the difference in the time it takes to cover the 1,000 feet is .27 seconds. If we make a pair of passes at 100 and 101 mph, the elapsed time difference is .07 seconds. For your statement to be true, the distance covered must be proportionally larger. Your math allows for this, but the opening statement doesn't seem to. That statement is only true in the context of the situation you presented, a constant radius turn at a defined speed and lat acceleration. These two values determine the radius and length of the turn. Of course, on a straight road the distance is the same for different speeds, thus the difference in elapsed time varies inversely to the speed.
3. ## Fast Turns Vs. Slow Turns - Where To Make Time?

The time difference for a small increment in speed is independent of the initial speed. Let v = initial speed, A = max. lat acceleration, theta = turn angle (90deg in your example) The radius for the given speed and lat acceleration is r = v^2 / A. The length of the circ segment L = (pi) (theta / 180) r = (pi) (theta / 180) (v^2) / A The time for the segment t = L / v = (pi) (theta / 180) (v) / A Take the derivative, dt / dv = (pi) (theta / 180) / A For a small change in speed dv, the change in time dt = (pi) (theta / 180) dv / A. Example (Turn 1): theta = 90 deg A = 0.8g = 7.848 m/s^2 dv = 1 mi/h = 0.447 m/s dt = (3.14159) (90 / 180 ) (0.447 m/s) / ( 7.848 m /s^2 ) = 0.0895 s

It happened for me while taking level 3... it had started to rain and I got distracted by that and a passing rider (first mistake) and misjudged the entry speed for a slow corner (second mistake) causing a scary feeling rear slide... which worked out OK and actually looked rather cool on video after. The only things I remember doing are keeping the throttle steady and relaxing my arms.
5. ## Which Tyre?

I heard Will mention to another student last week that the school currently is using GPA's instead of Q2's because Dunlop apparently can't produce and distribute Q2's fast enough... Interesting... Perhaps we'll get some real-world comparisons of the Q2 v. GP-A, e.g. wear rate, qualitative impressions from students, etc.
6. ## Which Tyre?

Dunlop Q2s on the student bikes; Coach bikes, IIRC, race take-offs.
7. ## New Guy

Superbike School student (age: 75) review
8. ## Know The Track

This is good advice regarding sighting laps; I would think some homework reviewing track maps, pictures or videos might help too. The track entry and exit is another thing to consider, to anticipate other riders entering and exiting. This is more troublesome at certain tracks where riders enter on higher speed sections.
9. ## Why Are We Weighting The Outside Peg?

Is this a pic from your recent track day following CSS L1? For simplicity let's assume the knee and elbow sliders are just hovering above the surface. As a consequence of Newton's Third Law, it is not possible for the rider to exert a net lateral force on the bike as there is no counter force. That is, nothing to press against. The rider's weight (vertical force of gravity, downward) is countered by an equal force of the bike pressing against the rider vertically (upward). There is a net torque which results from the offset of the point of application of the vertical forces.
10. ## Why Are We Weighting The Outside Peg?

I agree with the technique of one steering input and then relax through the corner.
11. ## Why Are We Weighting The Outside Peg?

If the bike and rider form a rigid body (rider is locked-on the bike) then the center of mass and inertial moment are independent of the attachment point. Now if the rider lays down on the tank instead of sitting upright this will get the rider's mass closer to the bike's center of mass. With the rider locked-on, the net forces on the bike are determined only by the rider's relative position w.r.t. the bike. A lateral displacement of the rider to the inside does have a a slight leaning effect, which is the equivelent of a streering torque of +1..4 Nm. Since most bikes have a small negative steering torque at typical lean angles and path curvatures, an optimal hang-off position will completely balance the steering torque, resulting in neutral steering. IMHO the only use of peg weighting is for pivot steering.
12. ## How Do You Use Your Rear Brake ?

I only use the rear brake at low (parking lot) speeds. I'm still perfecting the basic cornering skills, so rear brake is more a distraction at this point.
13. ## Cornering Style: Corner Speed Vs Square Off

... permits exact application of throttle control rule #1 ... (from memory, don't have my TOTW with me at the moment)

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