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RE: heavy crosswinds



Where:
Vmotorcycle=110 ft/sec
Mass=more than any M/C should actually weigh
Wind Vmax=Like God Himself puckered up to blow out a hundred birthday
candles.
DROPvert= "See that little tiny speck down there? That is a white van."

There is a certain bridge on Rt 1 up the CA coast just north of Monterey
upon which it is possible to have a full-dress Venture physically lifted and
deposited several feet to the right.

When that wind hit I slammed the bars right. As we started to lean it felt
like Dorothy leaving Kansas.  All I can figure is that at enough lean, the
side of the bike became an airfoil.  It wasn't just the front wheel being
pushed... the whole enchilada came off the ground.

We went back and saw the "landing" chirp marks.  About 16 inches from the
outer edge of the bridge!

So when your CG _is_ your roll center... how much effort does it take to
land the damn thing!  While traveling in a straight line of course....
Ker-unch!  8O

> -----Original Message-----
> From: owner-st@xxxxxxxxxx [mailto:owner-st@xxxxxxxxxx]On Behalf Of Jim
> Huber
> Sent: Monday, May 01, 2000 9:17 AM
> To: ST@xxxxxxxxxxxxxx
> Subject: re: heavy crosswinds
>
>
> >  From: Eric Sheley <eric@xxxxxxxxxx>, on 05/01/2000 09:58:
> >  At 12:21 AM 5/1/00 -0700, don draper wrote:
> >  >  The amount of effort required to correct for a crosswind is DIRECTLY
> >  >proportional to the height of the CG above the roll center of the
> >  >bike.
> >
> >  Huh ? We weren't talking about the amount of effort required.
> >
> >  The original question was not why it was required to lean in,
> but rather
> >  why in addition to leaning into the force there was also a
> certain amount
> >  of force that was required, directly opposite to the lean
> angle (in this
> >  case in the direction of the wind).
> >
> >  Let me see if I can break it down a little simpler. I will
> forgo the vector
> >  mechanics discussions as I don't think that will make the scenario any
> >  clearer.
> >
> >  These are some broad generalizations:
> >
> >  1 - A bike travelling in a straight line will continue to do
> so unless some
> >  external force is applied (wind, input from the bars, throttle etc)
> >
> >  2 - A bike travelling at a lean angle will travel in the
> direction of that
> >  lean (again very basic - we are not concerned with cg here)
> >
> >  3 - If the lean angle is increased the radius of the turn will
> be decreased
> >
> >  Of course in addition to the angle, the radius can be changed
> by tire size,
> >  tire profile, wheelbase etc.... we are not concerned with these factors
> here.
> >
> >  So we now have our bike travelling in a straight line. We now
> add a force W
> >  (wind) from the right, perpendicular to the direction of
> travel (things get
> >  really fun when the angle is <> 90). So in order to compensate for a
> >  quickly shifting motorcycle the rider begins to "lean into"
> our force W.
> >
> >  Now - abruptly remove force W. What does the bike do ? It immediately
> >  begins to turn to the right. Now add the force W again. That
> same tendency
> >  to turn is still present - only now, force W is acting against
> it. However,
> >  if the combination of the lean angle and the profile of the
> tire at the
> >  contact point are such that a turn in the direction of travel is still
> >  being caused (even at a minute level), some force must be
> introduced into
> >  the equation to compensate (to cancel out this tendency to
> turn). In this
> >  case a constant pressure to the left would be required - which
> is why even
> >  though leaned to the right, Patti had to exert a constant
> pressure on her
> >  left bar to maintain straight travel.
> >
> >  There - clear as ...... mud ?
> >
> >  Respectively submitted
> >
> >  - Eric
>
> 	Yep, clear 'nuff. And I agree with you, too. Pilots call it
> "cross-control". It's the same thing you do when landing a plane in a
> crosswind: bank into the wind but use the rudder to counteract
> the actual turn.
>
> 	"Jedi Dawg"  Jim Huber
> 	Spring, Texas, USA
> 	'00 blue ST  "Rhiannon"
>
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