REFLECTOR:NG strut angelf
Alexander Balic
reflector@tvbf.org
Tue, 17 Jun 2003 18:37:00 -0500
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Al,
I think that the reason why the center of the front wheel of the
bicycle/motorcycle is in front of the steering pivot point is so the cg is
behind the wheel, ( and so it resists throwing you over the handlebars every
time you apply the front brake) and also so you can fit a wheel in front of
the frame, the wheels on a bike are not stable by themselves- push a bike
with no rider slowly and you will see this, but when the speed increases,
the wheels stabilize due to gyroscopic action. If you pull a bike backwards,
then you will see that the front wheel (now the trailing caster) is more
stable.
I think that it is reasonable to say that the centrifugal and gyroscopic
forces acting on the wheel at certain speed might cause it to destabilize,
but I think you might need some sort of dynamics modeling program to figure
it out......
-----Original Message-----
From: reflector-admin@tvbf.org [mailto:reflector-admin@tvbf.org]On Behalf Of
Al Gietzen
Sent: Tuesday, June 17, 2003 9:39 AM
To: reflector@tvbf.org
Subject: RE: REFLECTOR:NG strut angelf
Subject: REFLECTOR:NG strut angelf
Hey, just looking at it one would think that the strut angle should be
negative, that is the lower end of the strut is REARWARD (NOT REWARD!) of
the upper end. But, take a look at a bicycle, a motorcycle front end.
There's more to this problem than meets the causal observer's eye.
nolan.
Nolan;
Take another look at the bike - the axle is forward of the pivot
centerline; therefore the vertical loading on the wheel produces a restoring
force; i.e., trys to turn the wheel back to straight. The castoring wheel
on the Velocity the axle is behind the pivot. A positive castor angle on
the pivot means the vertical load on the wheel produces a force away from
straight ahead - unstable condition.
Seat of the pants engineering (sope) tells me that the castor angle should
be neutral or negative; but I'll admit the gyroscopic forces are a factor
that makes the issue more complex. Like would you expect that leaning into a
left turn on a motorcycle you pull on the right handle bar?
That same sope tells me that a round profile tire could change the
dynamics considerable by changing the timing and magnitude of the restoring
force due to the forward rolling of the wheel; I think for the better.
Still waiting for someone to try that. I've got enough "firsts" on my
airplane, I don't want to add another.
Al
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<DIV><SPAN class=005132923-17062003><FONT face=Arial color=#0000ff
size=2>Al,</FONT></SPAN></DIV>
<DIV><SPAN class=005132923-17062003><FONT face=Arial color=#0000ff size=2>I
think that the reason why the center of the front wheel of the
bicycle/motorcycle is in front of the steering pivot point is so the cg is
behind the wheel, ( and so it resists throwing you over the handlebars
every time you apply the front brake) and also so you can fit a wheel in
front of the frame, the wheels on a bike are not stable by themselves- push a
bike with no rider slowly and you will see this, but when the speed increases,
the wheels stabilize due to gyroscopic action. If you pull a bike
backwards, then you will see that the front wheel (now the trailing caster) is
more stable.</FONT></SPAN></DIV>
<DIV><SPAN class=005132923-17062003><FONT face=Arial color=#0000ff size=2>I
think that it is reasonable to say that the centrifugal and gyroscopic forces
acting on the wheel at certain speed might cause it to destabilize, but I think
you might need some sort of dynamics modeling program to figure it
out......</FONT></SPAN></DIV>
<DIV><SPAN class=005132923-17062003></SPAN><FONT face=Tahoma><FONT size=2><SPAN
class=005132923-17062003><FONT face=Arial
color=#0000ff> </FONT></SPAN></FONT></FONT></DIV>
<DIV><FONT face=Tahoma><FONT size=2><SPAN
class=005132923-17062003></SPAN></FONT></FONT> </DIV>
<DIV><FONT face=Tahoma><FONT size=2><SPAN
class=005132923-17062003> </SPAN>-----Original Message-----<BR><B>From:</B>
reflector-admin@tvbf.org [mailto:reflector-admin@tvbf.org]<B>On Behalf Of </B>Al
Gietzen<BR><B>Sent:</B> Tuesday, June 17, 2003 9:39 AM<BR><B>To:</B>
reflector@tvbf.org<BR><B>Subject:</B> RE: REFLECTOR:NG strut
angelf<BR><BR></DIV></FONT></FONT>
<BLOCKQUOTE dir=ltr style="MARGIN-RIGHT: 0px">
<DIV class=Section1>
<P class=MsoNormal style="MARGIN-LEFT: 0.5in"><B><FONT face=Tahoma
size=2><SPAN
style="FONT-WEIGHT: bold; FONT-SIZE: 10pt; FONT-FAMILY: Tahoma">Subject:</SPAN></FONT></B><FONT
face=Tahoma size=2><SPAN style="FONT-SIZE: 10pt; FONT-FAMILY: Tahoma">
REFLECTOR:NG strut angelf</SPAN></FONT></P>
<P class=MsoNormal style="MARGIN-LEFT: 0.5in"><FONT face="Times New Roman"
size=3><SPAN style="FONT-SIZE: 12pt"></SPAN></FONT> </P>
<DIV>
<P class=MsoNormal style="MARGIN-LEFT: 0.5in"><FONT face=Arial size=2><SPAN
style="FONT-SIZE: 10pt; FONT-FAMILY: Arial">Hey, just looking at it one
would think that the strut angle should be negative, that is the lower end of
the strut is REARWARD (NOT REWARD!) of the upper end. But, take a look
at a bicycle, a motorcycle front end. There's more to this problem than
meets the causal observer's eye.</SPAN></FONT></P></DIV>
<DIV>
<P class=MsoNormal style="MARGIN-LEFT: 0.5in"><FONT face=Arial size=2><SPAN
style="FONT-SIZE: 10pt; FONT-FAMILY: Arial">nolan.</SPAN></FONT></P></DIV>
<DIV>
<P class=MsoNormal style="MARGIN-LEFT: 0.5in"><FONT face="Times New Roman"
size=3><SPAN style="FONT-SIZE: 12pt"></SPAN></FONT> </P>
<P class=MsoNormal><FONT face=Verdana color=blue size=2><SPAN
style="FONT-SIZE: 11pt; COLOR: blue; FONT-FAMILY: Verdana">Nolan;</SPAN></FONT></P>
<P class=MsoNormal><FONT face=Verdana color=blue size=2><SPAN
style="FONT-SIZE: 11pt; COLOR: blue; FONT-FAMILY: Verdana"></SPAN></FONT> </P>
<P class=MsoNormal><FONT face=Verdana color=blue size=2><SPAN
style="FONT-SIZE: 11pt; COLOR: blue; FONT-FAMILY: Verdana">Take another look
at the bike – the axle is forward of the pivot centerline; therefore the
vertical loading on the wheel produces a restoring force; i.e., trys to turn
the wheel back to straight. The castoring wheel on the Velocity the axle
is behind the pivot. A positive castor angle on the pivot means the
vertical load on the wheel produces a force away from straight ahead –
unstable condition.</SPAN></FONT></P>
<P class=MsoNormal><FONT face=Verdana color=blue size=2><SPAN
style="FONT-SIZE: 11pt; COLOR: blue; FONT-FAMILY: Verdana"></SPAN></FONT> </P>
<P class=MsoNormal><FONT face=Verdana color=blue size=2><SPAN
style="FONT-SIZE: 11pt; COLOR: blue; FONT-FAMILY: Verdana">Seat of the pants
engineering (sope) tells me that the castor angle should be neutral or
negative; but I’ll admit the gyroscopic forces are a factor that makes the
issue more complex. Like would you expect that leaning into a left turn on a
motorcycle you pull on the right handle bar?</SPAN></FONT></P>
<P class=MsoNormal><FONT face=Verdana color=blue size=2><SPAN
style="FONT-SIZE: 11pt; COLOR: blue; FONT-FAMILY: Verdana"></SPAN></FONT> </P>
<P class=MsoNormal><FONT face=Verdana color=blue size=2><SPAN
style="FONT-SIZE: 11pt; COLOR: blue; FONT-FAMILY: Verdana">That same sope
tells me that a round profile tire could change the dynamics considerable by
changing the timing and magnitude of the restoring force due to the forward
rolling of the wheel; I think for the better. Still waiting for someone
to try that. I’ve got enough “firsts” on my airplane, I don’t want to
add another.</SPAN></FONT></P>
<P class=MsoNormal><FONT face=Verdana color=blue size=2><SPAN
style="FONT-SIZE: 11pt; COLOR: blue; FONT-FAMILY: Verdana"></SPAN></FONT> </P>
<P class=MsoNormal><FONT face=Verdana color=blue size=2><SPAN
style="FONT-SIZE: 11pt; COLOR: blue; FONT-FAMILY: Verdana">Al</SPAN></FONT></P></DIV></DIV></BLOCKQUOTE></BODY></HTML>
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