Re: Tire Changing questions

["Clive Liddell" <cliddell@xxxxxxxxxx>]

Paige,
It's a pity I threw away my calcs when I was trueing up the 20mm shaft that
I use but, in the interests of science here goes:

Assume the shaft has 0.04 mm "runout" at it's midpoint.
Assume wheel and tire around 15kg.

The "assembly" would have a moment of:

15 x 9.8 x 0.04/1000 = 0.00588 Nm

Just to balance the above at 18"/2 or 229 mm (0.229 m):

0.00588/0.229 = 0.0256 N
or a mass of say 0.00261 kg ( ie some 2 to 3 grams).

Now this would either add or subtract from the actual out of balance
measured (or, with luck, be neutral).  Thus a tire needing typically 8 g
could get 11 or 5, and that is for a runout of only 0.04 mm or 1/1000".

My previous statement then, admittedly quoted from memory, was fairly "on
the money" I think...

Regards
Clive Liddell
Pietermaritzburg, South Africa.
'96 R850R   72k.km
'01 R1100RT 45k.km


- ----- Original Message -----
From: <PLPKLT@xxxxxxx>
To: <oilheads@xxxxxxxxx>
Sent: Tuesday, June 22, 2004 3:29 AM
Subject: Re: Tire Changing questions


> In a message dated 6/21/2004 10:24:51 AM Eastern Daylight Time,
cliddell@xxxxxxxxxx writes:
>
> > If you are a only a couple of thou's out then the whole
> > weight of the wheel acting at that tiny distance can easily equate to
around
> > 10 or 20 g - and, of course, you are only un-balancing the
> > wheel :(
>
>
> I'd like to see you provide some calculations on that!   1/1000 of an inch
times 1 gram  = 10 or 20 G ( does this stand of Gravity or something
else....
>
> Jess with such obvious dangerious physics just waiting to jump on our poor
unbalanced wheels , it is a wonder were not all laying somewhere in a
ditch....
>
>
> Paige

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