Let's throw in a  ripple into this...
the hypothenuse is really only the point from the radiator (effectively the
front of the bus) to the pivot point (rear axle, not the back of the bus).
You could use the length of the bus if you took all four wheels off and
pivoted on the back bumper.

Sooo... 12' x 12" x .258819=37.27"=3.1'

So, now you could do all the measurements and lift the front end the
differential of bumperheight(original) to bumper height(raised)...
or you could go get a set of ramps and drive the thing up on them and know
it's gonna work.
15 degrees is not the critical part of this operation.

tim o'brien

>Ok, i'll help out there-
>1) how long is the bus (the hypothenuse)?
>
>the height to raise the bus is like this
>Height=(length of bus in inches)*sine(angle)
>
>So for a bus that is 16' long and 15 degrees up...
>16x12xsin(15)=49.6932"
>(or -> 16' x 12" x .258819=49.6"=4.14'
>
>this is 4'!
>lets go backwards- lets rais the bus 30" instead... what angle?
>sin(X)=30'/192"
>sin^-1(30/192)=X
>X=8.98degrees
>-->sin^1 is the inverse sine. on a calculator is is usually shift-sin
>(the funciton that is inverse of sine)
>
>got it?
>cheers
>Matthew
>SOH-CAH-TOA
>
>Matthew Pollard          "Racing with the wind and flirting with death
>Dept. Of Chemistry        So have a cup of coffee and catch your breath"
>University of Idaho
>www.uidaho.edu/~poll7356
>
>
>
>On Sat, 24 Nov 2001, Jere Hawn wrote:
>
> > Hi All,
> >
> > I'm improving my "bleeding the cooling system" flow chart and I am at
>the
> > raise the front by 15 degrees.  Being very weak in trigonometry and
> > spelling... how many inches does the front end need to come up to make
>15
> > degrees.
> >
> > Sorry I don't have the wheel base dimensions.
> >
> > Once I've finish with this one part it will be available for
>distribution
> > for those wishing a copy please p-mail
> >
> > Thanks,
> >
> > Jere
> > 90 GL
> > 88 GL
> >

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