Scooterotica

piston moves on centa line between big end nlittle end; big end moves off centa line as crank spins; crank spins from bdc rod n pistn go up; crank spins to tdc piston is at top; this next bit shud get y ed round it; rod moves from bdc to 90 degree of crank spin piston goes up big end goes out from centa line; after 90 degrees to 180 degrees big end moves back t centa line; piston moves more or less for each degree cus pin moves to n from centa line,

muppet,

The big end pin must move round the circumference of a circle so there must be a pi involved somewhere

Mmmmmm pi,what kind
My favourites are killie pi and curry pi

like what Muppet said, at BDC, the crank pin is moving sideways (very slightly) so little movement of piston, at 90degs to barrel the crank pin will be moving back to front (almost) giving more distance of piston travel, TDC will be like BDC.

does that help any ? think it's got summat to do with chords, I find 'D-flat major sus' doesn't help at all x

no pies here

no, all that only describes piston linear movement in relation to degrees , moving more mm per degree at mid stroke and 20 0r 30 degree at tdc and bdc where it hardly moves.

what im talking about is specifically, the position of the piston at mid stroke.

If we call the position of the big end at tdc 9 o clock and bdc 3 oclock , mid stroke would be 12 oclock and 6 oclock.
So, with a 60mm stroke, and with the big end at 9 oclock (tdc) , if i rotate the crank 90 degrees, so big end is at 12 oclock the piston should of moved 30mm from tdc, yes ? no, its 33mm ?

I prefer hairy pie

law of co-sine applies

russell you have remembered to put the hour on? think its something to do with linear chords? am sure an adult will come along soon and educate us

I think it's because the big end describes the circumference of a circle which is equivalent to stroke. The rod pulls the piston horizontally in the barrel but the movement of the big end is in relation to the centre of rotation of the crank perpendicular to the linear movement of the piston, it will increase as the big end approaches the top and decrease as it approaches BDC.

I'm sure the angle of the rod must come into play here as well, but it probably needs a picture...

It's also entirely possible I've had a stroke

Carlos aka non believer

all im gona say here is...........


P2+R2-2 RP cos(q) = L2 (Law of Cosines)

So P = R cos(q) + (L2 - R2 sin (q)2) 1/2



simples............ maybe not