Where r is the offset from crank centreline. A is the angle and l is the length of the rod.
Let's say I want to know the distance the piston has moved from TDC to when crank is 10 degrees BTDC.
Crank throw is 77.6mm, conrod length is 125.0 mm.
Well?
From wikipedia:
Where r is the offset from crank centreline. A is the angle and l is the length of the rod.
Or try this programme
http://www.myvirtualnetwork.com/mklotz/files/crod.zip
Adrian
So to use that formula you would have to calculate r by using the crank displacement and the degrees rotated?
so = 13.48mm
[Edited on 6/8/09 by tegwin]
See the original article here...
http://en.wikipedia.org/wiki/Piston_motion_equations
r is the distance of the centre of rotation of the bottom of the rod from the crank centreline.
[Edited on 6/8/09 by SeaBass]
Can't get that to run properly sounds exactly what you need though.
quote:
Originally posted by AdrianH
Or try this programme
http://www.myvirtualnetwork.com/mklotz/files/crod.zip
Adrian
quote:
Originally posted by SeaBass
See the original article here...
http://en.wikipedia.org/wiki/Piston_motion_equations
r is the distance of the centre of rotation of the bottom of the rod from the crank centreline.
[Edited on 6/8/09 by SeaBass]
Don't know what he means by throw @ 77.6 mm ? If thats the crank radius or diameter?
r would either be 77.6 or 38.8?
Or does he mean the piston throw is 77.6mm??
[Edited on 6/8/09 by SeaBass]
Im confused... I give up!
My assumption for r is incorrect!
[Edited on 6/8/09 by tegwin]
great, thanks.
i worked it out as 0.8 mm; wiki link formula is 0.77 mm so good enough-ish.
quote:
Originally posted by SeaBass
Don't know what he means by throw @ 77.6 mm ?
r (crank radius) is half the stroke, simple.
quote:
Originally posted by 02GF74
great, thanks.
i worked it out as 0.8 mm; wiki link formula is 0.77 mm so good enough-ish.
Used to do that sort of thing in Higer Teccy
Using Autocad I get 1.89mm
[img]c:2009-08-06_131408.jpg[/img]
[Edited on 6/808/09 by mad4x4]
I work it out to be 0.7712mm from using the sin rule.
0.7712mm here too from excel assuming 77.6 is the stroke (which sounds typical for a car engine). Here's the whole mess (with a few more
brackets than are probably needed just to be safe (a bit like this post 
))...
((77.6/2)+125)-(((77.6/2)*COS(10*(PI()/180)))+SQRT(125^2-((77.6/2)^2*SIN(10*(PI()/180))^2)))
Mike - would this equation also apply to the conrod in your avatar?
quote:
Originally posted by Liam
Mike - would this equation also apply to the conrod in your avatar?
