|
Re: Scrub path length calculation?
[Re: polyspheric]
#1126187
12/03/11 04:31 PM
12/03/11 04:31 PM
|
Joined: Oct 2006
Posts: 527 alaska,usa
9secondsatellite
mopar
|
mopar
Joined: Oct 2006
Posts: 527
alaska,usa
|
Quote:
I've gotten a few questions about this, and noticed some commentary regarding "how large should this be?" (in addition to where it stops & starts). Some very large numbers have been reported with no shock & awe. In general, the scrub should be smaller than you would think, meaning some of those big numbers are really danger signals - the geometry isn't what you want. If you can't find another way, reducing the scrub length is a good direction to go. All you need to predict what the minimum scrub should be is: 1. set the valve-side lever to mid-lift, yatta yatta 2. what's the effective length of the long lever (shaft center to roller axle or pad radius) 3. what's the net valve lift?
Let "S" = the minimum scrub length Let "R" = rocker's long lever effective length Let "L" = net valve lift
The formula is: S=R-(R^2-(L/2)^2)^.5
Example 1: R (lever) = 1.75", L (lift) = .600" S=1.75-(1.75^2-(.600/2)^2)^.5 S=1.75-(3.0625-(.09)^.5 S=1.75-1.724 S=.026"
Example 2: R (lever) = 1.50", L (lift) = .700" S=1.50-(1.50^2-(.700/2)^2)^.5 S=1.50-(2.25-.1225)^.5 S=1.50-1.459 S=.041"
As you can see, scrub figures like .080" show something isn't working...
You can also calculate the rocker angle at zero lift and full lift (which will be the same with mid-lift, duh!). The total arc (1/2 up, 1/2 down) is: 2×arcsin(L/2R)
Example 1 (R (lever) = 1.75", L (lift) = .600"): L/2R=.600÷3.50=.171 sin=.171 Total arc = 19.74°
thanks for the post. good info.
|
|
|
|
|
|