Bore

2kuik4u

New Member
Nov 23, 2008
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I just bored out my head to a 67.5. Will it make me go any quicker?? And does it increase the cc's??
 
Hello
Let’s do the math for displacement and make it simple
Displacement=Bore squared times the stroke times .7854
Move the decibel place over by one to convert mm to cc
So here we go
6.75*6.75*5.7*.7854=203.97cc
Will the bike running faster=no
Why; you just dropped your port heights down
 
well thats a weird way of doin it... volume of (any) cylinder v=(pi)(r^2)(h)
thats v=(3.1415926535.....)((67.5/2)^2)(57)=203972.8 mm^3
to get that in cc's (cm^3) divide by 1,000 b/c there are 1,000 cubic millimeters in one cubic centimeter or multiply by (1cm/10mm)^3

so you get 203.9728 cc's.

i guess the weird way works too haha
 
Hello
Let’s do the math for displacement and make it simple
Displacement=Bore squared times the stroke times .7854
Move the decibel place over by one to convert mm to cc
So here we go
6.75*6.75*5.7*.7854=203.97cc
Will the bike running faster=no
Why; you just dropped your port heights down

iight so i did the math and with my 72.5 over bore on my 240 kit it came out to 235.3cm^3 so is that right? what gets me is when i put 60 in to do the math for a +3 stroker crank, comes out to 247.7cm3 so that seems like a huge jump. am i not getting the math right? and were does that .7854 come from?
 
the ports come up to the cylinder at an angle, so theoretically if you bore it you move the face of the cylinder out and therefore farther down the path of the transfer ports. but Angus, on a sleeve, don't the ports intersect the face of the cylinder at a 90 degree angle? however, if you did a 240 kit, you would be removing some of the material from the jug, and that would lower the ports.

yeah, those are the same numbers i'm getting 89custom. the jump makes sense tho, if you look at the difference in stroke between the 2 it is 3mm. use that as the stroke. you get (pi)((72.5/2)^2)(3)= 12.384 cc's, which is the same as the difference, 247.7-235.3=12.4 (the first one rounds to 12.4)