B
Backyard Lumberjack
Guest
Yep, there's a reason it is called a 'flow bench' and not a 'velocity bench'.
OK, let's talk transfers. Back in the days when Timberwolf used to post on AS he put some transducers in the transfers to see what the velocity was and in what direction. When the saw was running at max power there was a slight reversion as the residual exhaust gas pressure in the cylinder would push back down in the transfers. You can see this on your engines by noticing the bit of exhaust carbon near the top of the transfer ports.
Then as the piston continued to drop (and the crankcase compression increased) the flow changed and headed into the cylinder. The maximum flow occurred just before BDC (5-10 degrees), the flow would continue into the cylinder even as the piston was now moving upward (there was still pressure in the crankcase). In the case of maximum power, the flow would STILL BE FLOWING INTO THE CYLINDER as the piston closed off the port.
There is a thing called 'maximum delivery ratio' - this occurs when the transfer flow stops just as the piston closes the transfer ports. For all practical purposes we can call the maximum delivery ratio point - maximum torque.
So, what happens when we drop below maximum torque? The rpm is lower and there is more time for the mixture in the crankcase to get into the cylinder. In fact, it gets in there too soon - with the piston now rising, it starts sucking the mixture BACK OUT OF THE CYLINDER.
Thus, you can see there is only one point where the maximum delivery ratio will be at the maximum. Above that rpm and there isn't enough time/area to get all the mixture into the cylinder - and below that rpm the mixture is getting sucked back out of the cylinder.
When you increase the time/area of the transfers you will raise the point of maximum delivery ratio. In other words, you move the powerband up higher.
Now about raising the transfers on a chainsaw - if it is a quad port, why in the Hell would you do it? This is not a high performance two-stroke with multiple transfers wrapped around the cylinder. The required time/area of transfers on a saw is the LOWEST of all the two-strokes. There is ample room on the side of the cylinder to WIDEN the ports.
So why widen them? - Well, you can either make the transfer wider to flow the mixture, or you can increase the time they stay open - either way you will increase the time/area and can flow the same amount. HOWEVER, if you raise the transfers you are making the engine SMALLER.
A two-stroke only begins to suck in mixture when the transfers are closed. The swept volume of the engine is from the point of TRANSFER CLOSING to TDC. If you raise the transfers you are effectively shortening the STROKE.
interesting post, to say the least! very interesting. I like knowing, and assume it's true since measured and I am firm believer one can believe all he reads on the internet... that mx flow from crankcase due to pressurization/compressed effect of piston coming down... happens just before BDC. at that point the bottom of the piston, at least on my 044, that is the lower area of casting encompassing the bottom ring land is just above the top edge of bottom transfer port. ie, a mill ridge. I had planned on some contouring there to augment piston window reshaping, but now I have a better idea of where the max flow out of crankcase is taking place and just how much the piston's postion in the bore affects that mill ridge height...
*A two-stroke only begins to suck in mixture when the transfers are closed. The swept volume of the engine is from the point of TRANSFER CLOSING to TDC. If you raise the transfers you are effectively shortening the STROKE.
I would like to know more about the sweep volume to better understand it based upon your comment. I understand stroke, etc... and in general sweep volume... actually to include cam timing specs effects on it effectively in a V8... but with my 044 when I put a light in from the exhaust port, it lites up well the cylinder and the piston in it. when I look thru the plug hole I can see inside very well. when I place the piston so that the top ring would have just closed top transfer... the piston skirt has completely still blocked off the intake port. and the bottom of the piston skirt is only 1/16th or so up from flange of cyl base... and when I place the bottom ring land so as to have just sealed the upper cylinder... the upper transfer port is still cracked open, with an OE intake port would be even wider... and when the top ring is above the transfer port, the exhaust port is still open albeit closing... and even once the exh is closed the intake port is not open, although the rings are above top top of intake port and it still is a few more degrees rotation before the bottom of the piston skirt begins to open the intake port.
am I missing something here? seems to me the intake charge is not going to enter the engine until the intake port begins to open. and then as far as intake charge is concerned, all h*ll breaks loose and crankcase begins to fill due to atmospheric pressure augmented by piston signal strength as it rises.
I do not see how a 2-stroke begins to 'suck in mixture' the moment the transfer closes.... since when the transfers r closed the exh is open and the intake closed.
and I would like to fully understand it....especially if I am missing some important event or action.