Let me try and back this up with theory
I apologise in advance if I make a mistake and this will be very rough
Say we are traveling along in a Jinty at 20mph ( speed is Irrelevant ) and we shutoff steam. We then place the reverser in notch 1. The valve events in that notch are as follows approximately ( forward stroke ) Cutoff 22%, release 65% ( Backward stroke ) Compression 65% and lead steam at approx 3%. Now the following will not happen if steam is not reapplied, this is because without steam pressure slide valves will float off their seats so any a vacuum cannot occur, nor can compression since the cylinder valves would be constantly open so negating pretty much all the effect except at high speeds.
Lets look at the left hand front side of the cylinder first.
Backward strokeInitial at the start of the backward stroke the steam inlet would be open, steam will enter into the cylinders, but given the speed were traveling at a breath of steam means the steam chest pressure is still roughly zero ( atmospheric). At 22% of its backward stroke the port to the steam chest will be closed at this point pressure will be approximately atmospheric or 15psi ( ish ). Since there is no air entering from the steam chest as the it expands thereby maintaining the pressure as the piston continues to move back this causes the pressure on that side of the cylinder to reduce ( a vacuum ). At 66% the exhaust port will open but since the there is a partial vacuum vacuum air will actually flow into the cylinder through the blast pipe! that destroys the vacuum and negates the force on that side of the cylinder. The Cylinder will then remain at atmospheric for the remainder of the stroke.
Calculations:
Volume at 22% = 1450 cu inches, pressure = roughly 15psia
Volume at 66% = 4360 cu inches, pressure = ?
P1V1 = P2V2
so 1450 * 15 = 4360 * P2
P2 = 5psia
So at the moment before the exhaust port opens the pressure on the front side of the cylinder will be 5 psi absolute in other words a vacuum. Although this is rather negligible force in the grand scheme of things so we will effectively call this negligible.
Forward StrokeOn the return stroke the exhaust port remains open until 24% travel( relative to backward stroke ) when compression begins. At this point no ports are opened so any air on that side of the cylinder will begin to be compressed. At 5% the lead steam occurs, at this point no doubt the compressed air pressure will be greater than that of the steam chest so the air will flow out into the steam chest until pressure is equalised on both sides and the process then repeats.
Calculations
Volume at 24% = 1580 cu inches
Volume at 3% = 200 cu inchs
Pressure at P1 = 15psi
P1V1 = P2V2
1580*15 = 330*P2
P2 = 118Psia
Now that is a much bigger force! This one will actually have some effect since it is equivalent force to 103 psi of steam acting on that side of the cylinder.
At this point I should remind you there are two sides to the cylinders and 2 cylinders one being offset at 90 degrees. These events occur every 90 degrees, so the vacuum during some part of the travel will negate that of the compression so I'd estimate that you'd have about 80psi of compression at the end of each stroke. Granted at the end of each stroke the crank is now straight forward so there would be no force, but just before with that amount of pressure you'd definitely have some resistive force.
The graph of pressure against revolution I think would look something akin to this
http://www.atomic-album.co.uk/showPic.p ... ession.jpgBut at higher cutoff say full gear the compression point is 10% and the release at 3% ( with respect to backward stroke ), as such the pressure on the cylinder caused by compression is pretty much negligible hence why it is usual to coast slide valve engines at full cutoff.
It's probably pretty wrong but it's my best stab at it and someone may be able to improve upon it
anyway enough of me!
regards
Edward