About 2 yrs ago I found a Rover Wayfarer with a 4 stroke Briggs engine on it. The compression was hopeless( 25 pSi), so I put it away for another day). Well the time has come, and I have stripped it apart, taken out the piston, push rods, carby, flywheel, etc etc. My assumption is that the poor compression is due to the rings being worn out, or the valves not sealing well enough. I ordered new rings for it ( not the $90 Briggs ones), but the cheaper copy ones from China ($30) The photo below shows both the old rings in silver and new rings in black. The question is why are they significantly different in circumference? The rings are aligned in the top part of the image(beginning at the same spot), but by the time one passes around to the other end there is a difference of very close to 4 mm. I get that the wear would be where it rubs on the barrel wall, but why the wear at the ends?
Last edited by twostroker; 27/10/2106:13 PM. Reason: Grammar modifications
Plus it would hurt to have the engine numbers and the PN of the rings as some engine have multiple ring sets based on the date code. Then you still got to be careful of ordering after market parts.
And I do understand about the skyrocketing Briggs prices as they just raised them again this week at Power Distributors. This where I get my Briggs price file from for my auto manager system. They are also apparently having a hard time making parts available as several items I need are on long term back order. They may be stuck off the coast of California on a ship.
Comparing old rings to new ones is meaningless outside the engine. the only way to confirm if rings fit is to put them in the bore.
The old ones will be worn and will have lost heat treatment over the years of running. Similarly the new rings have not been heat treated to fit the new bore until it has run in so you are comparing an apple with a banana basically.
Also rings don't really wear at the gap, the gap increases as metal is worn off the circumference, and the ring gap grows as the spring opens further.