where does rotary power come from?
#26
Originally posted by xiaomingming50
oh yea i was thinking about boxer engines on my drive home from class
for those, if you had a force pushing the piston sideways, and a force holding the piston up horizontally, but no frictional force during its oscillation movement, then the rods would be doing all the centripetal acceleration on the piston
and remember (for those of you who've taken high school physics) centripetal acceleration is perpendicular to the velocity vector of the object it's acting upon, and work is force times displacement times cosine of the angle in between, and since they are at 90 degrees, the net work done is always zero, although the rods will suffer a lot of stress, and break if the stress is too great (where rod/stroke ratios come in)
oh yea i was thinking about boxer engines on my drive home from class
for those, if you had a force pushing the piston sideways, and a force holding the piston up horizontally, but no frictional force during its oscillation movement, then the rods would be doing all the centripetal acceleration on the piston
and remember (for those of you who've taken high school physics) centripetal acceleration is perpendicular to the velocity vector of the object it's acting upon, and work is force times displacement times cosine of the angle in between, and since they are at 90 degrees, the net work done is always zero, although the rods will suffer a lot of stress, and break if the stress is too great (where rod/stroke ratios come in)
power = work/time
#27
Originally posted by xiaomingming50
oh yea i was thinking about boxer engines on my drive home from class
for those, if you had a force pushing the piston sideways, and a force holding the piston up horizontally, but no frictional force during its oscillation movement, then the rods would be doing all the centripetal acceleration on the piston
and remember (for those of you who've taken high school physics) centripetal acceleration is perpendicular to the velocity vector of the object it's acting upon, and work is force times displacement times cosine of the angle in between, and since they are at 90 degrees, the net work done is always zero, although the rods will suffer a lot of stress, and break if the stress is too great (where rod/stroke ratios come in)
oh yea i was thinking about boxer engines on my drive home from class
for those, if you had a force pushing the piston sideways, and a force holding the piston up horizontally, but no frictional force during its oscillation movement, then the rods would be doing all the centripetal acceleration on the piston
and remember (for those of you who've taken high school physics) centripetal acceleration is perpendicular to the velocity vector of the object it's acting upon, and work is force times displacement times cosine of the angle in between, and since they are at 90 degrees, the net work done is always zero, although the rods will suffer a lot of stress, and break if the stress is too great (where rod/stroke ratios come in)
power = work/time
#28
bzzzt! No cigar for you either, rookie7. What you're saying is that the percentage of the revolution (crank/eccentric) during which the combustion products are pushing on the rotor/piston is all-important in determining power output. Why do you think that? What does the duration of the power "pulse", or the percentage of a rev that it occupies, have to do with power output, assuming a constant amount of fuel/air is burned? Nothing.
If anything, I would expect a longer power pulse to allow the combustion products a greater opportunity to lose heat to the engine surfaces, decreasing their pressure and hence, efficiency.
If anything, I would expect a longer power pulse to allow the combustion products a greater opportunity to lose heat to the engine surfaces, decreasing their pressure and hence, efficiency.
#30
Originally posted by Orange!FD
bzzzt! No cigar for you either, rookie7. What you're saying is that the percentage of the revolution (crank/eccentric) during which the combustion products are pushing on the rotor/piston is all-important in determining power output. Why do you think that? What does the duration of the power "pulse", or the percentage of a rev that it occupies, have to do with power output, assuming a constant amount of fuel/air is burned? Nothing.
bzzzt! No cigar for you either, rookie7. What you're saying is that the percentage of the revolution (crank/eccentric) during which the combustion products are pushing on the rotor/piston is all-important in determining power output. Why do you think that? What does the duration of the power "pulse", or the percentage of a rev that it occupies, have to do with power output, assuming a constant amount of fuel/air is burned? Nothing.
On top of that you have the rotary engine power cycle lasting 90 degrees of eccentric/crankshaft longer than that of a piston engine. That means you apply a force over a longer distance which translates into more work produced by the rotary which means more power.
#31
Old [Sch|F]ool
Originally posted by rookie7
That's where you're wrong OrangeFD, it is not "a constant amount of fuel/air", the power cycle of a 1.3L 4 cylinder engine uses the volume of 1 cylinder,that is 1.3L/4 = 325cc while the rotary power cycle uses the volume of 1 chamber which is 1.3L/2 = 654cc. In other words, in 2 revolutions a piston engine will have 4 power cycles of 325cc each cylinder pumping out 1.3L, while a rotary will have 4 power cycles using 654cc for each rotor pumping out 2.6L.
That's where you're wrong OrangeFD, it is not "a constant amount of fuel/air", the power cycle of a 1.3L 4 cylinder engine uses the volume of 1 cylinder,that is 1.3L/4 = 325cc while the rotary power cycle uses the volume of 1 chamber which is 1.3L/2 = 654cc. In other words, in 2 revolutions a piston engine will have 4 power cycles of 325cc each cylinder pumping out 1.3L, while a rotary will have 4 power cycles using 654cc for each rotor pumping out 2.6L.
So far so good...
On top of that you have the rotary engine power cycle lasting 90 degrees of eccentric/crankshaft longer than that of a piston engine. That means you apply a force over a longer distance which translates into more work produced by the rotary which means more power.
It's actually beneficial that the burn is 50% longer, because the chamber is so inefficient it takes forever to propagate. That's why, even with a 50% longer burn, it's STILL burning as it exits the exhaust port.
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