General Rotary Tech Support Use this forum for tech questions not specific to a certain model year

where does rotary power come from?

Thread Tools
 
Search this Thread
 
Old 03-07-02, 11:19 PM
  #26  
Junior Member

 
rookie7's Avatar
 
Join Date: May 2001
Posts: 38
Likes: 0
Received 0 Likes on 0 Posts
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)
by the way, there is no centripetal force acting on a piston. the piston has linear motion not a circular motion. and of course there is work produced by a piston, where do you think horsepower comes from??
power = work/time
Old 03-07-02, 11:25 PM
  #27  
Junior Member

 
rookie7's Avatar
 
Join Date: May 2001
Posts: 38
Likes: 0
Received 0 Likes on 0 Posts
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)
by the way, there is no centripetal force acting on a piston. the piston has linear motion not a circular motion. and of course there is work produced by a piston, where do you think horsepower comes from??
power = work/time
Old 03-07-02, 11:26 PM
  #28  
Full Member

 
Orange!FD's Avatar
 
Join Date: Mar 2001
Location: St. Louis
Posts: 100
Likes: 0
Received 1 Like on 1 Post
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.
Old 03-08-02, 02:55 AM
  #29  
Old [Sch|F]ool

 
peejay's Avatar
 
Join Date: May 2001
Location: Cleveland, Ohio, USA
Posts: 12,506
Received 416 Likes on 296 Posts
The longer power pulse is why the engines are so smooth, though. A longer, softer push.
Old 03-08-02, 08:39 PM
  #30  
Junior Member

 
rookie7's Avatar
 
Join Date: May 2001
Posts: 38
Likes: 0
Received 0 Likes on 0 Posts
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.
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.

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.
Old 03-09-02, 03:50 AM
  #31  
Old [Sch|F]ool

 
peejay's Avatar
 
Join Date: May 2001
Location: Cleveland, Ohio, USA
Posts: 12,506
Received 416 Likes on 296 Posts
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.


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.
This is where the snag is. Yes the power pulse lasts 50% longer, however that doesn't mean 50% more work done. Remember, there is only so much air and fuel in the chamber, and there is only so much power that can be exctracted from it. It works out to be roughly the same amount of "push" (power), just spread out over a longer time. I say roughly, because in fact less power is actually harnessed, due to various thermal problems mentioned previously.

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.
Related Topics
Thread
Thread Starter
Forum
Replies
Last Post
ItsJBohmzB
Build Threads
171
04-24-17 01:11 PM



Quick Reply: where does rotary power come from?



All times are GMT -5. The time now is 09:18 AM.