prop speed
01-08-2008 | 03:52 PM
#2
Platinum Member
Joined: Jun 2001
Posts: 21,346
Likes: 10
Very simply (and neglecting internal drivetrane considerations):
Power In = Power Out
and
Power Out = Thrust Power - power consumed by drag
The ratio of of Thrust power divided by applied power, is efficiency.
Your goal is to maximize thrust and minimize drag, thus maximizing efficiency.
To achieve increased thrust, you can either spin a given prop faster, or grow the size of the prop.
Now, with increases in a prop's angular velocity (rpms) and therefore the relative tangential velocity of a given radial element, you inherit torque drag in a directly proportional relationship to the 2nd power. Conversely, increasing blade area to achieve the same goal only increase torque drag linearly. Therefore, it would lend itself to reason that a larger prop turning slower would yield greater efficiency than a smaller prop turning faster - all with limits, of course.
As an additional note, here's a handy rule of thumb:
Doubling your blade area, doubles your drag.
Doubling your blade velocity, quadruples your drag.
Power In = Power Out
and
Power Out = Thrust Power - power consumed by drag
The ratio of of Thrust power divided by applied power, is efficiency.
Your goal is to maximize thrust and minimize drag, thus maximizing efficiency.
To achieve increased thrust, you can either spin a given prop faster, or grow the size of the prop.
Now, with increases in a prop's angular velocity (rpms) and therefore the relative tangential velocity of a given radial element, you inherit torque drag in a directly proportional relationship to the 2nd power. Conversely, increasing blade area to achieve the same goal only increase torque drag linearly. Therefore, it would lend itself to reason that a larger prop turning slower would yield greater efficiency than a smaller prop turning faster - all with limits, of course.
As an additional note, here's a handy rule of thumb:
Doubling your blade area, doubles your drag.
Doubling your blade velocity, quadruples your drag.
01-08-2008 | 05:15 PM
#4
Charter Member # 55
Joined: Oct 2000
Posts: 20,232
Likes: 2,482
From: Omaha/LOTO
You need to spin it so the engine is at its max recommended WOT range. As for spinning a smaller prop faster and changing gears, that depends on the boat and gear ratios. Some boats like it and some don't.
01-08-2008 | 05:38 PM
#5
Registered
Joined: Aug 2002
Posts: 5,646
Likes: 9
From: Apache Jct. AZ
Very simply (and neglecting internal drivetrane considerations):
Power In = Power Out
and
Power Out = Thrust Power - power consumed by drag
The ratio of of Thrust power divided by applied power, is efficiency.
Your goal is to maximize thrust and minimize drag, thus maximizing efficiency.
To achieve increased thrust, you can either spin a given prop faster, or grow the size of the prop.
Now, with increases in a prop's angular velocity (rpms) and therefore the relative tangential velocity of a given radial element, you inherit torque drag in a directly proportional relationship to the 2nd power. Conversely, increasing blade area to achieve the same goal only increase torque drag linearly. Therefore, it would lend itself to reason that a larger prop turning slower would yield greater efficiency than a smaller prop turning faster - all with limits, of course.
As an additional note, here's a handy rule of thumb:
Doubling your blade area, doubles your drag.
Doubling your blade velocity, quadruples your drag.
Power In = Power Out
and
Power Out = Thrust Power - power consumed by drag
The ratio of of Thrust power divided by applied power, is efficiency.
Your goal is to maximize thrust and minimize drag, thus maximizing efficiency.
To achieve increased thrust, you can either spin a given prop faster, or grow the size of the prop.
Now, with increases in a prop's angular velocity (rpms) and therefore the relative tangential velocity of a given radial element, you inherit torque drag in a directly proportional relationship to the 2nd power. Conversely, increasing blade area to achieve the same goal only increase torque drag linearly. Therefore, it would lend itself to reason that a larger prop turning slower would yield greater efficiency than a smaller prop turning faster - all with limits, of course.
As an additional note, here's a handy rule of thumb:
Doubling your blade area, doubles your drag.
Doubling your blade velocity, quadruples your drag.
01-08-2008 | 06:04 PM
#9
Registered
Joined: Aug 2002
Posts: 5,646
Likes: 9
From: Apache Jct. AZ
As I saw in another thread, "Some new guys are entertaining, some are azz hat's, and then there is Mark"
You're sense of humor is just as phucked as mine.
CigDaze just always floors me with his technical responses, and after I sit and analyze what the hell he said, it begins to make sense.
Jeez - It's like taking a Physics course all over again though.
Sorry for the sidetrack of the thread Gerard - my apologies.
Back to the tech stuff.
