Drum dia and torque

Bob Johnson wrote:
Hi Bill --

Today we towed in light wind and so turned 2300 engine RPM in the climb. The
454 c.i. torque/hp curves show the engine was generating about 430 lb-ft
torque and 200 hp at these revs.

Now here is where I tend to go off the rails. Just 75 hp is required to lift
a 1100 lb sailplane 1700 ft in 45 sec.

I know there are some aerodynamic and mechanical losses but it's hard to
believe they amount to some 125 hp.

The 200 hp rating is at full throttle - is that what you were doing? If
it is using an automatic transmission with an unlocked torque converter,
that could account for a lot of horse power. There is also the drag and
weight of the wire. I suspect wire drag is substantial when it's 2000'
long, but not enough to account for the 125 horses. The glider wings are
working at 2 G or so, which doubles the drag. Start adding these up, and
there are a lot of potential losses. Also, your engine may not be
running at it's rating.


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Eric Greenwell
Washington State
USA

"Bob Johnson" wrote in message
news:XtUnd.87541$%x.66152@okepread04...
Hi Bill --

Today we towed in light wind and so turned 2300 engine RPM in the climb.
The
454 c.i. torque/hp curves show the engine was generating about 430 lb-ft
torque and 200 hp at these revs.

Were you at wide open throttle? Those curves are for WOT and say nothing
about part throttle operation. They are also for a bare engine on a
dynomometer corrected to sea level at standard atmosphere.

Now here is where I tend to go off the rails. Just 75 hp is required to
lift
a 1100 lb sailplane 1700 ft in 45 sec.

I know there are some aerodynamic and mechanical losses but it's hard to
believe they amount to some 125 hp.

The horsepower at the winch drum is easy. It's the cable speed in FPS times
the tension in pounds divided by the constant 550. (or if you have drum
torque data, RPM x torque in ft lbs. divided by 5252) Drum torque is tension
times effective drum radius. Say, 1100 x 1.5 or 1650 ft. lbs.

Let me guess that the cable speed is 90 FPS at the fastest point and the
tension equals the weight of the glider or 1100 pounds. That computes to
180 HP at the drum for just that moment when the cable speed peaks. Since
power = speed x force, the power will drop off quickly as the cable speed
drops even as the cable tension remains constant.

Your figure of 75HP is the power required to lift the glider. Still more
power is required to accelerate it to flying speed. The power delivered to
the glider increases until the peak cable speed is reached and then
decreases quickly.

The new German winches seem to have turbo diesels in excess of 350HP with
some as high as 800HP. The old Opel Diplomat V8 (Think Chevy 350) powered
winches are on the used market cheap. They must know something. Maybe it's
launching those water logged ASH-25's.

Bill Daniels

On Sat, 20 Nov 2004 22:05:06 -0600, "Bob Johnson"
wrote:

Hi Bill --

Today we towed in light wind and so turned 2300 engine RPM in the climb. The
454 c.i. torque/hp curves show the engine was generating about 430 lb-ft
torque and 200 hp at these revs.

Now here is where I tend to go off the rails. Just 75 hp is required to lift
a 1100 lb sailplane 1700 ft in 45 sec.

I know there are some aerodynamic and mechanical losses but it's hard to
believe they amount to some 125 hp.

"The 454 c.i. torque/hp curves show the engine was generating about
430 lb-ft torque and 200 hp at these revs." ... your fallacy is that
"was generating" should read "was capable of generating". The engine
could have been running 2300 with no load (as it was right after the
hook was released)

Looking at the power required to rotate the drum, we were running 400 rpm,
6.28 ft/rev, 45 sec to release and estimating 1000 lb line pull. I calculate
that mix to equal 102 hp.

!02 hp minus 75 hp is 27 hp. Seems to be a fair if maybe somewhat high
approximation of aerodynamic drag of glider and Spectra tow rope.

The throttle was opened to the point where the pilot reported his airspeed
to be in the desired 55-60 kt.range and which corresponded to the 2300 rpm I
was seeing on my gauge.

Again, I'm finding it hard to explain most of the discrepancy between the
altitude-corrected
180 hp
454 c.i. dynamometer data and what I'm seeing with my own eyes. Can a water
pump, an alternator (the radiator is cooled by electric fan) and mechanical
friction
eat up 180 engine hp minus102 drum hp= 78hp?

Thanks for your input but still off the rails,

Bob Johnson




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"Bob Johnson" wrote in message
news:XtUnd.87541$%x.66152@okepread04...
Hi Bill --

Today we towed in light wind and so turned 2300 engine RPM in the climb.
The 454 c.i. torque/hp curves show the engine was generating about 430
lb-ft torque and 200 hp at these revs.

Now here is where I tend to go off the rails. Just 75 hp is required to
lift a 1100 lb sailplane 1700 ft in 45 sec.


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"Bob Johnson" wrote in message
news:XtUnd.87541$%x.66152@okepread04...
Hi Bill --

Today we towed in light wind and so turned 2300 engine RPM in the climb.
The 454 c.i. torque/hp curves show the engine was generating about 430
lb-ft torque and 200 hp at these revs.

Now here is where I tend to go off the rails. Just 75 hp is required to
lift a 1100 lb sailplane 1700 ft in 45 sec.

I know there are some aerodynamic and mechanical losses but it's hard to
believe they amount to some 125 hp.

Bob


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"Bill Daniels" wrote in message
newsxAnd.71122$5K2.10295@attbi_s03...

"F.L. Whiteley" wrote in message
...

"Jim Vincent" wrote in message
...
The drum diameter has nothing to do with the torque!!!

Sure it does. The rope has a certain amount of tension on it, usually
measured
in lbs. The rope is pulled off the drum at a certain distance from
the
center
or rotation. That distance is the moment arm. The torque is the
tension
X
moment arm, hence inch lbs or ft lbs.


Jim Vincent
N483SZ
illspam

And, in practice, constantly changing, generally increasing, depending
on
design and layup.

Frank Whiteley


The following is the summation of a couple of decades of thinking about
winch launch.

Winch drum torque is a complicated subject. It involves glider behavior,
engine torque and power curves and the winding characteristics of the
winch
drum. Drum torque cannot be described without understanding all the
other
variables.

The glider acts to demand both cable tension and cable speed. (Cable
Speed x
Tension = Power) The winch engine, controlled by the winch driver, tries
to
meet that demand while holding the glider airspeed at a value requested
by
the pilot.

Note: The winch driver can control either glider airspeed or cable
tension
but not both.

The torque on the drum shaft varies with demand and is limited by the
engines Wide Open Throttle (WOT) torque curve. (And, of course, the
breaking
strength of the weak link.)

If the winch is unable to meet cable tension demand, the glider airspeed
will decay with increasing pitch attitude. If the winch meets or exceeds
the demand, the glider airspeed will increase when the nose is raised.

If the gliders airspeed decays with increasing pitch angle, then the
glider
is rapidly approaching the stall AOA since the wing loading is also
increasing with pitch attitude. If the airspeed increases with
increasing
pitch, then the AOA will remain more nearly constant. The later is a
safer
condition.

The actual radius of the drum depends on the quantity of cable wound onto
the drum at any moment. If the instantaneous radius is one foot then the
torque in foot/pounds equals cable tension in pounds. This is a typical
mid-launch condition.

If the cable tension is to remain equal to the gliders gross weight
throughout the launch, which is desirable, then the torque at the drum
shaft
must increase with the increasing drum radius even as the drum RPM is
reduced, to maintain a constant glider airspeed.

This places heavy demands on the winch engine. Engines capable of very
high
torque at low RPM are desirable. Diesel engines typically have their
torque
peak just above idle. If the highest engine RPM is 2100 RPM then the
engine
torque capacity will increase even as the drum RPM decreases. In other
words, diesel WOT torque curves tend to match the demand of winch launch.
This explains why diesels are popular winch engines.

Spark ignition engines tend to have torque peaks closer to the max RPM
utilized by the winch. As the launch progresses, the torque capacity
declines rapidly with RPM even as the demand increases.

Sorry for the lecture and apologies to our metric friends. Now, lets
build
some winches.

Bill Daniels

Bad error, I used torque arm instead of diameter for the drum. We are
actually doing 12.7 ft/rev, 460 rpm, 1000 lb estimated line pull, and 45 sec
to release, making the power to rotate the drum 236 hp.

The 236 hp at the drum is compared to the 75 hp required to raise the 1100
lb glider 1700 ft in 45 sec. The difference must be due to aerodynamic drag
losses. We'll need an aerodynamicist to check that out.

The altitude-derated 180 hp provided by the 454 c.i. engine is compared to
236 drum hp. Either the rope pull was overestimated, or those 454 engines
are way stronger than we thought.

If we estimate the rope pull at 686 lb, the drum power drops to 162 hp which
equates to estimated mechanical losses through the drive train at 90
percent, which seems about right.

So the aerodynamic drag losses must be in the order of 162 minus 75, or 87
hp.

Nicht wehr?

Bob Johnson

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"Bob Johnson" wrote in message
news:jGcod.88080$%x.33211@okepread04...
Looking at the power required to rotate the drum, we were running 400 rpm,
6.28 ft/rev, 45 sec to release and estimating 1000 lb line pull. I
calculate
that mix to equal 102 hp.

!02 hp minus 75 hp is 27 hp. Seems to be a fair if maybe somewhat high
approximation of aerodynamic drag of glider and Spectra tow rope.

The throttle was opened to the point where the pilot reported his airspeed
to be in the desired 55-60 kt.range and which corresponded to the 2300 rpm
I
was seeing on my gauge.

Again, I'm finding it hard to explain most of the discrepancy between the
altitude-corrected
180 hp
454 c.i. dynamometer data and what I'm seeing with my own eyes. Can a
water
pump, an alternator (the radiator is cooled by electric fan) and
mechanical
friction
eat up 180 engine hp minus102 drum hp= 78hp?

Thanks for your input but still off the rails,

Bob Johnson




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"Bob Johnson" wrote in message
news:XtUnd.87541$%x.66152@okepread04...
Hi Bill --

Today we towed in light wind and so turned 2300 engine RPM in the climb.
The 454 c.i. torque/hp curves show the engine was generating about 430
lb-ft torque and 200 hp at these revs.

Now here is where I tend to go off the rails. Just 75 hp is required to
lift a 1100 lb sailplane 1700 ft in 45 sec.


--


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"Bob Johnson" wrote in message
news:XtUnd.87541$%x.66152@okepread04...
Hi Bill --

Today we towed in light wind and so turned 2300 engine RPM in the climb.
The 454 c.i. torque/hp curves show the engine was generating about 430
lb-ft torque and 200 hp at these revs.

Now here is where I tend to go off the rails. Just 75 hp is required to
lift a 1100 lb sailplane 1700 ft in 45 sec.

I know there are some aerodynamic and mechanical losses but it's hard to
believe they amount to some 125 hp.

Bob


--


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"Bill Daniels" wrote in message
newsxAnd.71122$5K2.10295@attbi_s03...

"F.L. Whiteley" wrote in message
...

"Jim Vincent" wrote in message
...
The drum diameter has nothing to do with the torque!!!

Sure it does. The rope has a certain amount of tension on it,
usually
measured
in lbs. The rope is pulled off the drum at a certain distance from
the
center
or rotation. That distance is the moment arm. The torque is the
tension
X
moment arm, hence inch lbs or ft lbs.


Jim Vincent
N483SZ
illspam

And, in practice, constantly changing, generally increasing, depending
on
design and layup.

Frank Whiteley


The following is the summation of a couple of decades of thinking about
winch launch.

Winch drum torque is a complicated subject. It involves glider
behavior,
engine torque and power curves and the winding characteristics of the
winch
drum. Drum torque cannot be described without understanding all the
other
variables.

The glider acts to demand both cable tension and cable speed. (Cable
Speed x
Tension = Power) The winch engine, controlled by the winch driver,
tries
to
meet that demand while holding the glider airspeed at a value requested
by
the pilot.

Note: The winch driver can control either glider airspeed or cable
tension
but not both.

The torque on the drum shaft varies with demand and is limited by the
engines Wide Open Throttle (WOT) torque curve. (And, of course, the
breaking
strength of the weak link.)

If the winch is unable to meet cable tension demand, the glider
airspeed
will decay with increasing pitch attitude. If the winch meets or
exceeds
the demand, the glider airspeed will increase when the nose is raised.

If the gliders airspeed decays with increasing pitch angle, then the
glider
is rapidly approaching the stall AOA since the wing loading is also
increasing with pitch attitude. If the airspeed increases with
increasing
pitch, then the AOA will remain more nearly constant. The later is a
safer
condition.

The actual radius of the drum depends on the quantity of cable wound
onto
the drum at any moment. If the instantaneous radius is one foot then
the
torque in foot/pounds equals cable tension in pounds. This is a
typical
mid-launch condition.

If the cable tension is to remain equal to the gliders gross weight
throughout the launch, which is desirable, then the torque at the drum
shaft
must increase with the increasing drum radius even as the drum RPM is
reduced, to maintain a constant glider airspeed.

This places heavy demands on the winch engine. Engines capable of very
high
torque at low RPM are desirable. Diesel engines typically have their
torque
peak just above idle. If the highest engine RPM is 2100 RPM then the
engine
torque capacity will increase even as the drum RPM decreases. In other
words, diesel WOT torque curves tend to match the demand of winch
launch.
This explains why diesels are popular winch engines.

Spark ignition engines tend to have torque peaks closer to the max RPM
utilized by the winch. As the launch progresses, the torque capacity
declines rapidly with RPM even as the demand increases.

Sorry for the lecture and apologies to our metric friends. Now, lets
build
some winches.

Bill Daniels

"Bob Johnson" wrote in message
news:cTpod.1733$3I.752@okepread01...
Bad error, I used torque arm instead of diameter for the drum. We are
actually doing 12.7 ft/rev, 460 rpm, 1000 lb estimated line pull, and 45
sec
to release, making the power to rotate the drum 236 hp.

The 236 hp at the drum is compared to the 75 hp required to raise the 1100
lb glider 1700 ft in 45 sec. The difference must be due to aerodynamic
drag
losses. We'll need an aerodynamicist to check that out.

The altitude-derated 180 hp provided by the 454 c.i. engine is compared to
236 drum hp. Either the rope pull was overestimated, or those 454 engines
are way stronger than we thought.

If we estimate the rope pull at 686 lb, the drum power drops to 162 hp
which
equates to estimated mechanical losses through the drive train at 90
percent, which seems about right.

So the aerodynamic drag losses must be in the order of 162 minus 75, or 87
hp.

Nicht wehr?

Bob Johnson


What your data says is that winch launch is a very lossy process
mechanically and aerodynamically. That's true.

However, since a 40 second launch only consumes about a quart of regular,
doubling or tripling the efficiency isn't going to decrease costs much.

The solution for the winch designer is to overcome these inefficiencies with
sheer power. That's why you see 350 to 800 HP winches in Europe. Most
winch fuel is consumed while the engine is idling. Diesels idle very
efficiently.

Bill Daniels