Drum dia and torque

[Bob Johnson]

Reducing the drum size has more effect on torque than I realized. My
spreadsheet shows 953 lb-ft for 48", 242 lb-ft for 36" and only 85 lb-ft for
24".

What do you think?

Bob

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[Eric Greenwell]

Bob Johnson wrote:

> Reducing the drum size has more effect on torque than I realized. My
> spreadsheet shows 953 lb-ft for 48", 242 lb-ft for 36" and only 85 lb-ft for
> 24".
>
> What do you think?

Maybe this was a private email that escaped onto the newsgroup, but I
suggest the torgue should be proportional to diamenter; i.e., the 24"
drum torque should be 475 lb-ft.
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Eric Greenwell
Washington State
USA

[GeorgeB]

On Wed, 17 Nov 2004 11:13:26 -0600, "Bob Johnson" >
wrote:

>Reducing the drum size has more effect on torque than I realized. My
>spreadsheet shows 953 lb-ft for 48", 242 lb-ft for 36" and only 85 lb-ft for
>24".
>
>What do you think?

I think that there is someting confusing here ... the torque will not
vary with the drum size if it is shaft driven ... the line tension
will vary inversly with diameter ... tension x diameter = constant
(torque)

Give us more to go on.

George

[Bill Daniels]

"Bob Johnson" > wrote in message
news:_DLmd.1161$3I.654@okepread01...
> Reducing the drum size has more effect on torque than I realized. My
> spreadsheet shows 953 lb-ft for 48", 242 lb-ft for 36" and only 85 lb-ft
for
> 24".
>
> What do you think?
>
> Bob
>

I think Bob's working on something like this.
http://www.lippmann.de/germany/produkte/hoistline.html

The Lippmann folks manufacture Dyneema "plastic" winch cable. They are
claiming a 950 meter (~3100' AGL) winch launch on a 2000 meter (~6500')
runway. They also note 3000 launches at one club without a break.

This puts the total per launch cost (fuel, maint. cable replacement etc...)
at about 1.5 Euro. Now that's cheap flying.

Bill Daniels

[Jacek Kobiesa]

"Bob Johnson" > wrote in message news:<_DLmd.1161$3I.654@okepread01>...
> Reducing the drum size has more effect on torque than I realized. My
> spreadsheet shows 953 lb-ft for 48", 242 lb-ft for 36" and only 85 lb-ft for
> 24".
>
> What do you think?
>
> Bob

Something is wrong here.....you need to read your info or your
spreadsheet doesn't make any sense....!!!??? Torque measured where?
The drum diameter has nothing to do with the torque!!! Is this for
which drum application? Need more and correct info in order to give
you some numbers.

[Jim Vincent]

>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

[F.L. Whiteley]

"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
>

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

Frank Whiteley

[Bill Daniels]

"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
> >
>
> 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

[Jacek Kobiesa]

"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
> >
>
> And, in practice, constantly changing, generally increasing, depending on
> design and layup.
>
> Frank Whiteley

Definition of Torque:
The Torque on an object about some pivot point is due to the action of
a force on the object. (So, if this is drum of a winch on a shaft I am
assuming)
Magnitude of the Torque:
t = "Force" times the "Lever Arm"
= (Component of the Force perpendicular to Lever Arm) x (Lever Arm -
the distance between the pivotal axis and the point where the force is
applied) = Fp l
= (Force) x (Perpendicular line of action lever arm - the shortest
distance between the pivot point and the line of action of the force
through the body) = F lp
= (Force) x (Lever Arm) x sin(angle between the two) = F l sin(q)


Direction of Torque:

· The more exact definition is that the torque is that the torque is
the cross product of the lever arm with applied force.
· The torque's direction is perpendular to both the direction of the
lever arm and the direction of the force. The direction can be found
using the right hand rule.
I Units: N m
* These are the same units as a Joule, but nobody uses Joules. There
must be some deep connection which nobody has yet discovered. Any
ideas ?

Torque on a Body
A force is applied to an irregular shaped body on a frictionless,
horizontal surface. The location the point of application of the force
can be altered by dragging the application-point. The magnitude and
direction of the applied force can be altered by dragging its tail.
Displayed are the torque, the magnitude and direction of the force,
the lever arm, and the angle between the force and the lever arm. Also
displayed are the perpendicular values of the applied force and the
lever arm.

Just my few thoughts....jk

[Bob Johnson]

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
news:DxAnd.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
>> >
>>
>> 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
>

[Eric Greenwell]

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

[Bill Daniels]

"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

[GeorgeB]

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)

[Bob Johnson]

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
>
>
> --
>
>
> ----------------------------------------------------
> This mailbox protected from junk email by MailFrontier Desktop
> from MailFrontier, Inc. http://info.mailfrontier.com
>
> "Bill Daniels" > wrote in message
> news:DxAnd.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
>>> >
>>>
>>> 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]

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$%x.33211@okepread04...
> Looking at the power required to rotate the drum, we were running 400 rpm,
> 6.28 ft/rev, 45 sec to release and estimating 1000 lb line pull. I
calculate
> that mix to equal 102 hp.
>
> !02 hp minus 75 hp is 27 hp. Seems to be a fair if maybe somewhat high
> approximation of aerodynamic drag of glider and Spectra tow rope.
>
> The throttle was opened to the point where the pilot reported his airspeed
> to be in the desired 55-60 kt.range and which corresponded to the 2300 rpm
I
> was seeing on my gauge.
>
> Again, I'm finding it hard to explain most of the discrepancy between the
> altitude-corrected
> 180 hp
> 454 c.i. dynamometer data and what I'm seeing with my own eyes. Can a
water
> pump, an alternator (the radiator is cooled by electric fan) and
mechanical
> friction
> eat up 180 engine hp minus102 drum hp= 78hp?
>
> Thanks for your input but still off the rails,
>
> Bob Johnson
>
>
>
>
> --
>
>
> ----------------------------------------------------
> This mailbox protected from junk email by MailFrontier Desktop
> from MailFrontier, Inc. http://info.mailfrontier.com
>
> "Bob Johnson" > wrote in message
> news:XtUnd.87541$%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.
>
>
> --
>
>
> ----------------------------------------------------
> This mailbox protected from junk email by MailFrontier Desktop
> from MailFrontier, Inc. http://info.mailfrontier.com
>
> "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
> >
> >
> > --
> >
> >
> > ----------------------------------------------------
> > This mailbox protected from junk email by MailFrontier Desktop
> > from MailFrontier, Inc. http://info.mailfrontier.com
> >
> > "Bill Daniels" > wrote in message
> > news:DxAnd.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
> >>> >
> >>>
> >>> 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
> >>
> >
> >
>
>

[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