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fly911
1st Jul 2010, 12:43
http://i135.photobucket.com/albums/q140/fly911/TQ.jpg

1. Hovering into the wind produces 100% engine torque.
2. Your helicopter is transmission limited on torque.
3. Increasing anti-torque pedal to hover crosswind produces a torquemeter reading of 105%.

To me, this is a deceptive torquemeter reading since your transmission is only subjected to 100%. The other 5% while going through the engine torquemeter is being generated by the tail rotor and not being absorbed by the main xmsn.

Anybody out there ignore that extra torque produced by the tail rotor?

topendtorque
1st Jul 2010, 13:00
That could be an interesting question, because we all tend to allow the aircraft to windvane, when utilising all the power avbl to pick up the maximum load and save a bit of power, but.

could I put it to you this way just for something right outside the square..

a sailing boat sail when sailing direct downwind will produce the most power with the least drag possible withe boat aligned withe downwind direction.

when / if, which you can't with a boat, turn the boat cross wind and continue direct downwind with the max power downwind component , will the mast bend forward more or less?

assuming of course that there is negligable water current in relation to the wind.

blackhand
1st Jul 2010, 13:02
Logic would seem to say yes.
But then for take off power you are allowed a margin of about 5% (for up to 5 minutes), so would probably allow for the scenario you proposed.

Cheers
BH

RVDT
1st Jul 2010, 13:03
Hovering into the wind produces 100% engine torque

Actually that would be transmission torque not engine torque unless the engine was flat rated. 100% engine torque on the torquemeter of a 206B3 would be about 130% on the dial.

On a Bell 206 and 350 Astar where the T/R drive actually comes from the input drive and does not go through the input pinion you would be correct. Not so on an MD500 for example that takes its T/R drive after the input pinion.

How do you determine exactly which component is the factor for the torque limitation?

Why not hover the other way across the wind then according to your calculation the hover Q required might only be 95%?

topendtorque
1st Jul 2010, 13:19
Why not hover the other way across the wind then according to your calculation the hover Q required might only be 95%?


another trick question.

fly911
1st Jul 2010, 13:28
Flying ENG news gathering in a 206 and AStar while hovering OGE, I was often required to turn crosswind to get a shot or keep the action in view. I always thought that if a helo was transmission limited for torque, then the torque reading should come off the transmission and not the engine.

RVDT
1st Jul 2010, 16:04
then the torque reading should come off the transmission and not the engine

which it does.

The torquemeter instrument whether it be in % or PSI or whatever is calibrated to indicate the limits of the input to the transmission. It is all part of the "transmission" of power it just happens to be in the engine for lots of good reasons too long to list here.

B222 230 430 etc though have a MAST Torquemeter.

The scenario you quote MAY only be correct on certain types of aircraft. I wouldn't go pulling that trick in a Bell medium twin. 2 engines through ONE input for everything.

Sounds like you need either:

a) a bigger helicopter
or
b) less weight in the one you have
or
c) a stabilised turret camera
or
d) all of the above.

Read your RFM regarding the use of TOP. It has speed limitations.

Shawn Coyle
1st Jul 2010, 16:09
Terminology is always an issue - rotor blades require / demand torque - they don't produce it.
And it depends where the torque is being measured - in the RR250 series engines, it's usually measured at the output from the engine, so it is the combination of main and tail rotor torque required / demanded.
Some helicopters measure torque at the transmission - but again you need to know if that measurement is mast torque or torque on a gear somewhere. And if it measures tail rotor torque as well.
And whatever the limitation is will be the limitation that must be obeyed for gauges installed - without trying to be clever and say that this can't be 'real' torque because it comes from hovering with a crosswind demanding more tail rotor power.

wmy
1st Jul 2010, 16:55
whatever the limitation is will be the limitation that must be obeyed for gauges installed
:D

Limitation means limitation. Stay below, and you (and the guy who will fly the ship after your mission) will live longer...

fly911
1st Jul 2010, 17:44
Just kidding.
Thanks for the input.

bugdevheli
2nd Jul 2010, 05:09
Out of interest could anyone tell me what the maximum ammount of tail rotor pitch is available and used on their particular machine, and a technical definition of the condition whee whaa Thanks Bug

fly911
2nd Jul 2010, 07:54
Shawn Coyle: "but again you need to know if that measurement is mast torque or torque on a gear somewhere. And if it measures tail rotor torque as well." Agreed Shawn. On my ship it's pretty easy to figure out. When I input pedal, the torquemeter changes.
Thanks for all the advise, but although my question had a basis in fact, It's pretty much theoritical as I'm not trying to fix a problem or figure out how to get something done or what's "by the book", but phrased the question the way I did to make it more ubderstandable. I was hoping someone would come up with "Oh, that's because...", and that I was missing something, but apparently it's not that simple. Thanks again.

Nf stable
2nd Jul 2010, 09:15
If we turn the machine 90 degrees and expose the length of the fusilage to the wind, we create more drag. Surely the disk needs to be tilted further in to wind in this configuration, just to hold us at zero groundspeed. If this tilt is at the expense of the vertical component of thrust, then will we not need to increase power to maintain height?
Surely this becomes more true in big slab sided machines. So how do you tell which extra Tq is to the Tail rotor and which is attributed to the Main rotor. Just my two cents and all.....

riff_raff
3rd Jul 2010, 02:13
fly911,

If that gauge is an engine torquemeter reading, then all of that torque is being transmitted through at least part of the MRGB before being split between the tail and main rotor drives. Depending upon the particulars of the MRGB design itself, the greater torque split going to the tail rotor may or may not allow safe limited operation above 100%. It all depends upon what component(s) and/or operating conditions in the MRGB establish that 100% torque limit. For example, if the torque limiting components (gears, bearings, shafts, etc) are downstream of the power take-off point for the tail rotor drive, then exceeding the 100% MRGB torque rating may not harm the gearbox itself, as long as all of that additional torque was transmitted to the tail rotor drive.

Conventional rotorcraft MRGB and drivetrain components are designed for fatigue life using an RMC derived value of various estimated load and speed conditions, combined with a statistical reliability factor (L10 or similar). This value is usually conservative to ensure safe and reliable operation of the drivetrain over its TBO. However, MRGB torque limits may not be based solely on mechanical loads. Sometimes the power that can be transmitted across a gear mesh is limited by a lubrication condition called "scoring". An overloaded gear mesh that encounters scoring will fail rapidly, but scoring failures are typically not catastrophic if handled timely and are easily detectable.

Current MRGB component engineering practice is to design for "On-Condition" life, with no set TBO. That means the gearbox continues in service until a life-related issue is detected. On-condition gearbox life is made possible through sophisticated gearbox Condition Based Monitoring (CBM) instrumentation, electronics and software. The CBM data acquisition constantly monitors operating conditions and actively computes how much fatigue life the MRGB system theoretically has remaining at any point. Thus, if your MRGB has a modern CBM system, it would allow for and take into account slight overtorque excursions like the one you describe, but it would also calculate for a reduced remaining gearbox life accordingly.

Hope that was helpful.
riff_raff

fly911
3rd Jul 2010, 10:44
Thanks rif_raf. You put out a lot of interesting info there, but.... I disagree with your statement: If that gauge is an engine torquemeter reading, then all of that torque is being transmitted through at least part of the MRGB before being split between the tail and main rotor drives.
http://i135.photobucket.com/albums/q140/fly911/RR250.jpg

In the gearbox section on the above RR 250 engine, you will note that the power output to the left goes to the main transmission. I added the arrow to the right to show that the same gear drives the tail rotor and both outputs affect the torquemeter reading, but only the power output shaft to the left goes through the main transmission. In other words, the tail rotor does not go through the main transmisssion (MRGB).

I'm sure that you are correct on many helicopters, especially most if not all multi-engine models, but the Bell 206/407 series powertrain using the above series engine would not apply. Thanks for the solid information.

victor papa
4th Jul 2010, 09:59
riff raff statement is also incorrect for the AS350 series and EC130. The TR is driven directly off the engine MO5 reduction gearbox rearwards and the MGB is driven forwards off the MO5 via the freewheel unit and 2 shafts. for the arriel 1 the torque reading is a function of the movement of the pinion gear in the engine MO5 and variable oil pressure in the chamber. The arriel 2 the torque reading is the displacement of the MGB driveshaft in the MO1 vs a uncoupled shaft. All the training manuals state clearly however that the tQ function is not to protect the engine but the MGB despite the fact that for all the tQ reading is taken on the engine and not the MGB.

riff_raff
5th Jul 2010, 04:27
fly911,

"In other words, the tail rotor does not go through the main transmission (MRGB)"

There's always exceptions, eh? Even on that Allison model 250 you show, I suppose I could argue that the full engine torque passes through part of the transmission gear train. The same issues would still be relevant, the only difference being that the affected gears happen to be within the engine case and not the MRGB case proper. Regardless, I'll choose my words more carefully next time.

victor pappa,

"All the training manuals state clearly however that the tQ function is not to protect the engine but the MGB despite the fact that for all the tQ reading is taken on the engine and not the MGB."

The engine and gearbox guys always point their fingers at each other when it comes to the issue of what hardware is the cause of the torque limit. The gearbox guys will say the torque limit protects the long, slender power turbine drive shaft. And the engine guys will claim that their engine can easily make more torque than the gearbox can ever handle. The truth is it's usually a little of both.

The torque meters themselves are almost always supplied as part of a turboshaft engine, since they must provide feedback to the FADEC. The old method of torque monitoring was via pressure measurement in a hydrostatic circuit as you describe. The mechanism was actuated via axial displacement of a helical gear or helically splined shaft. This system was originally developed for monitoring BMEP on large aircraft piston engines. Newer engines use electronic systems, usually employing hall effect sensors and trigger wheels on the engine output shaft that measure torsional displacement.

Regards,
riff_raff

fly911
5th Jul 2010, 11:42
rif_raf...
You make an interesting point...and it's making my brain hurt.
I suppose I could argue that the full engine torque passes through part of the transmission gear train. That kind of makes sense to me, but I'm not really sure. I'll have to give it some more thought....later.
Thanks, fly911

Hughes500
5th Jul 2010, 21:20
Surely the engine drives the main rotor gearbox, the main gearbox splits the power to mr and tail rotors. Does in a 500
If you are pulling 100% in the hover and press the power pedal ( left one in US helis) you power up the tail rotor, this drag causes the main rotor to slow, thus you descend, but you will pull more collective to maintain the same height. Therefore tq gague goes up. However if you push right pedal, tr depowers main rotor speeds up and you climb, to maintain same height lower collective and hey presto gague goes down

fly911
6th Jul 2010, 01:36
rif_raf and Hughes500 see if this makes sense.
rif_raf: I suppose I could argue that the full engine torque passes through part of the transmission gear train.
Hughes500: If you are pulling 100% in the hover and press the power pedal ( left one in US helis) you power up the tail rotor, this drag causes the main rotor to slow, thus you descend, but you will pull more collective to maintain the same height.

In a Bell 206... You are pulling 100% torque in a hover. You push in the power pedal. Your engine is kept from drooping by the governor and engine output speed remains the same. Your MRGB is going the same speed and wasn't even aware that the tail rotor demanded more torque from the engine. To the main xmsn, nothing has changed. Yes? No?

VeeAny
6th Jul 2010, 09:30
The freewheel unit on a 206 is not part of the engine its part of the helicopter.

Any torque demanded by the drivetrain goes through the freewheel unit be it , from the Tail Rotor or the Main Rotor.

If the Tail Rotor where bolted to an engine power output it would either not turn in the event of an engine failure or the drivetrain would be turning the engine over in autorotation which kind of negates the need for a freewheel.

The splines at the back to the engine gearbox in a 206 are driven from a shaft in the freewheel unit not the engine.

Who knows what the torque limits are for the freewheel unit ? I don't, so I have a good reason to stay within the limits other than common sense / because the manufacturer says so and any legal requirements.

A great topic that got me thinking for more than 5 minutes !

sycamore
6th Jul 2010, 12:07
It is a transmission `system; if you apply left pedal you will increase torque and engine rpm/TiT;

fly911
6th Jul 2010, 17:58
VeeAny, I'm glad that you took that 5 minutes to give it some thought. Thanks. You gave my brain something to work on before it exploded.
The splines at the back to the engine gearbox in a 206 are driven from a shaft in the freewheel unit not the engine.


So true! For those that find that a little confusing, please see the attached drawing. The freewheeling unit shaft outer race (#14) is driven by the engine. The inner race shaft (#1) attaches to the main drive shaft and is driven by #14. The spline adapter shaft (#28) leads to the tail rotor.
http://i135.photobucket.com/albums/q140/fly911/scan0003.jpg

So, we're in agreement that all the torque goes through the freewheeling unit. What I'm not sure about is that it all goes through the main transmission, counting against the torque limit.
Please note that I'm not saying to exceed any limits! Don't do that. I don't do that. I'm just using that as an example. I should have said 90% instead of using 100%. I was just trying to figure out if the engine torque demanded by the tail rotor goes through.... oh you know.

.

Graviman
7th Jul 2010, 11:56
1. Hovering into the wind produces 100% engine torque.
2. Your helicopter is transmission limited on torque.
3. Increasing anti-torque pedal to hover crosswind produces a torquemeter reading of 105%.


Is that torque just as heli starts to turn (acceleration), once yaw rate established (velocity), or once established in crosswind? Just interested.

fly911
7th Jul 2010, 13:53
Graviman,
Actually, I was referring to any increase in torque attributed to the tail rotor. Thanks.

riff_raff
9th Jul 2010, 01:36
fly911,

Let's take a look at the x-section of the Allison model 250 engine you posted (the Bell 206 powerplant):

http://i135.photobucket.com/albums/q140/fly911/RR250.jpg

The engine power travels from the two stage power turbine (N2) along a shaft to a pinion gear. The pinion gear drives an idler gear, which also incorporates the engine torquemeter. The idler gear then drives a power output gear. The power take-off for the tail and main rotors is split at this power output gear.

So based on your example given, the following can be deduced:
1) There are at least 3 gear meshes and 1 shaft that must transfer 100% of the torque produced by the power turbine, minus any efficiency losses in those 3 gearshafts and minus the parasitic losses incurred by driving the engine accessories.
2) The torquemeter measures 100% of the torque produced by the power turbine, minus the mechanical losses in the pinion/idler gear mesh, its bearings and the bearings on the power turbine shaft (these losses likely total less than 1.5%). Since the torquemeter is ahead of the tail/main rotor power take-off point, the torquemeter should be indicating the total torque transmitted to both tail and main rotors.
3) The overrun clutch is not shown, but is likely part of the main rotor gearbox proper. Since the overrun clutch is located after the tail/main rotor drive split, it is only subjected to the main rotor torque loads.

riff_raff

RVDT
9th Jul 2010, 06:33
Since the overrun clutch is located after the tail/main rotor drive split, it is only subjected to the main rotor torque loads.

Really? :rolleyes:

fly911's diagram.

Item 1 Shaft Inner Race and item 28 Adapter are connected. Item 28 drives the TR. The OR clutch is BEFORE the shaft inner race at item 11 and is subject to ALL the torque.

From the Shaft Inner Race forward to the main rotor is only subject to main rotor torque and aft is subject to TR torque. The torquemeter measures ~ TOTAL torque applied.

Explain in your scenario what would drive the TR in autorotation?

That skinny bit of Item 1 Shaft Inner Race leading to item 28 Adapter is what is driving the TR. It is probably about as thick as your little finger.