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Converting Shaft Horsepower to Torque


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Converting Shaft Horsepower to Torque

Old 2nd August 2006 | 10:32
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From: Brunei
Converting Shaft Horsepower to Torque

I am trying to calculate the Torque which should be required to hover a S61 helicopter in various conditions. Unfortunately the graphs which I have access to give the answer in Shaft Horsepower.
Is there an easy (or even difficult) way to convert this SHP requirement to the torque which should be seen on the pilots guages ?
Any help would be greatly appreciated.
C.King is offline  
Old 2nd August 2006 | 13:49
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'India-Mike
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Torque = Power / rotorspeed.

Power in W, rotorspeed in radians/sec, torque in Nm.

If you're going to use hp and rpm in the equation, it'd become

Torque = ( Power / 1.341 ) / ( rotorspeed x 0.1047 ) x 1000

Still gives you torque in Nm though. If your cockpit gauges are in Nm you're in business but I bet they're in %. Will check a cab next time I'm over to move the Chippy....

If gauges are in % you'd need to know the 'datum' torque ie how many Nm of torque = 100%. That'll be a question for Westlands - unless on-site engineering has that figure.
 
Old 2nd August 2006 | 18:35
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From: Philadelphia PA
If you know the transmission rating in shp, and the rotor RPM, then you can figure out the torque, but why?
I'd be interested in the power required to hover the helicopter, not the torque.
All the theory and non-dimensional analysis uses power (or some variant of it) but I've never seen torque used, except for presenting a power-required curve in terms that the pilot can use. That would be a good reason to use torque.
Unless there's another reason....
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Old 3rd August 2006 | 11:29
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Gentlemen,
Many Thanks for your replies. I will use the information next time I am in work to see if I can get he answers I am looking for.

There is no great drama in why I am looking for the information. The guages are calibrated in %, and the company procedures determining what we can and cannot do are also given as torque.
I am trying to confirm (or not) that the figures I am seeing in varying profiles (displayed in the cockpit as torque) are as expected by Sikorsky (given in the flight manual as shp). The reason for this is that at first glance it could appear that the tq figures for one aircraft appear significantly higher than others in similar conditions. Rather than making assumptions I am trying to find something to hang my hat on !
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Old 3rd August 2006 | 12:05
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The RFM Part 1 Section 4 has the answer.
1. Power required to hover OGE Figure 4-8 (in shp)
2. Power available CT58-140 engines max continuous or takeoff power Figure 4-100 (shp)
3. Indicated torque vs engine shaft horsepower 100% Nf Figure 4-102

So assume sea level standard day, nil wind, 19000 lb, power required 2260 shp (1130 per engine).
Min spec engine will provide 1240 shp continuous or 1390 shp at takeoff power so that's OK, although of course the transmission is limited to 1250 shp/103% torque per engine at takeoff power and 1050 shp/86.5% torque per engine continuous.
Indicated torque will be (1130 shp) 93% per engine.

For ops above 19000 lb go to Supp 15, where the engine power available charts 15-4-8 to 15-4-12 show torque not shp, and Fig 15-4-14 has the shp to torque conversion for both 100% and 103% Nr.
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Old 3rd August 2006 | 12:25
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From: Philadelphia PA
Ah- if you're seeing differences in torque between aircraft, there are several possible causes - torquemeters not calibrated (I assume they're oil pressure type torquemeters, not electrical shaft-twist types), different blade cleanliness, different blade balancing setups, etc.
I was able to prove to the maintenance folks that the torquemeter pistons on a particular Bell 212 weren't right because we had a prototype power required to hover calculator that showed we were off in torque required by about 5% - they didn't believe me until they really checked.
Remember that torque is drag on the rotor blades and any differences between blades, such as bugs, or other build up will change the drag.
I think the civil S-61 doesn't have trim tabs, but if there is a large difference in the way blades are set up for tracking and balance, (are pockets moved?) that can make a difference.
Good luck!
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Old 6th August 2006 | 02:42
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The universal SHP equation is:

SHP = rpm x torque (lb-ft) / K

where K = 33000/6.2832 = 5252

Note: RPM & torque must be measured on the same shaft - e.g. gearbox input

What you need to know is -
What is 100% gage rpm?
What is 100% gage torque?

Once you know these you can do the numbers.
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Old 6th August 2006 | 12:26
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From: Brunei
Gentlemen, Once again many thanks.

steve_oc,

The graphs you describe are just what I'm looking for.

Many Thanks again.
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