wip Single engine performance bug

[exil]

Hey,

something must be a bit faulty with single engine performance. In helicopters, if troque is the limit, you can take double engine performance and multiply it times two to get your single engine power. So i compared some numbers while taking one engine to idle in flight (tried with engine 1 and engine 2):

double engine power = 2 x 30 % TRQ
single engine power = 1 x 80 % TRQ (should be 1 x 60 % TRQ)

double engine power = 2 x 40 % TRQ

single engine power = 1 x 112 % TRQ (should be 1 x 80% TRQ)

Find the attached trackfile to see what i mean.

 

SINGLE ENGINE PERFORMANCE BUG.trk

Edited March 26, 2022 by exil
Typo in numbers

[BIGNEWY]

Hi, 

I will check with the team but before I check what are your assumption based on? do they reflect the charts available to the public?

thanks

[exil]

Thanks for the answer!

Well, not that it matters much, but I'm a real life pilot flying commercially.

It's just a matter of physics and load distribution. 

With both engines operative the load both engine have to bear (expressed in torque) is equally distributed across both engines. 

When one fails, the remaining one has to bear the total load alone. That means the double amount of torque (which is usually taken from the input shaft from the engine to the transmission). 

It is common for every multi-engine helicopter. The Apache should be no exception here.

As a rule of thumb you learn that during your hover check, double the amount of current torque displayed to know your single engine performance. 

I am pretty sure the SMEs will confirm that. 

 

Edited March 26, 2022 by exil
Added additional remark

[TED]

It's not strictly true that the oei torque required will be exactly twice that of the aeo (all eng op) tq. 

The coefficient of drag effect on total drag is not linear. 

The drag equation is - 

D = Cd * A * .5 * r * V^2

Drag = coefficient of drag x area x 1/2 rho V squared. 

Very simply put - So with the reduction of thrust to weight ratio a higher relative aoa will be required to compensate for the increased drag. The net result is that the required single engine - oei tq, will be higher the aeo tq x 2. How much more depends on the drag coefficient of the particular aircraft. 

I don't have access to the ah64 perf charts but this is common to all helicopters to some degree. Whether it's correctly modelled in dcs or not, I can't say, but it's fair to say it should be more than simply twice aeo tq. 

Hope that makes sense. 

Edited March 26, 2022 by TED

[exil]

16 minutes ago, TED said:

It's not strictly true that the oei torque required will be exactly twice that of the aeo (all eng op) tq. 

The coefficient of drag effect on total drag is not linear. 

The drag equation is - 

D = Cd * A * .5 * r * V^2

Drag = area x 1/2 rho V squared. 

Very simply put - So with the reduction of thrust to weight ratio a higher relative aoa will be required to compensate for the increased drag. The net result is that the required single engine - oei tq, will be higher the aeo tq x 2. How much more depends on the drag coefficient of the particular aircraft. 

I don't have access to the ah64 perf charts but this is common to all helicopters to some degree. Whether it's correctly modelled in dcs or not, I can't say, but it's fair to say it should be more than simply twice aeo tq. 

Hope that makes sense. 

 

No not really. Thrust doesn't really change, because the remaining engine will deliver the same amount of thrust as when aeo (so basically twice the amount) . To deliver that, you sort of need twice the amount of power from the engine. 

What does change is NR. It will decrease slightly and therefore you would have to compensate with a bit more collective for that lift loss. But that's just a tiny bit. 

Basically what you are describing would be an decrease in Rpm. For a decreased RPM you would need to compensate with a higher aoa (increased collective) to compensate for the lift loss. 

But RPM doesn't decrease (significantly) when in oei. 

 

Edited March 26, 2022 by exil

[NineLine]

Our SME's agree, but it has not been fleshed out enough yet, so we can color this one WIP at the moment. But thanks for the detailed report, good work guys.

[Swift.]

When operating in Single Engine, the torque indicated for the remaining engine is more than double the torque indicated when both engines are running.

This indicates there is extra resistance in the rotor when one engine is shut down. Likely indicating the lack of modelling of the sprag clutches.

Single Enginer Torque.trk

[bradmick]

This is normal and expected. With two engines operating, each engine is carrying 50% of the load. When one engine fails, the one operating engine has to take over for both. The end result is a doubling of the torque provided by that one operating engine. So if you were cruising at 60% torque dual engine, and an engine fails you will now be at 120% single engine. 
 

This has nothing to do with clutches whatsoever.

Edited June 25, 2022 by bradmick

[Swift.]

22 minutes ago, bradmick said:

This is normal and expected. With two engines operating, each engine is carrying 50% of the load. When one engine fails, the one operating engine has to take over for both. The end result is a doubling of the torque provided by that one operating engine. So if you were cruising at 60% torque dual engine, and an engine fails you will now be at 120% single engine. 
 

This has nothing to do with clutches whatsoever.

 

Yes exactly, you would expect the torque to double. What we see is over double.

So in the track, dual engine reads 18% in both. One expects 36% when on a single engine then, but instead we see 42%. 

So the question is then, where is that extra 6% coming from.

[exil]

I already reported it as a bug and it's set to 'wip'. 

 

 

[BIGNEWY]

threads merged