Merlin negative G cutout too quick?

[Moggy]

I don't know if this relevant or helpful but there's a section about negative G cut outs in the RAF Pilot's Notes General from 1943 (apologies if this has already been posted);

[klem]

Guys I have been fortunate to get a reply from a current Hurricane MkI display pilot. He asks to remain anonymous but gives me the following:

"I have the privilege of flying and displaying Hurricane Mk1, [serial deleted]. It will not surprise you to know that in deference to it's age and historical importance we do not fly the aircraft as aggressively as it would have been flown during combat. Particularly, we avoid negative g so I am not well placed to answer your question specifically. However, I can give you some clues.

First, I can tell you that it does not require negative g to make the engine suffer from a shortage of fuel supply; a significant reduction of g down to, say, 0.3g can be enough to make the engine misfire. This can be experienced towards the top of a wing-over but I would estimate that the reduction in g needs to be maintained for 2 seconds or more before there are any effects. Undoubtedly, if the reduction in g was greater (to less than zero g) and particularly if the bunt was abrupt then the effect could be instantaneous. I have never, though, experienced any misfiring in turbulence; albeit, were the turbulence severe enough to produce g spikes to less than zero g, I would not rule out the possibility of the odd cough from the engine. Of interest to you I am sure is that on recovery from an episode of fuel starvation the engine recovers through a short period of over-richness shown by, I would estimate, up to a second of black, sooty exhaust before normal combustion is resumed.

Good luck with your simulation."

It's not a complete profile of the problem but it gives some useful check points for whatever 1C come up with so I hope they will use the information for that purpose.

I suppose the thing the Merlin flyers want to know is that the effect isn't overmodelled to an unrealistic and irritating degree in level flight or modest pushovers into what might be called 'normal' descents.

What the Axis flyers want to know is that an aggressive pushover in combat will have the Melin cutting out and preserving their escape maneouvre.

The truth will lie somewhere in the middle but is probably not too critical if the two situations are preserved.

I think what the report tells us is that for modest maneouvres there is little if any effect and for sustained reduced G of perhaps 0.3G there is something like a 2 second delay or more before it occurs, perhaps due to the carburettor trying to keep up on the knife-edge of the problem. Recovery seems to be in the order of a second once positive G is restored.

A question remains, for an aggressive pushover causing that level of negative G that first causes virtually instantaneous fuel cutoff at the carburettor inlet, exactly how fast would the fuel cutoff at the carburettor inlet occur and how quickly would the carburettor fuel be used up and the engine be affected? If 1C were able to decide on a G figure for a near instantaneous cutout they would I suppose still be left looking at the 'curve' for intermediate values and times.

Hope this has helped.

[Viper2000]

Ok guys, several things here.

Firstly, the dive was only posted for the g history; there would have been no risk of cutout because the aeroplane was a PR.XI, which would have had a 2 stage engine, and all the 2 stage engines had later carburettors.

The point was just that you can get the nose down quite smartly without recourse to negative g, which is probably why people didn't immediately realise that a negative g capability was required for a military engine.

Secondly, I've gone digging through my archives and have found some relevant RRHT books.

This means that I can now hopefully shed some light both on the engine ratings and FTHs (which are slightly more conservative than given in some Pilots Notes) and also the negative g cut.

Appendix VI
The Merlin S.U. Carburettor Under Negative-G
This Appendix expands the comments made about the effects of negative-g on the S.U. carburettor on page 46 of the third edition of "The Merlin in Perspective - the combat years".

A characteristic of the Merlin which has led to some comment was the cut that occurred if negative-g was applied when entering a dive. The momentary power loss was in two stages; an initial loss of power which some pilots regarded as only a very brief hesitation, followed by a rich cut of about 1½ secs duration. Full power returned once the negative-g was removed. The tactical disadvantage it caused could be reduced by pilot technique, such as entering the dive from one wingtip. Wing commander 'Roly' Beaumont, apart from undertaking the first flights of the Canberra and Lightning fighter, flew Hurricanes in the air fighting in 1940 prior to and during the Battle of Britain, in No. 87 Squadron. I am grateful to him for allowing me to use his quote at the Hurricane 50th Anniversary evening, which while dealing with the abilities of the aeroplane encompasses the engine.

"Another thing we did was to devise a manoeuvre which was aimed at getting us out of a difficult corner if we ever got into one. This may sound very extraordinary, probably, to practising pilots today, but it consisted of putting everything into the left-hand front corner of the cockpit. If you saw a 109 on your tail, and it hadn't shot you down at that point, you put on full throttle, fine pitch, full left rudder, full left stick and full forward stick. This resulted in a horrible manoeuvre, which was in fact a negative-g spiral dive. But you would come out of it with no Me 109 on your tail and your aeroplane still intact."

Comparisons have been drawn with the direct injected DB600 series engines in the Me 109, which did not suffer in this way. The problem was known before the war, but it was thought that British fighters would be operating as bomber destroyers and the situation would not arise, which shows how wrong one can be. The Daimler-Benz engine was good, but direct injection provided no charge cooling due to the latent heat of evaporation of fuel, which was worth 25ºC reduction in charge temperature. Like all engines, it had its own, problems, including failures of injector pipes. Direct injection had been considered by Rolls-Royce, and had been turned down in favour of the carburettor. An example of its use by the Company was the direct injected sleeve valve diesel Kestrel, powering Captain George Eyston's Flying Spray, which had for a time held long distance records and the speed record for diesel cars.

Most of the credit for reducing the negative-g problem to negligible proportions goes to the late Miss Shilling (Mrs Naylor) - Tilly to her friends, who was in charge of the carburettor section of RAE at Farnborough. This remarkable woman saw further than most through a barn door, recognising that the engine fuel pump was part of the problem. It had been designed as two separate gear pumps with individual quill drives, so if one pump were to fail the other would keep the engine running. Each pump was capable of meeting maximum engine demand plus 20%, a relief valve controlling outlet pressure. Miss Shilling saw that if the float needle was not controlling, both pumps could suddenly put a large excess of fuel into the carburettor. She developed the RAE restrictor, which was fitted in the fuel line to the carburettor and which limited the flow to just above the maximum engine demand. It seems unlikely that any were fitted during the Battle, except for a trial quantity of six. Later retrospective action was taken and it was a useful palliative. The scheme finally adopted was for an anti-g version of the S.U. carburettor developed by her section, which allowed the existing carburettor to be modified. The weak cut, which was caused by the fuel in the float chamber rising to the top and starving the jets when negative-g was applied, was cured by fitting jet shrouds which continued to draw fuel from halfway up the chambers. Ball valves were added to prevent the fuel from flooding out of the air vents under this condition. The ensuing rich cut was prevented by forming a restrictor on the end of the float needle, so that when forced fully open by the fuel pressure, the floats then having lost control, the restrictor, like that fitted earlier limited the flow to slightly above maximum engine demand. This carburettor continued to be fitted to various marks of Merlin until production ceased. The changes to the S.U. carburettor to give it its anti-g characteristics are shown in the illustration.[attached]

As mentioned in the main text a negative-g version of the S.U. carburettor was developed, in parallel with the anti-g version, to the point of undergoing service trials, but was abandoned as its mechanism could not sense zero g. On certain marks of engine the RAE restrictor was retained, leaving the carburettor unmodified, but where the anti-g features were embodied the RAE restrictor was removed. Two-stage engines in the range of 60 to 100 series, depending upon mark number, could have either an S.U. or Bendix carburettor, the latter not being subject to g effects. The Bendix was fitted to the 65, 66, 70, 76 and 85. Above 100 series single point S.U. injection was featured and again was not sensitive to g effects. Packard engine were fitted with Bendix carburettors, with the exception of some 100 series equivalents built at the end of the war with Bendix units and Simmonds power control units.

Harvey-Bailey (1995).
I have preserved the original formatting as far as possible, which means that the paragraphs are rather long. I have therefore highlighted some important sections in red so that they aren't missed.

The illustration is of interest because if you remove the modifications then you're back to the standard S.U. carburettor fitted during the Battle.

You can see that the lean cut was caused by the exhaustion of the fuel in the small chamber; whilst the rich cut was caused by the big chamber filling up with fuel and subsequently flooding this small chamber, as well as by fuel leaking through the air line which the mod protected using the ball valve.

Recovery from the rich cut requires that the engine consume the excess fuel in the big chamber. We know that this took about 1½ seconds.

This allows us to make several observations about the behaviour of the system.

1. The maximum rate of fuel pump delivery was about 2.4 times the rate at which the engine could consume fuel.
2. The amount of time taken to flood the big chamber would therefore be about 1.5/2.4 = 0.625 seconds in the worst case.
3. The small chamber is something like half the size of the normal space above the fuel level in the big chamber. We can therefore infer that full power engine demand would empty the small chamber in something like 0.75 seconds.
4. However, under negative g, fuel can also escape the small chamber through the holes that admit it under positive g. Since these holes were at least big enough to supply full engine demand, this would at least double the rate at which the small chamber could empty; therefore the lean cut would be expected within 0.3 seconds of negative g onset.

This means that we can probably safely say that when reduced or negative g is applied, nothing much will happen for at least say ¼ a second, because there is certainly enough fuel in the small chamber to supply the engine for that amount of time even if it's flowing out of the entry holes. And under reduced positive, close to zero g, it will take more like ½ a second.

Once the fuel in the small chamber is exhausted, the engine will start to suffer a lean cut.

However, we have already calculated that the big chamber is likely to be flooded in about the 0.6 seconds.

Once the big chamber is full of fuel, an unregulated supply of fuel will be forced into the small chamber under pressure.

It will take perhaps another ¼ second or so to fill the small chamber, at which point it will then proceed rapidly into the carburettor and cause the rich cut.

So the sequence of events was probably:

ZERO G Onset:

1. t = 0 s, pilot pushes to zero g
2. t = 0.6 s, small chamber empties; engine suffers weak cut
3. t = 0.75 s, fuel flow into small chamber resumes
4. t = 1 s, rich cut begins

NEGATIVE G Onset:

1. t = 0 s, pilot pushes to negative g;
2. t = 0.3 s, small chamber empties; engine suffers weak cut
3. t = 0.75 s, fuel flow into small chamber resumes
4. t = 1 s, rich cut begins

Once the rich cut has started, both chambers are full of fuel; therefore the recovery from negative g would be identical to the recovery from reduced positive g.

Recovery:

1. t = 0, pilot returns positive g
2. t = 1.5 s, excess fuel in float chamber consumed; float resumes controlling

The final piece of the jigsaw is the fact that it was felt tactically advantageous to roll the aeroplane into a dive rather than to suffer the cut.

If you look at the roll rate diagrams here you can see that the worst-case for high speed roll rate would have been about 60º/s, and therefore it would take about 3 seconds to roll inverted for dive entry.

The alternative, of pushing through the cut would result in the engine going on strike for roughly the push time plus 1¼ seconds (because the engine would keep supplying full power for roughly the first ¼ second of the manoeuvre which is therefore subtracted from the 1½ second rich cut recovery after positive g is restored).

If we assume -20 m/s^2 acceleration and constant 150 m/s TAS, since
a = v^2/r,
r = v^2/a = 1125 m
The turn circumference is therefore about 7 km, and so the time to execute a complete outside loop would be about 47 seconds; the time taken to push through 90º would therefore be something like 11.75 seconds, and so the total duration of the loss of power would be about 13 seconds.

So it's fairly obvious that in this sort of situation you'd win by rolling and pulling rather than pushing, because you'd lose a lot of distance in 13 seconds.

The critical case would be a pitch change of about 20º, because at -2 g you'd get there in about 2.5 seconds, for a total cut duration of 3.75 seconds or so, which is of the same order as the amount of time lost in the roll.

This all seems pretty reasonable to me.

The cut duration lines up with the current reports, and the calculated 0.3 second grace period explains the lack of misbehaviour in turbulence.

But what about the reduced positive case?
Well, that's been puzzling me for a while, because the float position is still defined by the float position, and therefore it's not immediately obvious why there would be a problem.

But if you look at the diagram, you'll see that the fuel has to flow down through the holes in the bottom of the big chamber into the small chamber in order to supply the jet. The driving force for this is the head of fuel in the big chamber, which is of course gz.

So when g reduces close to zero, the force driving the fuel through the holes into the small chamber is dramatically reduced, and therefore it follows that the flow rate reduces.

If the flow rate is less than engine demand then the small chamber will gradually empty and starve the jet. This explains the fact that it takes such a long time for reduced positive g to induce a cut. Indeed, it implies that if you waited long enough at say 0.75 g you'd probably get a lean cut eventually; it's just that in reality this never happens because people don't fly like that.

Of course, under reduced positive g the float is still controlling. However, because the flow rate through the holes into the small chamber is less than would normally be the case, whilst the pump delivery rate remains normal, the big chamber starts to over-fill. The float moves up and reduces the rate of fuel supply, but the equilibrium under reduced positive g will be a higher float position such that the progressive reduction in fuel flow into the float chamber balances the reduced rate at which fuel leaves to enter the small chamber.

This means that when 1 g flight is restored, there is going to be too much fuel in the system until equilibrium is restored.

This will happen somewhat more quickly than in the zero or negative g cases because the equilibrium point for reduced positive g is reached when the float chamber is only partially (albeit still excessively) filled rather than totally filled.

Therefore the duration of the rich cut recovery time should be expected to progressively increase towards the 1½ second maximum as g tends to zero.

Finally, what about g onset rate?
I've been thinking about this, and I suspect that it wouldn't make a lot of difference. If the sudden negative g was applied then the float would rise at the same rate as the surface of the fuel in the float chamber; this would momentarily cut off the fuel flow into the float chamber. However, as soon as the fuel hits the top of the float chamber, the float will instantly float downwards, re-opening the valve and admitting fuel at the full pump delivery rate. It won't bounce around because buoyancy would just peg it to its stop.

Therefore any misbehaviour is likely to simply be a function of g and duration.

Reference:
Harvey-Bailey, A. 1995. The Merlin in Perspective - the combat years. Derby: Rolls-Royce Heritage Trust.

[klem]

Hi Viper,

well I think I followed that (amazed).

Just two points.

1. You estimate it takes 1/4 second for the small chamber to refill as it heads towards Rich cut but previously you said the entry holes were sized to permit the full power demand which you felt would contribute an inlet hole emptying under -G in 0.75 seconds (along with the emptying due to engine demand). So wouldn't it take 0.75 seconds to refill the small chamber through the inlet holes as it heads towards Rich cut?

2. I didn't understand the part under sudden -G where you said "as soon as the fuel hits the top of the float chamber, the float will instantly float downwards". Why would the float float downwards when it is being held to the top by the raised fuel surface?

I think your estimates fit in with the info I was given by the MkI Hurricane pilot at reduced G where he felt it did not cause a problem down to about 0.3G with around a 2 second delay before engine response. His rich cut recovery from that he estimates to take about 1 second which is pretty close to your 1.5.

btw for interest and on a parallel topic, the Spitfire MkIa/Ib pilots notes say that it is quicker to make a turning dive onto a target passing below you in the opposite direction than to roll inverted and pull through, presumably because of the fairly slow roll rate. Its not the tail chase pushover case we are talking about but based on that and the earlier barrel roll comments I have found it very effective to barrel or corkscrew down into the dive in a tail chase with the slightest stick-back as it maintains +ve G and I think is quicker than the pushover.

[Viper2000]

Originally Posted by klem

Hi Viper,

well I think I followed that (amazed).

Just two points.

1. You estimate it takes 1/4 second for the small chamber to refill as it heads towards Rich cut but previously you said the entry holes were sized to permit the full power demand which you felt would contribute an inlet hole emptying under -G in 0.75 seconds (along with the emptying due to engine demand). So wouldn't it take 0.75 seconds to refill the small chamber through the inlet holes as it heads towards Rich cut?

The holes must be bigger than required for max engine demand otherwise the rich cut couldn't happen. The fuel pump is providing fuel at roughly 2.4 times maximum engine demand (2 pumps each sized for 120% of the requirement). I rounded to ¼ second because the 0.75 second figure was also an estimate, so there's not much point being over-precise; in the end the diagram is somewhat schematic, and only shows one of the 2 jets. There's also obviously the float adding depth to the float chamber, so the error bars on the calculation are quite large. But having said that, I think that the results are in pretty good agreement with the available data, so the chances are that it's not a million miles away from the truth.

Originally Posted by klem

2. I didn't understand the part under sudden -G where you said "as soon as the fuel hits the top of the float chamber, the float will instantly float downwards". Why would the float float downwards when it is being held to the top by the raised fuel surface?

Once the fuel hits the top of the float chamber, the situation looks like the "Negative G conditions" part of the figure I uploaded; remember that local acceleration is pointing upwards and so buoyancy pushes the float down.

To put it another way, if the float chamber was huge and you were sat in an inflatable boat inside it when negative g was applied, your experience would be:

1. weightlessness, with the ceiling appearing to rush towards you
2. headache from hitting the ceiling!
3. then you'd start floating "up" towards the surface of the liquid again; but upon reaching the surface you'd realise that the floor and ceiling had changed places...

Originally Posted by klem

I think your estimates fit in with the info I was given by the MkI Hurricane pilot at reduced G where he felt it did not cause a problem down to about 0.3G with around a 2 second delay before engine response. His rich cut recovery from that he estimates to take about 1 second which is pretty close to your 1.5.

The 1.5 second figure is from RR; it's approximate and it's predicated upon the assumption of negative g rather than reduced positive.

In case of reduced positive g the rich cut won't last as long because the equilibrium is just that the float will sit higher. So there's more fuel than the 1 g equilibrium, but it's not like the negative g case where the float chamber is literally filled to overflowing. As such, recovery would be quicker, because the rate at which the engine can suck away the excess fuel is fixed for any given rpm and OAT (since the supercharger is supersonic and therefore the non dimensional flow passing though its diffuser is fixed if you want to be technical about it).

So the figures line up quite nicely.

Originally Posted by klem

btw for interest and on a parallel topic, the Spitfire MkIa/Ib pilots notes say that it is quicker to make a turning dive onto a target passing below you in the opposite direction than to roll inverted and pull through, presumably because of the fairly slow roll rate. Its not the tail chase pushover case we are talking about but based on that and the earlier barrel roll comments I have found it very effective to barrel or corkscrew down into the dive in a tail chase with the slightest stick-back as it maintains +ve G and I think is quicker than the pushover.

In the end, this sort of thing turns into a parametric study because there are quite a lot of factors involved.

But if you're unconstrained by other threats and have the necessary energy then it's probably best of all to loop and roll off the top, because you'll exit the manoeuvre with at least as much energy as you came into it with.

But in the end this is just another way of restating the energy vs angles/ lead vs lag tradeoff, and the best option will always be a function of the geometry.

The bigger the height difference, the more attractive the idea of going straight into the vertical becomes.

[klem]

Got it

For the float I forgot the G had reversed to upside down.

its too sensitive definitely...even a very gentle push forward with the stick and engine cuts...

[IvanK]

The documented value in RAE documentation specifically investigating this problem is cutout onset at 0.1G ...."i.e. at accelerometer readings of less than 0.1g"





The evolution of the cutout and time taken for recovery is also well documented in AVIA 18/1281 Tests of RAE devices for the reduction of "Negative G" engine cutting on merlin engined fighter aircraft" Though specifically looking at various cutout reduction methods some good info on cut duration and recovery in there with various amounts of negative G application.

Both these documents are available at the UK National Archives. The devs have copies of both these documents.

[Crumpp]

http://forum.1cpublishing.eu/showpos...0&postcount=36

[Scavenger]

Hi! what about Tilly orifice? Do you have any data about minimal G with this improving?