Shortround6
Major General
Actually Kestrels were available with .632, .553 and .477 gears depending on intended use.
How many were sold with each ratio I have no idea.
Advance Ratios and Gear Ratios
- Thread starter Zipper730
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How many were sold with each ratio I have no idea.
In many cases, the upgrade for the DeHavilland 2 pitch prop ( Which was basically a Ham-Stan, IIRC) was fairly simple - removing the 2 position pitch change mechanism, adding the Constant Speed Governor, adding the plumbing for the engine oil used to run the prop, and adding the controls.
The Book Time for the upgrade was 20 man-hours.
By Adlertag, All frontline Spits, and nearly all Hurricanes had been upgraded.
The performance improvement, particularly in climb, made them act like all new airplanes
GregP
Major
Horespower = (torque x rpm) / 5,252
So, at 5,252 rpm, torque = HP. Below 5,252 rpm, torque rules. Above 5,252 rpm, HP rules.
Most of the V-12s of WWII were 3,000 rpm engines, and torque is what counted. You cannot measure HP. You measure torque and calculate HP.
So, at 5,252 rpm, torque = HP. Below 5,252 rpm, torque rules. Above 5,252 rpm, HP rules.
Most of the V-12s of WWII were 3,000 rpm engines, and torque is what counted. You cannot measure HP. You measure torque and calculate HP.
- Thread starter
- #24
Was there any possibility of the Merlin-C being built for a ratio of 0.420? I'm curious because it seemed to be mentioned on a website that describes the first flight of the Hawker Hurricane.
Here it is: Hawker Monoplane F.36/34 K5083 Archives - This Day in Aviation
Looking at WWII Aircraft Performance, it seems that I have the following figures for the merlin variants
Here it is: Hawker Monoplane F.36/34 K5083 Archives - This Day in Aviation
Looking at WWII Aircraft Performance, it seems that I have the following figures for the merlin variants
- XX/20: 0.42:1
- 24: 0.42:1
- 45: 0.48:1 (not sure if this is a round up of 0.477:1)
- 61: 0.42:1
- 63 & 64: 0.477:1
- 66: 0.477:1
- 68-69 & 266P: 0.42:1 or 0.477:1
- 70-71: 0.477:1
- 72-73: 0.42:1
- 76-77: 0.42:1
- 102: Unspecified in chart
- 113-114: 0.42:1
- 120: 0.375:1 (co-axial propellers)
- 130: 0.42:1
Shortround6
Major General
Why???
The 0.477 gears (yes the 0..48 is rounded up) were used on fighters and single engine planes,
The O.42 gears were used on bombers or multi engine planes to swing larger diameter propellers.
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- #26
Not sure honestly, but there were several variants of the Merlins and I was trying to whip up an advance ratio chart for as many fighter planes as I can find data on as you can use the data to produce thrust figures.
I need to know the speed, the propeller diameter, the engine RPM, and the gear-ratio to produce the advance ratio figures.
The reason I was curious about the 0.420:1 ratio was this link here: Hawker Monoplane F.36/34 K5083 Archives - This Day in Aviation
The early Hurricanes used a two-bladed watts propeller, so I assume things were pretty similar.
- Thread starter
- #27
This is probably going to be a dumb question, but where do you find torque values for the Merlin variants I listed. World War II Aircraft Performance doesn't list these figures...
Shortround6
Major General
Why do you need them?
HP = Torque x RPM ÷ 5252
If you know HP you can calculate for torque.
Engine dynamometers measure torque directly and the readings have to be converted to horsepower ( some radials had torque meters built into the reduction gears and flight engineers could read torque on a meter.
1030hp Merlin III is making 1796.32 ft lbs of torque at 3000rpm.
At take-off it is making 880hp so the torque is 1534.72ft/lbs. Cylinders aren't getting the full effect of the 6 lbs of boost because the throttle is closed down so much choking the engine.
And so on, if you know the HP at a certain rpm then you know what kind of work the engine can do, effort over time.
HP = Torque x RPM ÷ 5252
If you know HP you can calculate for torque.
Engine dynamometers measure torque directly and the readings have to be converted to horsepower ( some radials had torque meters built into the reduction gears and flight engineers could read torque on a meter.
1030hp Merlin III is making 1796.32 ft lbs of torque at 3000rpm.
At take-off it is making 880hp so the torque is 1534.72ft/lbs. Cylinders aren't getting the full effect of the 6 lbs of boost because the throttle is closed down so much choking the engine.
And so on, if you know the HP at a certain rpm then you know what kind of work the engine can do, effort over time.
A couple of quick thoughts -
When you're calculating torque at the propeller, don't forget to divide by the gear ratio - the Horsepower is constant, but the rotational speed, (and thus torque) is different,
It's not so much that you're choking the engine per se, but the supercharger is compressing to what would be, at full airflow, an intake manifild pressure of anout 10-11 lbs, and you're paying the cost in horsepower to drive the supercharger.
As for the previous question on the selection of gear ratios - what it boils down to is that they are tryiing to optimize peak propeller efficiency at a particular Advance Ratio (Ratio between rotational speed of the propeller, and the forward speed of the airplane) to suit the flight regime expected for the airplane. By varyiing the propeller RPMs at, say, takeoff,cruise or climb speeds, you can supply more thrust per unit of power at those speeds, either getting more thrust (Takeoff and Climb), or requiring less engine HP (Cruise)
Shortround6
Major General
They were also trying to suit the propeller slip stream to the job, that may be badly stated.
for a high speed fighter a lower volume of air at a higher speed works very well for high speed flight (small diameter prop, high advance ratio, high rotational speed)
using the same reduction gear and prop on a much slower and much heavier bomber doesn't work so good. A larger diameter prop (with different gearing to keep the prop tip speed nearly the same) will move a large mass of air per revolution of the prop and give more "thrust" or at least the larger amount of air at a slower speed gives a better match to the speed of the aircraft.
I hope that makes a bit more sense.
for a high speed fighter a lower volume of air at a higher speed works very well for high speed flight (small diameter prop, high advance ratio, high rotational speed)
using the same reduction gear and prop on a much slower and much heavier bomber doesn't work so good. A larger diameter prop (with different gearing to keep the prop tip speed nearly the same) will move a large mass of air per revolution of the prop and give more "thrust" or at least the larger amount of air at a slower speed gives a better match to the speed of the aircraft.
I hope that makes a bit more sense.
Shortround6
Major General
In the case of the Merlin III (and it is one of the worst at this) it could have gone to 17-18lbs of boost at sea level if the throttle was open fully, so they were choking it down quite a bit.
The supercharger really didn't care, power wise, it was turning the same rpm regardless of how opened or closed the throttle plate was. Yes it was moving more air weight wise with the throttle open. The killers to power were the pumping losses, the supercharger trying to suck air through the throttle assembly and heat, the warmer air at sea level meant they had to restrict things to avoid detonation.
The Merlin VIII in the Fulmar picked up just about 200hp for take-off with a lower supercharger gear because, as you say, less power was need to drive the supercharger (it needed about 54% as much power), the air was heated proportionally less meaning it was denser and the throttle was a lot wider open., less pumping loss.
- Thread starter
- #32
Sorry, I realized it after I posted...
So if I know the HP at one point across the performance envelope, I can calculate all the others?
While this is probably not necessary, but a pure hypothetical: The prototype listed a critical altitude of 16200' which is also where RPM peaked. For some reason, which I cannot understand, 1029 HP @ 11000 feet is listed as being the horsepower the engine is rated for. Did it gain 5200 feet of ram? Or is there something I don't get?
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wuzak
Captain
No, you can get the torque at that point.
Typically, a WW2 aero engine graph does not show power against rpm, but altitude. There may be several lines depicting different operational modes, engine speeds and boost.
wuzak
Captain
You didn't answer his question: why do you need them?
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- #36
Okay, I have the following figures available for K-5083, it's just a test run to make sure I'm getting numbers right.
From what I've got, we should see the
Horsepower = Torque * RPM / 5252
1029 =.....x..... * 2600(0.477) / 5252
1029 =.....x..... * 1240.2 / 5252
1029 = 1240.2x / 5252
1029 = 0.236138613861386x
x = 4357.610062893084319 ft-lbs
I showed how I arrived at the result, but something seems wrong: They look hinky to me
Shortround6
Major General
first, keep things simple. get rid of the reduction gear number, while it might tell how much torque is being delivered to the prop it is a needless complication at this point.
Torque is the twisting force acting on the Crankshaft at any given moment in time. Horsepower is the force over a period of time.
Torque is the result of pressure in the cylinders pushing down on the pistons and acting through the connecting rods to turn the crank.
at 11,000ft (which is immaterial at the moment)
1029hp = Torque x 2600rm /5252
1029 x 5252= torque x 2600
5,404,308= torque X 2600
5,404,308/2600= torque
2,078.58 = torque.
Now that you know what the engine is doing you can figure in the prop reduction gear to get the torque going to the propeller.
now please note that the Merlin at 16, 250 ft making 1030 hp (identical for our purposes) was only making 1796.32 ft lbs of torque at 3000rpm.
It wasn't developing quite the same pressure per square in of piston area in the cylinders. but the increase in rpm will allow it to do the same amount of work per minute.
Torque is the twisting force acting on the Crankshaft at any given moment in time. Horsepower is the force over a period of time.
Torque is the result of pressure in the cylinders pushing down on the pistons and acting through the connecting rods to turn the crank.
at 11,000ft (which is immaterial at the moment)
1029hp = Torque x 2600rm /5252
1029 x 5252= torque x 2600
5,404,308= torque X 2600
5,404,308/2600= torque
2,078.58 = torque.
Now that you know what the engine is doing you can figure in the prop reduction gear to get the torque going to the propeller.
now please note that the Merlin at 16, 250 ft making 1030 hp (identical for our purposes) was only making 1796.32 ft lbs of torque at 3000rpm.
It wasn't developing quite the same pressure per square in of piston area in the cylinders. but the increase in rpm will allow it to do the same amount of work per minute.
- Thread starter
- #38
Okay, that definitely makes more sense out of things, I computed the same results as you did.
How do you do that?
Now I'm curious -- how do you calculate torque when the RPM and horsepower are different? It's usually easier to solve for one variable than shitload of them, and math was never my strongpoint -- this might be obvious to you
Shortround6
Major General
The problem is that, like I said, the torque is the result of the pressure in the cylinders (pounds per sq in acting on the piston tops) puhsing down through the connecting rods that turn the crank. So we have a crap load of variables. If you are turning 2600rpm and using 4lbs of boost you get one pressure in the cylinders,if you are using 2600rpm and 6lbs boost you get a different pressure in the cylinders (roughly and that is the problem) 10% more pressure and if you are using 3000rpm and the same boosts the pressures will be slightly different.
It gets even worse because this only works at one altitude or air density. Pressure in the manifold is related to the mass of air (weight of the air) but it is not a 100% correlation.
6lbs boost at sea level is not the same mass of air per minute as 6lb boost (or the same intake pressure ) at 10,000ft.
I still don't know what you are trying to do with this, The best props in WW II were only about 80% efficient which means 20% of the power going to them (no matter how you measure it) was wasted. Unless you know that prop was 79% efficient at a certain speed and altitude and not 77% efficient 20-30 hp can get lost real easy. bigger discrepancies just get worse.
Prop efficiency was all over the place as you need different prop diameters for different altitudes (generally larger diameters) and you need smaller diameter propellers for high speed.
It gets even worse because this only works at one altitude or air density. Pressure in the manifold is related to the mass of air (weight of the air) but it is not a 100% correlation.
6lbs boost at sea level is not the same mass of air per minute as 6lb boost (or the same intake pressure ) at 10,000ft.
I still don't know what you are trying to do with this, The best props in WW II were only about 80% efficient which means 20% of the power going to them (no matter how you measure it) was wasted. Unless you know that prop was 79% efficient at a certain speed and altitude and not 77% efficient 20-30 hp can get lost real easy. bigger discrepancies just get worse.
Prop efficiency was all over the place as you need different prop diameters for different altitudes (generally larger diameters) and you need smaller diameter propellers for high speed.
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- #40
