davparlr
Senior Master Sergeant
Apparently, water injection also improves mass flow by cooling the air, which, by itself, improves thrust.
Bf-109 vs Spitfire vs Fw-190 vs P-51
- Thread starter Soren
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kool kitty89
Senior Master Sergeant
Yes, the "intercooler" effect, mentioned above, in post 456 and 457. (an effect still useful at higher altitudes, where increased boost isn't obtainable)
drgondog
Major
Those numbers look good to me, Max - now stick 60 quarts of oil and ?? of coolant (CRS), plus 170 pounds of pilot and equipment (Chute/dingy pac/heavy boots, huge watch and fighter pilot ego) into the ship
drgondog,
You bring up some good points. The source I used states basic weight. Basic weight usually consists of all oils, coolant and anything else which is not consumed or disposed of during flight.
Add in the guns, ammo, wing racks and fuel and you get what I posted. Your right I need to add the weight for the pilot and his garb. This usually is taken as 200 lbs., so we're up to 10,000 lbs. Normal take off weight is usually given at 10,100 lbs. So this is in the ballpark. I was just trying to show the weights for different amounts of guns, ammo and fuel.
If you want to add the other items, which I'm postive are already included in the above figure:
There is 21.2 gal. of oil which aircraft oils now a days is 7.5 lb/gal. = 161.25 lbs., coolant for the engine is 16.7 gal at 8 lb/gal. = 139.4, for the after cooler 4.8 gal at 8 lb/gal. = 40.1. Total would be 340.75 pounds.
You bring up some good points. The source I used states basic weight. Basic weight usually consists of all oils, coolant and anything else which is not consumed or disposed of during flight.
Add in the guns, ammo, wing racks and fuel and you get what I posted. Your right I need to add the weight for the pilot and his garb. This usually is taken as 200 lbs., so we're up to 10,000 lbs. Normal take off weight is usually given at 10,100 lbs. So this is in the ballpark. I was just trying to show the weights for different amounts of guns, ammo and fuel.
If you want to add the other items, which I'm postive are already included in the above figure:
There is 21.2 gal. of oil which aircraft oils now a days is 7.5 lb/gal. = 161.25 lbs., coolant for the engine is 16.7 gal at 8 lb/gal. = 139.4, for the after cooler 4.8 gal at 8 lb/gal. = 40.1. Total would be 340.75 pounds.
drgondog
Major
Looks like we still have to nail difference between empty and basic for the build up... but should have similar facts for all the other birds as well. 300-400 pounds 'extra' can be huge at the edge of a stall.
drgondog
Major
Thanks Juha - what was your source for the detail 'delta'between the two states? And I would assume trapped hydraulic fuids and coolant also fit in the basic profile?
- Thread starter
- #469
Davparlr,
I don't believe the P-51D could do 4,200 ft/min for the simple reason that the P-51B managed 4,380 ft/min at 9,350 lbs. The P-51D weighes roughly 400 to 500 pounds more, which will atleast rob away 400 ft/min or more. The climb rate of the P-51D at 75" Hg is ~4,000 ft/min.
Note that because only of the extra drag caused by the ETC-504 rack the Fw-190 Dora-9 lost 1.5 m/s in climb rate, climbing at 22.5 m/s without the ETC rack and 21 m/s with it.
I don't believe the P-51D could do 4,200 ft/min for the simple reason that the P-51B managed 4,380 ft/min at 9,350 lbs. The P-51D weighes roughly 400 to 500 pounds more, which will atleast rob away 400 ft/min or more. The climb rate of the P-51D at 75" Hg is ~4,000 ft/min.
Note that because only of the extra drag caused by the ETC-504 rack the Fw-190 Dora-9 lost 1.5 m/s in climb rate, climbing at 22.5 m/s without the ETC rack and 21 m/s with it.
davparlr
Senior Master Sergeant
The basic weight of the P-51B is 7325 lb. and the basic weight of the P-51D is 7673 lbs or a difference of 348 lbs. However, on closer examination of the calculated charts showing difference of performance with weight, 4000 ft/min would be a valid estimate.
All the performance data used for the P-51B and P-51D is with wing racks also.
All in all, I think we are carving a tooth pick with an axe. I am sure all variances are within overall calculations, system performance, and test errors. Both planes could climb very well. I think the P-51D and the Fw-190A-5 were pretty well evenly matched from SL up to 25k ft., with the Fw-190 having a better roll rate and a slight edge in climb, while the P-51D was slightly faster and could dive faster. The P-51B had a better performance edge.
- Thread starter
- #471
drgondog
Major
Soren - the first P-51B-1 was tested against the P-47D-10 (and P-38J-5, P-29N-0 and P-40N) at Eglin, using 61" @ 3000rpm in September, 1943...
The report states that
"The diving charcteristics of the P-51B are far superior to those of any other fighter type airplane..it is exceptionally easy to handle and requires very little trimming. The P-51B dives away from all other fighters except the P-47D, against [which] the P51B jumps several hundred feet ahead in the initial pushover and then holds that position,neither gaining nor losing distance"
source - ARMY AIR FORCES BOARD PROJECT NO. (M-1) 50
TITLE: TACTICAL EMPLOYMENT TRIALS ON NORTH AMERICAN P-51B-1 AIRPLANE
What is the source you are using that a 190A-5 had an initial dive faster than the P-47(D-10 and above) or any version of the 51?
- Thread starter
- #473
drgondog
Major
It would be fair to say the initial acceleration in the dive was pretty equal.. but one of the reasons the 47 didn't out dive the 51 is simply the 51 was a cleaner airframe (than both the 190 and the 47). I would imagine the 190D and 51 were closer than the 190A and 51 - all the way from push over.
The RAE tests also had the 51B out diving everything in that test - Allied and LW -
- Thread starter
- #475
Gentlemen, I am new to the 'aeronautics field' and was examining this 'thread' as a mechanism to understand some of the basics, mainly to establish a more simplistic model for the comparison of aircraft in this time period. The following is an anomaly that perhaps you help clarify for my understanding.
1) In post #158,
"Anyway I did the calculations on L/D ratio for you (Added the Ta-152H for comparison):
Ta-152 H-1:
(1.62^2) / (pi * 8.94 * .83)
1.62 / 0.112580856
_______________
L/D ratio = 14.38"
the L/D ratio appears to be only the ratio of the coefficient of Lift vs the coefficient of Drag for induced drag only (evaluating the mathematics as presented). I had expected the L/D ratio to include parasitic drag so as to present a more complete representation of the performance of the wing.
Your assistance in helping me with my confusion on this item would be greatly appreciated.
1) In post #158,
"Anyway I did the calculations on L/D ratio for you (Added the Ta-152H for comparison):
Ta-152 H-1:
(1.62^2) / (pi * 8.94 * .83)
1.62 / 0.112580856
_______________
L/D ratio = 14.38"
the L/D ratio appears to be only the ratio of the coefficient of Lift vs the coefficient of Drag for induced drag only (evaluating the mathematics as presented). I had expected the L/D ratio to include parasitic drag so as to present a more complete representation of the performance of the wing.
Your assistance in helping me with my confusion on this item would be greatly appreciated.
drgondog
Major
L/D ratio's are typically discussions about wings
When you move into a force diagram to discuss the aircraft as a whole, you need to describe all the forces acting on the airframe.
So, the Horizontal Forces look like Thrust = Induced Drag + Parasite Drag for an airframe in static balance horizontally. Thrust> Induced+Parasite Drag for an a/c with excess power for the starting velocity and is in acceleration
Induced Drag is highest at low speeds (in comparison with Parasite Drag), and Parasite Drag is highest at high velocities. The huge component of 'Parasite' drag at high speeds is the Propeller, but at top speed the Cd0 which is the zero lift drag of the airfoil, plus other effects like wake drag in compressibility regimes also become major contributions.
Another factor with using the CD values from the tables is that the CD is a nice stable constant value until some, as low as .5 Mach, start to approach compressibility effects of air. The MCrD is the value at which the Cd increases by .002 over the steady Cd
The Vertical Forces are Weight and Lift of the total Wing/Body combination.
For WWII fighters the Wing Lift is by far the dominant value in the Vertical Free Body diagram.
If we ever get around to agreeing how we get accurate Thrust vales for the Hp/Prop values for each aircarft - at different altitudes.. we are then positioned to calculate both rate of turn when entering into a horizontal (no altitude loss) turn at max speeds, but for the case of one a/c which has a higher speed capability, demonstrate the energy availablity remaining when the slower a/c has maxed it's turn and bank angle for level flight.
We can actually do a decent job at sea level for all of these ships but it gets complicated as you start to look at the Hp curves for each ship as a function of altitude and design boost/blower stages.
Another member (Crumpp) put together a series of profiles much earlier which does a really nice job for one set of Hp values, then converting them to Thrust before proceeding on to the bank angle and G force calcs before the a/c begins to stall and lose horizontal flight.
The problem with all of these ships in turns of 'ease of calc' Is that while Most Propeller designs were pretty efficient and comparable to each other the values of .80-.85, while reasonable for efficiencies in these calculations -they ALL degrades with Mach > .5 - and all of theses fighters were capable of those velocities at max Hp.
Thanks for your quick response and the outline of the 'project' goals. After 'chasing the math' a little longer, and reading more on the subject, I was able isolate the problem. For what it is worth, I'll share my 'insight'.
L/Dmax, being defined as when Drag is at its minimum, I plotted the graph with the assumptions that L=W. At this point apparently, the 'curves' for Induced drag and Parasitic Drag 'cross'. This leaves Di = Do (I 'cheated' a bit here, in that I am using the publish Cd0 of 0.0163, assuming all other factors are small enough to be insignificant at the '4 significant decimal' level). The results are as follows:
Given W=9200, Cd0=0.0163, b=37, S=235, L/Dmax=14.6, rho=0.002378 (i.e. @SL)
Using:
L/D=.5*Sqrt(pi*A*e/Do) to calculate e as 0.7455
Di=2L^2/(pi*rho*e*b^2v^2)
Do=.5*rho*S*Cd0*v^2
At the intersection of the graphs, Do=Di=316.64 at a velocity of 262.4 ft/sec (178.9 mph)
Using:
CL=W/(.5*rho*S*v^2)
At the same intersection, CL=0.4783.
I don't see a value of CL=1.35 on the graph until v gets much closer to stall (approx 100mph). So, I am assuming the values posted, that were part of my confusion, are CLmax.
On the subject of turning rate, if I understand you to say you were planning to calculate that for no altitude loss at maximum speed, would that not simply be a function of maximum speed at maximum G load. At maximum speed would not your power already be maxed out so that it would have to be a steady state turn?
L/Dmax, being defined as when Drag is at its minimum, I plotted the graph with the assumptions that L=W. At this point apparently, the 'curves' for Induced drag and Parasitic Drag 'cross'. This leaves Di = Do (I 'cheated' a bit here, in that I am using the publish Cd0 of 0.0163, assuming all other factors are small enough to be insignificant at the '4 significant decimal' level). The results are as follows:
Given W=9200, Cd0=0.0163, b=37, S=235, L/Dmax=14.6, rho=0.002378 (i.e. @SL)
Using:
L/D=.5*Sqrt(pi*A*e/Do) to calculate e as 0.7455
Di=2L^2/(pi*rho*e*b^2v^2)
Do=.5*rho*S*Cd0*v^2
At the intersection of the graphs, Do=Di=316.64 at a velocity of 262.4 ft/sec (178.9 mph)
Using:
CL=W/(.5*rho*S*v^2)
At the same intersection, CL=0.4783.
I don't see a value of CL=1.35 on the graph until v gets much closer to stall (approx 100mph). So, I am assuming the values posted, that were part of my confusion, are CLmax.
On the subject of turning rate, if I understand you to say you were planning to calculate that for no altitude loss at maximum speed, would that not simply be a function of maximum speed at maximum G load. At maximum speed would not your power already be maxed out so that it would have to be a steady state turn?
drgondog
Major
The comparisons, to be rational, need to be done at different altitudes - say SL, 10K, 20K and 30K. Each of the a/c mentioned have different top speeds at different altitudes. Some are superior in speeds at all altitudes - say a P-51B-15 versus a Me 109G-6 or and Fw190A-5.
So to further complicate the comparisons I would think that once you have the Thrust for a P-51B-15 with a 1650-7 Merlin at 75" boost at 22,500ft we are running at ~ 442mph.. but a 190A-5 at that same altitude does perhaps 410mph.. so to compare the two the first set of comparisons have both ships entering at 410 which gives the 51B-15 excess Thrust available when the turn starts.. (it doesn't need to run at full boost to keep up) and consquently more potential energy as the Horizontal turn starts.
So you start the engagement at a Given Velocity and Given Altiude and a Given HP/Boost condition from Flight Test, then an empirical calculated Thrust based on Mfr Spec and an assumed Efficiency - and you solve for Parasite Drag (= Thrust - Induced Drag) in level flight. That presumably will be the maximum Parasite Drag attainable in level flight because the airplane can go no faster in level flight.
Between that specific Level Flight/Zero Bank Force Diagram and the final set of conditions (Each airplane has reached a speed and maximum bank angle in which the Lift Vector exactly matches the Weight vector.. you have a specific solution... but because the relative angle of attack, the CL and the velocity have all changed to that equilibrium point both the Induced Drag and Parasite Drag WILL have changed while Thrust theoretically remained 'constant' (we know it changes from relative angle of attack also - but this is too complicated to go there)
So, at the end of the day - if you really want to yank some hair remember that a Prop is a mass flow device designed to maximize the difference in Pressure behind the fan interface, from the Free stream Pressure in front of the fan interface, and it depends on the Rho at that altitude. The actual Force equation is (F=.5 x Rho x Aprop (V>>2-V0>>2) where V=velocity of accelerated flow aft of Prop and V0 = Velocity of Free Steam in front of Prop... This is the True Thrust but we can't easily get to V0.
Got it. I can see the difficulty with the need to map over the entire velocity/altitude scale. Am I correct in assuming that boost figures are only necessary when the HP required includes it (i.e., to simplify absurdly, you don't push the RED GO FAST button at 50% throttle)?
