LLv34_Flanker Posted May 22, 2014 Author Posted May 22, 2014 S! Finnish tests can be read on Kurfürst's page: http://kurfurst.org/Performance_tests/109G_MT215/109G2_MT215_en.html 1
JtD Posted May 22, 2014 Posted May 22, 2014 But you say your data is from March, Helsinki, and that's not on Kurfürsts page. Unfortunately I don't have the report myself so I can't look it up.
SCG_Neun Posted May 22, 2014 Posted May 22, 2014 A very interesting read guys.....keep it coming and thanks for the hard work...
LLv34_Flanker Posted May 22, 2014 Author Posted May 22, 2014 S! Hmm..I have to cross check the data. Give me a few.
Holtzauge Posted May 22, 2014 Posted May 22, 2014 (edited) I think you are being a bit pessimistic about not knowing the conditions: If you look at what Flanker posted earlier we have most of the conditions: The aircraft was new or close to new. It was fully loaded with ammo and fuel. The trials took place in March. So based on the above, given that the aircraft was new it seems reasonable to assume that the engine was performing as expected and that the surface finish was according to specs. The second piece of info ties down the weight and from the third we can make a reasonable assumption (Helsinki in March) that the temperature was most likely 5-10 deg C and not a balmy 20 deg C when the trials took place. Now remember that we have previously talked about a temperature difference from a standard atmosphere 15 deg C to a Stalingrad winter temp of -15 deg C, i.e 30 deg lowered temp and still come up short in terms of power. I agree with JtD that the likely power boost for a 30 deg lowered temp is most likely closer to 10 than 30%: Don't know how JtD came to 10% but I came to a similar conclusion based on a power/alt curve I came across for the DB601N which showed the effects of ram pressure both in terms of increased critical altitude and also lowered power due to the temp increase of the air fed to the supercharger. From the diagram you can read that the power is reduced around 40 Ps if the ram pressure is 1200 Kp/m**2. The delta PS in the diagram is roughly constant with altitude (around 40 Ps) so I used this to calculate a SL speed of around 500 Km/h i.e. around M= 0.407. From this I calculated the stagnation temp T0/T=(K-1)/2*M**2 from which we can conclude that the stagnation temp is roughly 9.55 deg C higher than ambient. So a ballpark figure for the effects of temp on power would be 40/9.55=4.19 Ps/deg C. So if we now assume that the delta T is 30 deg C we come up with a power increase of 125.7 Ps. Given the SL power of the DB601N is 1200 Ps this gives us an increase of around 10.5%. But since the test was in March it was most likely not done at 15 deg C but probably at a 5 to 10 deg lower temp so the power increase when going to -15 deg C compared to the tests is most likely in the range 7 to 9% and not 30%........ While not a cast iron proof it certainly raises some questions and based on the evidence so far I would on the basis of the Finnish tests at 27 s be inclined to label 18 s 300 to 440 Km/h acceleration time ingame as overly optimistic but that is just my 2 cents....... Edited May 22, 2014 by Holtzauge
Holtzauge Posted May 22, 2014 Posted May 22, 2014 Found this in the Finnish report translation: Malmi 5 April 1943 at 12.15-13.05 Pilot : Kapt.P. Kokko At airfield : Atmospheric pressure = 747,6 mm Hg Temperature = +11,2° C Atmospheric pressure measured with onboard manifold pressure gauge = = 1,018 ata
Crump Posted May 25, 2014 Posted May 25, 2014 The Finnish data is converted to a standard model and is not the issue. CINA in the 40's has some differences but for subsonic incompressible flow it is close enough to many of the standard models used today. It is the games atmospheric conditions that are unknown. Without known atmospheric conditions.....there is nothing to conclude outside of general trends.
Crump Posted May 25, 2014 Posted May 25, 2014 Found this in the Finnish report translation: Malmi 5 April 1943 at 12.15-13.05 Pilot : Kapt.P. Kokko[/size] [/size] At airfield :[/size] Atmospheric pressure = 747,6 mm Hg[/size] Temperature = +11,2° C[/size] Atmospheric pressure[/size] measured with onboard[/size] manifold pressure gauge = [/size] = 1,018 ata[/size] That is the conditions the raw data was measured under......yes. It is NOT the conditions the performance listed in the report will be realized under....... That raw data was converted to performance under a CINA standard atmosphere. That is kind of standard in flight testing. While there a examples of raw datum listed in some flight reports from the World War II era it is rare. Raw datum is inconvenient and easily leads to gross misconceptions.
RAY-EU Posted May 31, 2014 Posted May 31, 2014 Seems very Good historical data review Interesting: I think they will repas and if is wright will be corrected in a later edition , Like happens in IL2 Sturmovik : Forgotten Battles and them there was Il2 sturmovik: Forgotten Battles 2 . Correct ... Remember that right now is in mode Alfa... IL2 Sturmovik BOS is the best and most realistic comercial airplane simulator in the world , only military simulators can compare but are not Comercial and you can not buy or to use you have to be a pilot in the top gun of Nevada ...Imagine and you are inmerse in The WW2. IL2 Sturmovik BOS nothing Compears is the Best I have never seen somethig as much realistic . But in this Word nothing is Perfect ...
Holtzauge Posted May 31, 2014 Posted May 31, 2014 (edited) I agree that the IL2 series is a great simulator and while I don't have BOS yet I fly the IL2 4.12 mods sometimes. So my input here is in no way a critizism but an effort to help improve the sim and get the performance numbers as right as possible. I have a double interest in this: One is to verify my C++ modelling and that is where the info like the one on the Me109G2 acceleration comes in handy since I can use this to see if the C++ modelling is correct. Right now it's looking pretty good! The other is to help get sims (BOS and others) as accurate as possible because the closer it is to reality the more fun it is to fly. I mostly fly 190´s and 109's so I have no interest in porking them. However, I don't want to be flying some uber plane because the fun of flying a sim is enhanced if the planes are as historically correct as possible right? ;-) Anyway, will be interesting to see if some new info about the test conditions on MT-215 will be posted here to see if this can be used to deduce some more facts about why the performance figures are as they are. Especially the 24.4 m/s climb rate seems strange...... Edited May 31, 2014 by Holtzauge
Crump Posted June 8, 2014 Posted June 8, 2014 (edited) agree with JtD that the likely power boost for a 30 deg lowered temp is most likely closer to 10 than 30%: It is very easy to estimate using basic aerodynamic atmospheric rules. We will JUST look at a temperature change of 30 degree Celsius and not consider the higher pressure such conditions produce.... Therefore delta(Pressure ratio) = 1 Theta is simply temperature at altitude in Rankin / Temperature at sea level Rankin. 465 degree Rankin = -15 degrees Celsius 519 degrees Rankin = 15 degrees Celsius Theta = 465 / 519 = .896 Density ratio(sigma)= delta/theta = 1/.869 = 1.1163 Increase in Brake Horsepower = Bhp(altitude) = Bhp(sealevel) *(sigma-.1)/.9 Bhp(alt)= 1400*(1.163-.1)/.9 = 1581 Bhp a 13% increase in power... In terms of thrust production at 100 KEAS... {(1400 *.85)*325}/100 = 3867lbs of thrust {(1580 *.85)*325}/100 = 4365lbs of thrust Ballpark the velocity increase; 4365lbs - 3867lbs = 468lbs SQRT(468) = 22 Knots velocity increase Which represents ~22 Knot or 41 Kilometer per hour increase in Vmax under standard conditions. With actual pressure conditions, you can expect that increase in thrust to rise. Edited June 8, 2014 by Crump
Crump Posted June 8, 2014 Posted June 8, 2014 a 13% increase in power... BTW, Cold Dry airmasses typically produce barometric pressures in excess of 31.00inHg. At just 30 degrees below standard at 31inHg, represents a 20% increase in power. Bump that up according as the pressure increases over 31..... In otherwords, a 30% increase in power is not out of the ballpark at all depending on the atmospheric conditions.
qtStamphth Posted June 8, 2014 Posted June 8, 2014 (edited) You all forget one thing: The tests done from a captured plane cannot be trusted entirely because it is done on one aircraft only! This one aircraft might have an engine defect or whatever other issue/slight mechanical disarray that keeps it from performing like the majority of it's fellow planes. I guess ya just have to hope the captured plane has been in good keep maintenance wise and production wise so as to be representative of the wider fleet of the same exact plane model. Edited June 8, 2014 by qtStamphth
Crump Posted June 8, 2014 Posted June 8, 2014 The tests done from a captured plane Allied power not captured. The Finns purchased Bf-109's from Mtt. Came with a full support package although it was an export variant.
LLv34_Flanker Posted June 9, 2014 Author Posted June 9, 2014 S! Crump. The Finnish Bf109G-2's were no special export version, identical to airframes supplied to Luftwaffe. Some of the Bf109G-6's received later were new, but some were rebuilds and quality varied.
Cpt_Branko Posted June 9, 2014 Posted June 9, 2014 In terms of thrust production at 100 KEAS... {(1400 *.85)*325}/100 = 3867lbs of thrust {(1580 *.85)*325}/100 = 4365lbs of thrust Ballpark the velocity increase; 4365lbs - 3867lbs = 468lbs SQRT(468) = 22 Knots velocity increase Which represents ~22 Knot or 41 Kilometer per hour increase in Vmax under standard conditions. At 285 knots (representative of the 109's vmax at sea level), using also .85 efficiency for lack of better data, it would be something along the lines of 1357 lbs versus 1531 lbs, and expected velocity increase being around 13 knots. Using MW50 to get about a 30% increase in power resulted in a 12% increase in speed; for a 12% increase in power 4% increase in speed is more plausible. 1
Crump Posted June 9, 2014 Posted June 9, 2014 (edited) At 285 knots (representative of the 109's vmax at sea level), using also .85 efficiency for lack of better data, it would be something along the lines of 1357 lbs versus 1531 lbs, and expected velocity increase being around 13 knots. Using MW50 to get about a 30% increase in power resulted in a 12% increase in speed; for a 12% increase in power 4% increase in speed is more plausible. The percentage power does not change but that has little to do with the fact a 30% increase in acceleration is not out of the ballpark. Just the temperature alone increases the power 13% not including pressure changes or the fact the engine itself is just going to make more power. Over the course of the thread, the 30% increase in acceleration somehow morphed incorrectly into requiring a 30% increase in power. Again, no conclusions can be made without knowing the atmospheric conditions represented in the game. Edited June 9, 2014 by Crump
Crump Posted June 9, 2014 Posted June 9, 2014 At 285 knots (representative of the 109's vmax at sea level), using also .85 efficiency for lack of better data, it would be something along the lines of 1357 lbs versus 1531 lbs, and expected velocity increase being around 13 knots. Using MW50 to get about a 30% increase in power resulted in a 12% increase in speed; for a 12% increase in power 4% increase in speed is more plausible. Think of it this way.... Force = Mass x Acceleration Acceleration is in units of time squared. For a 13% increase in force: 1.13^2 = 1.27 or a 27% decrease in time. In otherwords, the game is right and the Players just do not understand the effects of atmospheric conditions on aircraft performance.
Cpt_Branko Posted June 9, 2014 Posted June 9, 2014 (edited) Think of it this way.... Force = Mass x Acceleration Acceleration is in units of time squared. For a 13% increase in force: 1.13^2 = 1.27 or a 27% decrease in time. You can't do this, though - squaring the difference in force is not legitimate. The difference is linear with increase in net force. The kicker is that increase of power by 10% is not an increase in net force accelerating the plane by 10%, but rather by a higher percentage. Since acceleration is about (T-D) / mass, you are right - acceleration will increase more rapidly if thrust increases and mass is constant. For instance, plane 1 has, say: (100 - 50) / 10 = 5 Increase T by 10%: 110-50 / 10 = 6 (a 20% increase in acceleration) But yes, it's not really possible to judge FM performance without either knowing atmospheric conditions or having test flight mode with standard atmosphere. Edited June 9, 2014 by Cpt_Branko 1
Crump Posted June 9, 2014 Posted June 9, 2014 (edited) You can't do this, though - squaring the difference in force is not legitimate. The difference in acceleration would hypothetically be linear with increase in force. Sure you can. You are just using the math to ballpark what you should see. It is done all the time. No it does not represent the actual values nor is it a detailed analysis. It just simply explains the trend you should see. A small increase in force equates to a large reduction in time because the relationship is squared. It won't be practically because at speeds below maximum, acceleration is about (T-D) / mass, and acceleration will increase more rapidly then linear. Exactly. So we are back to the original conclusion. In otherwords, the game is right and the Players just do not understand the effects of atmospheric conditions on aircraft performance. A 30% increase in acceleration is well within the confines of reality depending on the atmospheric conditions. Edited June 9, 2014 by Crump
Cpt_Branko Posted June 9, 2014 Posted June 9, 2014 It's possible, yes. Ability to testfly planes in standard atmosphere settings would be nice
Crump Posted June 9, 2014 Posted June 9, 2014 acceleration will increase more rapidly then linear. BTW and FYI....... The rectilinear motion equations work well for aircraft by using the derivative of the acceleration values. A 30% increase in acceleration is well within the confines of reality depending on the atmospheric conditions. This should be a 30% decrease in time not increase in acceleration, btw. Ability to testfly planes in standard atmosphere settings would be nice I agree. It would make life easier.
Cpt_Branko Posted June 9, 2014 Posted June 9, 2014 Well, how to put what I mean, using my meager paint skills (it's much easier to talk about this with a piece of paper): Anyway, since atmospheric condition impact both drag and power, I agree that it's impossible to say without knowing atmospheric conditions. 1
Crump Posted June 9, 2014 Posted June 9, 2014 (edited) Well, how to put what I mean, using my meager paint skills (it's much easier to talk about this with a piece of paper): Anyway, since atmospheric condition impact both drag and power, I agree that it's impossible to say without knowing atmospheric conditions. I understand exactly what you are saying. You are referring to what is termed the "speed stable" region of the power curve vs the "power stable" region of the power curve. Do you understand what I am saying? It is a very common technique to use the derivative of the acceleration values plugged into the rectilinear motion equations to determine aircraft acceleration distances covered and time required. Therefore the original conclusions of F=ma hold true with the value of time being squared in relation to force. Edited June 9, 2014 by Crump
Crump Posted June 9, 2014 Posted June 9, 2014 Simply put Cpt_Blanko, Given any two points of velocity on the curve, we can expect our time to be reduced by ~27% if we have a 13% increase in force.
DD_bongodriver Posted June 9, 2014 Posted June 9, 2014 (edited) so 13% force increase = 150% speed? Edited June 9, 2014 by DD_bongodriver
Crump Posted June 9, 2014 Posted June 9, 2014 (edited) so 13% force increase = 150% speed? Bongodriver, this claim is complete nonsense. You have obviously misunderstood something when you read the thread. Assuming you mean to constructively participate in the discussion, please point out where you think this claim is being made so we can get you on track. Edited June 9, 2014 by Crump
DD_bongodriver Posted June 9, 2014 Posted June 9, 2014 (edited) Please point out where you think this claim is being made. Read the thread please. I have been reading, that's my first post is number 66. I meant 130% speed i.e. a 30% increase Anyway I'm really wondering why you are carful to separate the time taken from acceleration when the relationship is direct, half the time means twice as fast etc. Edited June 9, 2014 by DD_bongodriver
Crump Posted June 9, 2014 Posted June 9, 2014 (edited) I meant 130% increase in speed. Certainly. The velocity increase occurs at 100 Knots which well within the power stable region of the curve. You can expect to see large increases in velocity for small changes in power. If we take it to the velocity stable portion of the power curve as Cpt_Blanko points out, our velocity change becomes only 13 Knots which is only 4.5% of our 285Knot Vmax. So do you see how aircraft performance works and why the speed change is a much larger percentage in the power stable region? The is discussion is about the reduction in time noted in koko's quick test of acceleration times over a velocity range. Given any two points of velocity on the curve, we can expect our time to be reduced by ~27% if we have a 13% increase in force. Force = Mass x Acceleration Acceleration is in units of time squared. For a 13% increase in force: 1.13^2 = 1.27 or a 27% decrease in time. Edited June 9, 2014 by Crump
Cpt_Branko Posted June 9, 2014 Posted June 9, 2014 (edited) Do you understand what I am saying? It is a very common technique to use the derivative of the acceleration values plugged into the rectilinear motion equations to determine aircraft acceleration distances covered and time required. Therefore the original conclusions of F=ma hold true with the value of time being squared in relation to force. Using square of net force differences to approximate, say, distance needed to reach a specific speed (I've seen similar used for takeoff distance correction in practice, just squaring the ratio of weights) is totally fine but, but not for time to reach a specific speed. Simply put Cpt_Blanko, Given any two points of velocity on the curve, we can expect our time to be reduced by ~27% if we have a 13% increase in force. Or more, or less, but yes, I agree. It's quite obvious from the graph that acceleration (directly tied with excess power) can vary dramatically between two points, especially when one or both are close to maximum speed. Excess power varies a lot with just a 10-15% power increase. Plotting the graph to scale with actual 109 data and acceleration differences could be read off the chart very easily. I mean especially acceleration between 410-510km/h can vary dramatically with just 15% more power, the second speed is close to maximum level flight speed. The difference in excess power is a lot more then 15% at 510km/h! Edited June 9, 2014 by Cpt_Branko
Crump Posted June 9, 2014 Posted June 9, 2014 Anyway I'm really wondering why you are carful to separate the time taken from acceleration when the relationship is direct, half the time means twice as fast etc. Well, the relationship is not direct. The math tells us that. This might help you to understand the basic physics: http://www.physicsclassroom.com/mmedia/kinema/acceln.cfm
DD_bongodriver Posted June 9, 2014 Posted June 9, 2014 Force = mass x acceleration since when was Force^2 = acceleration? Mass is going to be the constant here lets make it nice round figure of 100Kg lets make acceleration 100 flobnerbels (just needs to be a figure units are not relevant) 100kg X 100flobnerbels = 10000N 10000N @113% =11300N 11300N = 100Kg x 113 flobnerbels Acceleration = Force / mass ..........not F^2 113 flobnerbels = 11300N /100kg All very directly related to make the formula work but if we decrease the time by 30% we have increased the acceleration by 43%, lets say something took 100 seconds to reach 100metres = 1m/s, now it takes 70 seconds (30% less) 100/70 = 1.428m/s 1.428 x 100 =142.8 or 143 rounded up 143 143 flobnerbels = 14300N/100..........doesn't seem like a 13% increase in force to me
Crump Posted June 9, 2014 Posted June 9, 2014 Using square of force differences to approximate, say, distance needed to reach a specific speed (I've seen similar used for takeoff distance correction, just squaring the ratio of weights) is totally fine but, but not for time to reach a specific speed. Acceleration is a unit of distance over time squared. The technique not only works, it is what is taught in the aerodynamic departments of all major universities in the United States. It is where I learned it.
MiloMorai Posted June 9, 2014 Posted June 9, 2014 Is what was also taught in high school physics classes back in the day. It is probably being taught even earlier these days.
Crump Posted June 9, 2014 Posted June 9, 2014 Force = mass x acceleration since when was Force^2 = acceleration? Force = time squared..... Further analysis of the first and last columns of the data above reveal that there is a square relationship between the total distance traveled and the time of travel for an object starting from rest and moving with a constant acceleration. http://www.physicsclassroom.com/class/1DKin/Lesson-1/Acceleration
DD_bongodriver Posted June 9, 2014 Posted June 9, 2014 Force = Mass x Acceleration Acceleration is in units of time squared. For a 13% increase in force: 1.13^2 = 1.27 or a 27% decrease in time. Feel free to start making sense.
LLv34_Flanker Posted June 9, 2014 Author Posted June 9, 2014 S! Just a dumb question..When speed is increased the power needed also increases as friction greatly increases as well? Take Bugatti Veyron, it took 1000hp for it to reach 407km/h and 1400hp to reach 430km/h. So wouldn't this apply to planes as well? You get power with colder air but the power is not enough to give a substantial increase in speed as the friction also increases with speed? Am I right or totally lost? Math and physics a bit rusty for me
JtD Posted June 9, 2014 Posted June 9, 2014 An increase in engine power due to lower temperature and thereby higher density of the air will result in about the same true air speed (TAS), the speed with which the aircraft moves through the air. It will also result in a significantly higher indicated air speed (IAS), which is actually a pressure difference between the dynamic ram pressure and the static environmental pressure, which is used for the dials in the cockpit. 1
DD_bongodriver Posted June 9, 2014 Posted June 9, 2014 (edited) But the instruments are what the pilot sees, therefore he would fly to indicated airspeeds, at lower density altitudes he would get a reduction in TAS. Edited June 9, 2014 by DD_bongodriver
Crump Posted June 10, 2014 Posted June 10, 2014 (edited) S! Just a dumb question..When speed is increased the power needed also increases as friction greatly increases as well? Take Bugatti Veyron, it took 1000hp for it to reach 407km/h and 1400hp to reach 430km/h. So wouldn't this apply to planes as well? You get power with colder air but the power is not enough to give a substantial increase in speed as the friction also increases with speed? Am I right or totally lost? Math and physics a bit rusty for me The aircraft produces more thrust. Things like acceleration, take off and landing performance, and climb performance realize the largest benefits. Yes drag increases but remember lift and drag are connected. Lift increases as well and the wing can produce a given amount of lift at a lower velocity due to the higher density. It is all about the Lift to Drag ratio which does not change and is fixed by design so the power curve simply shifts closer to the origin as density increases. Yes, the power available curve is a Euler approximation so it is not linear. That is why we deal in dimensionless ratio's such as density ratio, temperature ratio, pressure ratio, coefficient's of pressure, and Standard Means of Evaluation. Those ratio's are simply another way to express a percentage which is of course, another non-dimensionless ratio. Edited June 10, 2014 by Crump
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