La-5 vs FW 190A-3 roll rate comparison

January 20, 2016

Yes, that is what the "time to bank" means.

January 20, 2016

But doesn't Fig. 6 say "Steady Rate of Roll" in both the y-axis and in the caption? That's also what I read from the report itself (Section 4): "The best method of comparison of the rolling performance of different aircraft is based on the steady rate of roll a pilot can generate using a definite stick force, say 50 lb. This course has been adopted in Fig.6".

January 20, 2016

But doesn't Fig. 6 say "Steady Rate of Roll" in both the y-axis and in the caption? That's also what I read from the report itself (Section 4): "The best method of comparison of the rolling performance of different aircraft is based on the steady rate of roll a pilot can generate using a definite stick force, say 50 lb. This course has been adopted in Fig.6".

Yes it does.

The report clearly states what was measured and how. Time to bank was measured and from that "steady rate of roll" was calculated.

Neither the measurement technique nor the calculation fit the NACA definition of either instantaneous or steady rate of roll.

January 20, 2016

Yes it does.

The report clearly states what was measured and how. Time to bank was measured and from that "steady rate of roll" was calculated.

Neither the measurement technique nor the calculation fit the NACA definition of either instantaneous or steady rate of roll.

And what can be concluded with this information regarding how the BOS devs modeled the rollrate of the FW190?

January 20, 2016

The instantaneous rate in BoS is too slow. That pretty much negates the FW190 agility advantage.

January 20, 2016

You can get an idea of the relative combat effectiveness of the FW-190's agility from this information:

This includes instantaneous rate, steady rate, control hinge moments/deflection, and human reaction time to control input from a pursuing fighter. It is the whole enchilada on relative agility against measured point of reference....the normal wing Spitfire.

January 21, 2016

Thanks for the reply Crump.

First I must correct my initial post. Sorry if it was a bit confusing. I wrote:

Do you mean that the times recorded in the Fig. 6 of the RAE report are from level flight (stick more or less centred) to a certain bank angle?

The figure of course does not record any time (directly), but roll rate.

But even then, to me it seems that you are contradicting yourself in this thread, but maybe I've misunderstood something.

Yes it does.

The report clearly states what was measured and how. Time to bank was measured and from that "steady rate of roll" was calculated.

Could you point out where that calculation is done, because the way I read the paper is opposite. I.e. steady rate of roll was measured, while time to bank to 45 degrees was calculated from the data in order to compare it to A.D.M.295 criterion -- what I suspect to be a some minimum recommended initial roll rate for fighter plane manufacturers. The empirical data obviously includes the acceleration portion of the roll, as they give an average delay of 0.5 seconds from (what I assume is) stick centre (0 lbs) to 50 lbs of stick force.

What I'm saying is that the initial roll rate (if I'm interpreting Za_Hairy's post correctly) overlaid on the Fig 6. in the first post of this thread is not directly comparable to the measured data shown in that figure and that Han's claim stands that after acceleration the roll rate is correct.

However, the main question then seems to be how fast that roll rate can be achieved (i.e. acceleration to the steady roll rate). Looking at the report, they state that the time to bank for 45 degrees at 400 mph can be estimated well by adding a delay of 0.25 seconds (!) to the steady roll rate (giving time to roll of 0.85 s for 45 degrees at 400 mph) . To me that hints that the acceleration to the steady roll rate happens indeed very quickly. Maybe this could contribute as indirect evidence, if something needs to be proven?

There is one more problem with the comparison between real world roll data and BoS: The tests were done using a regulated stick force. I believe we do not have details on how exactly the stick forces are modelled in BoS (Only thing I know that if my MS Sidewinder II FFB is able to produce 50 lbs of force -- which I believe is about 30 light years in the metric scale -- I must be not in too bad shape, as I can easily push it to its extremes at any speed. With one hand ).

Neither the measurement technique nor the calculation fit the NACA definition of either instantaneous or steady rate of roll.

I assume the NACA report is this paper: http://naca.central.cranfield.ac.uk/reports/1947/naca-report-868.pdf

Just had a quick look at it, but I can't really see that their definition for steady rate of roll differs per se (could you point me to the appropriate location if it is otherwise). More than that they seem to suggest a different criterion to be used as a guideline to compare roll manoeuvrability of different airframes.

As you have also hinted earlier, it clearly states that their data come from other sources (I guess the roll rate is combination of RAE report data among with other sources. Unfortunately they don't seem to specify their sources for that figure) and that the paper is more of a "survey" than a test report, but I wouldn't judge the figure they give as erroneous or misleading on that basis, and definitely wouldn't go as far as call their paper as "not the most scientific piece of work".

You can get an idea of the relative combat effectiveness of the FW-190's agility from this information:

This includes instantaneous rate, steady rate, control hinge moments/deflection, and human reaction time to control input from a pursuing fighter. It is the whole enchilada on relative agility against measured point of reference....the normal wing Spitfire.

The problem I see with interpreting those results is that we have no information about the initial conditions for any of those lines. While the report suggests that these lines represent times to bank and speculates about the speed of the planes, information about the altitude, if the planes are in coordinated flight, if the planes are already rolling at the point where the measurement starts (I've understood that the gun camera is synchronised with the gun trigger?) , what're the aileron angles / stick forces during the roll, etc. is not included.

Edited January 21, 2016 by PitbullVicious

January 21, 2016

The figure of course does not record any time (directly), but roll rate.

Rate includes an element of time.

steady rate of roll was measured

Yes and No. Rate of Roll was measured but not as we define stead rate of roll today. Steady rate of roll in the modern definition is the peak rate after instantaneous roll rate acceleration. The RAE lumped Instantaneous rate + Steady Rate into Figure 6 and labeled it "Steady Roll Rate". Figure 6 is simply time to bank angle as they measured it.

The RAE Data points were measured and calculations done to produce a meaningful chart that includes the entire envelope and not just a few single data points of performance.

The measurements were actually taken at speeds between 200 mph IAS and 400 mph IAS.

They measured angle of bank with a free gyro, aileron deflection angle with a "rat", and a henschel type force gauge to measure stick forces.

All of the instruments were linked into an electric timer that recorded the input data.

Rolls were made to left (port) and right (starboard) at aileron deflections of 1/4 deflection, 1/2 deflection, and 3/4 deflection.

These few data points represent the entirety of the measured performance.

What was then calculated.....

From those measured data points, aircraft performance math was used to construct figure 6 showing a 50lbs control force input at 10,000 feet from 150 mph Equivalent Airspeed to 400 mph Equivalent Airspeed.

What I'm saying is that the initial roll rate (if I'm interpreting Za_Hairy's post correctly) overlaid on the Fig 6. in the first post of this thread is not directly comparable to the measured data shown in that figure

Ahh but actually it is directly comparable. You can tell that by the assumptions listed in the RAE determination of ADM 295 standards which you confuse with the steady rate determination. If Figure 6 was just stead rate of roll, our process to achieve agreement would be:

Figure 6 rate + PIlot reaction times + Instantaneous rate = ADM 295 standard

That is not the case. Instead we must add too ADM 295 standards to reach agreement with Figure 6. That definitively proves Figure 6 is not just Steady Roll rate in isolation. It includes pilot reaction times and instantaneous roll rate. In other words, Figure 6 is just raw recorded data for roll rate information.

ADM 295 standard + Pilot reactions time + Instantaneous rate = Figure 6 rate

Here is the time assumptions made to determine ADM 295 standards.

First of all, as a universal correction for pilot input times all the measured times were reduced by 1/2 second.

ADM 295 is an lateral control criteria being used in the report to evaluate the high speed lateral control.

In order to make the ADM 295 data match figure 6, we must ADD back in the 1/4 second for our instantaneous roll performance which is removed in the ADM 295 time to 45 degree standard.

ADM standard for the FW-190 = .85 seconds to 45 degrees of bank at 400 MPH EAS @ 10,000 feet.

45 seconds/.85 seconds = 52.94 seconds

That is far short of the ~67 degrees Figure 6 shows us as the roll rate!!

If we add back in the 1/4 of second the RAE removed to account for instantaneous roll rate "steady state roll" as defined by the NACA we get:

52.94 seconds * .25 = 13.235 seconds

Now that represents the roll rate accelerating from zero to maximum stead roll rate. Our actual instantaneous rate will not be 13.235 seconds but will be less than that. We can crudely ballpark it by using the derivative of 6.5 seconds.

Now lets account for the 1/2 second it takes for the pilot to go from zero aileron deflections to the 50lbs of input mark...

52.94 seconds * .5 seconds = 26.47 seconds. Well that is not all rolling velocity because our aileron angle is changing from zero seconds to 26.47 seconds at its maximum rate per second. Using the derivative we get ~13.235 seconds added to the rate of change to account for the pilot input time.

Now let's add back in our time for pilot reaction, instantaneous roll rate, and ADM 295 standard to see if it matches figure 6, RAE "Steady Roll Rate".

53 seconds + 13 seconds + 13 seconds = 79 seconds. This is a best case scenario and too high an estimate. It gives us a good upper limit.

Now let's add back the factored times for instantaneous roll rate:

53 seconds + 13 seconds + 6.5 seconds = 72.5 seconds. This represents a good lower boundary of our actual rate of roll from wings level.

In terms of significant digits and factoring in our instantaneous rate is not, that is spot on agreement with figure 6.

Considering the accuracy of their time measuring equipment and the small amounts of time being recorded, I would say it is a closed case.

Since one must add back in the rate information for pilot reaction times, instantaneous roll rate, and steady roll rate information to get close to the Figure 6 rates, it is safe to say Figure 6 represents what they recorded as time to bank from wings level.

It is NOT stead roll rate as defined by the NACA, the math proves it. Figure 6 is the raw time to bank simply converted to 50 lbs of input at 150 mph EAS to 400 mph EAS at 10,000 feet.

January 21, 2016

The problem I see with interpreting those results is that we have no information about the initial conditions for any of those lines. While the report suggests that these lines represent times to bank and speculates about the speed of the planes, information about the altitude, if the planes are in coordinated flight, if the planes are already rolling at the point where the measurement starts (I've understood that the gun camera is synchronised with the gun trigger?) , what're the aileron angles / stick forces during the roll, etc. is not included.

Seperate issue.

Yes, there is data missing from the combat film investigation. Most prominently is the speed the FW-190 was traveling at the time the film recorded it.

In terms of relative performance, this missing data is not important.

The combat film investigation shows the actual relative performance RAF pilots were experiencing in dogfights.

Understand, RAE 1231 is not an absolute measurement. In fact, it is very poorly represented the FW-190's agility in these games because it is just a baseline that becomes the pinnacle of performance.

It is 50lbs of input over a Frise aileron design. Unlike some of the other design tested, the FW-190 was not at maximum aileron deflection in Fig 6. Pilots can input as high as 80 lbs into lateral control.

165 degrees per second at 15 degrees aileron deflection is what the control system was capable of producing at 360 mph EAS at 10,000 feet with an 80lbs input.

That is what the charts produced in the report tell us.

Unfortunately because of the way your game models stability and control, the FW-190 is limited to only 90 degrees per second and that is all it can ever do.

January 22, 2016

Thanks for the reply, Crump. However, I found some problems with your mathematics, so if you could kindly elaborate.

Yes and No. Rate of Roll was measured but not as we define stead rate of roll today. Steady rate of roll in the modern definition is the peak rate after instantaneous roll rate acceleration. The RAE lumped Instantaneous rate + Steady Rate into Figure 6 and labeled it "Steady Roll Rate". Figure 6 is simply time to bank angle as they measured it.

<snip>

Ahh but actually it is directly comparable. You can tell that by the assumptions listed in the RAE determination of ADM 295 standards which you confuse with the steady rate determination.

The paper itself seems to omit this. Do you have an external source on this, or why do you assume so?

I didn't find any clear definition for the steady rate of roll in the paper, so I assumed that it is the literal definition of "steady rate of roll", i.e. roll rate without any acceleration with constant stick force. I will also rather hold on to this definition unless you are able to show otherwise. Furthermore, the paper explicitly states that the "times to 45 deg bank have been calculated from the results shown in Fig.6" (emphasis mine).

NB: the following was edited after Crump's initial reply due to erroneous quote, which made my question seem out of place.

The RAE lumped Instantaneous rate + Steady Rate into Figure 6 and labeled it "Steady Roll Rate". Figure 6 is simply time to bank angle as they measured it.

Ummm.. why would they do that and leave it out of the paper? With their instrumentation the steady rate of roll is easily measurable.

End edit

That is not the case. Instead we must add too ADM 295 standards to reach agreement with Figure 6. That definitively proves Figure 6 is not just Steady Roll rate in isolation. It includes pilot reaction times and instantaneous roll rate. In other words, Figure 6 is just raw recorded data for roll rate information.

ADM 295 standard + Pilot reactions time + Instantaneous rate = Figure 6 rate

Sorry, where does this formula come from? I cannot find it in the paper.

In order to make the ADM 295 data match figure 6, we must ADD back in the 1/4 second for our instantaneous roll performance which is removed in the ADM 295 time to 45 degree standard.

ADM standard for the FW-190 = .85 seconds to 45 degrees of bank at 400 MPH EAS @ 10,000 feet.

45 seconds/.85 seconds = 52.94 seconds

I find your mathematics incredibly difficult to follow as your units seem to be all over the place.

But I guess what you meant to write above is:

45 deg / 0.85 s = 52.94 deg / s

Or where did you get the time 45 seconds and what does it mean?

That is far short of the ~67 degrees Figure 6 shows us as the roll rate!!number 1/4 s in

I believe the roll rate read from Fig. 6 @ 400 mph is very close to 75 deg / s. And yes, I am ashamed to admit it, but I did actually count pixels for that :D

If we add back in the 1/4 of second the RAE removed to account for instantaneous roll rate "steady state roll" as defined by the NACA we get:

52.94 seconds * .25 = 13.235 seconds

Here is where you completely lost me. If we take my assumption that you meant 45 degrees above, this comes as:

52.94 deg / s * 0.25 s = 13.235 deg

Do you mean that the accelerating part of the 45 degree roll is 13 degrees?

Now that represents the roll rate accelerating from zero to maximum stead roll rate. Our actual instantaneous rate will not be 13.235 seconds but will be less than that.

That also does not make sense as the paper clearly states that it took in average 0.5 seconds for a pilot to achieve the required stick strength. I think we can agree that the steady roll rate at certain stick force can not be achieved before the force is achieved.

We can crudely ballpark it by using the derivative of 6.5 seconds.

Sorry, what are you taking the derivative of here (looks like you are just dividing the 13 degrees by 2)? Why derivative?

Now lets account for the 1/2 second it takes for the pilot to go from zero aileron deflections to the 50lbs of input mark...

52.94 seconds * .5 seconds = 26.47 seconds. Well that is not all rolling velocity because our aileron angle is changing from zero seconds to 26.47 seconds at its maximum rate per second. Using the derivative we get ~13.235 seconds added to the rate of change to account for the pilot input time.

Again I assume that you get 26.47 degrees (as above) not seconds (or sec^2 as your formula says the units of the result should be). What does this figure represent and doesn't it overlap with the 13.235 degrees above?

Now let's add back in our time for pilot reaction, instantaneous roll rate, and ADM 295 standard to see if it matches figure 6, RAE "Steady Roll Rate".

53 seconds + 13 seconds + 13 seconds = 79 seconds. This is a best case scenario and too high an estimate. It gives us a good upper limit.

Now let's add back the factored times for instantaneous roll rate:

53 seconds + 13 seconds + 6.5 seconds = 72.5 seconds. This represents a good lower boundary of our actual rate of roll from wings level.

In terms of significant digits and factoring in our instantaneous rate is not, that is spot on agreement with figure 6.

Considering the accuracy of their time measuring equipment and the small amounts of time being recorded, I would say it is a closed case.

Since one must add back in the rate information for pilot reaction times, instantaneous roll rate, and steady roll rate information to get close to the Figure 6 rates, it is safe to say Figure 6 represents what they recorded as time to bank from wings level.

I really can't even start to comment this as your units do not make sense before you elaborate on the questions above. Could you kindly do that?

The way it is stated on the paper is "the time to bank is then fairly accurately estimated by assuming the aircraft commences to roll (at the steady rate) 1/4 seconds after the pilot starts to move the stick."

I.e. 45 deg / 75 deg/s + 0.25 s = 0.85s,

Which is the number they have in the table on page 3.

What I don't quite understand is how exactly they come to their estimate, as this seems to hint that the roll rate accelerates to the steady roll rate really fast. In fact, BEFORE the full stick force is achieved, but maybe someone can explain the logic behind this (or maybe the ADM295 document sheds some light on this, if someone has this)...

Edited January 22, 2016 by PitbullVicious

January 22, 2016

the math is correct but I did mislabel some units and make it much more confusing that it should be...

Forgive me PitBull...I will edit the post and try to make it clearer.

Honestly, I think a review of some basic math might be in order because at first glance that is what I see.

For example:

As for the derivative it looks like 26.47 divided by two for reason.

Try the change in deflection over the change in time. That is the basic definition of a derivative.

Starting point = zero degrees deflection at the zero second

Ending point = 26.47 degrees deflection at the 2nd second

(26.47-0)/(2-0)

Simply that and looks a lot like 26.47/2 because it is......lol.

Edited January 22, 2016 by Crump

January 22, 2016

It is late and I will fix it tomorrow night when I finish flying for the day.

January 22, 2016

It is late and I will fix it tomorrow night when I finish flying for the day.

I'm looking forward to it.

January 26, 2016

Maybe you read the news about the winter storm hitting the east coast of the US. I just finished a four day flying. You think it was busy with all the delays from every major airport on the East Coast being backed up? Yep...sure was and the performance of any gameshape was the farthest thing from my mind.

First of all, as a universal correction for pilot input times all the measured times were reduced by 1/2 second.

ADM 295 is an lateral control criteria being used in the report to evaluate the high speed lateral control.

In order to make the ADM 295 data match figure 6, we must ADD back in the 1/4 second for our instantaneous roll performance which is removed in the ADM 295 time to 45 degree standard.

ADM standard for the FW-190 = .85 seconds to 45 degrees of bank at 400 MPH EAS @ 10,000 feet.

45 DEGREES/.85 seconds = 52.94 DEGREES PER second. That is far short of the ~67 degrees PER SECOND Figure 6 shows us as the roll rate!!

If we add back in the 1/4 of second the RAE removed to account for instantaneous roll rate "steady state roll" as defined by the NACA we get: 52.94 DEGREES PER second * .25 = 13.235 DEGREES PER second.

Now that represents the roll rate accelerating from zero to maximum stead roll rate. Our actual instantaneous rate will not be 13.235 DEGREES PER second but will be less than that. We can crudely ballpark it by using the derivative of 6.5 seconds.

Now lets account for the 1/2 second it takes for the pilot to go from zero aileron deflections to the 50lbs of input mark...

52.94 Degrees Per second * .5 = 26.47 Degrees Per second.

Well that is not all rolling velocity because our aileron angle is changing moving our rate of roll from zero Degrees Per second to 26.47 Degrees per second at its maximum rate per second.

Using the derivative we get ~13.235 Degree Per seconds added to the rate of change to account for the pilot input time.

Now let's add back in our time for pilot reaction, instantaneous roll rate, and ADM 295 standard to see if it matches figure 6, RAE "Steady Roll Rate". 53 Degrees Per second + 13 Degrees Per second + 13 Degrees Per seconds = 79 Degrees Per second.

This is a best case scenario and too high an estimate. It gives us a good upper limit. Now let's add back the factored times for instantaneous roll rate: 53 Degrees Per seconds + 13 Degrees Per second + 6.5 Degrees Per second = 72.5 Degrees Per second. This represents a good lower boundary of our actual rate of roll from wings level.

In terms of significant digits and factoring in our instantaneous rate is not, that is spot on agreement with figure 6. Considering the accuracy of their time measuring equipment and the small amounts of time being recorded, I would say it is a closed case. Since one must add back in the rate information for pilot reaction times, instantaneous roll rate, and steady roll rate information to get close to the Figure 6 rates, it is safe to say Figure 6 represents what they recorded as time to bank from wings level.

Let's check out that last statement again:

Since one must add back in the rate information for pilot reaction times, instantaneous roll rate, and steady roll rate information to get close to the Figure 6 rates, it is safe to say Figure 6 represents what they recorded as time to bank from wings level.

January 26, 2016

Ummm.. why would they do that and leave it out of the paper? With their instrumentation the steady rate of roll is easily measurable.

Not really. Steady rate of roll is defined constant acceleration rate of roll at a specific control surface deflection. In other words, our rate of acceleration is no longer increasing.

Instantaneous rate of roll is very difficult to estimate and is most often just measured because of the small periods of time being represented.

Seems like it would be inconsequential but that is not the case. Aircraft with high steady rates can be seem much less agile if they have a low instantaneous rate.

In other words, the way the RAE did it was not only the most likely way it was performed at the time given the state of lateral control technology, it is the most useful way to measure it with the technology of the day.

I believe the roll rate read from Fig. 6 @ 400 mph is very close to 75 deg / s. And yes, I am ashamed to admit it, but I did actually count pixels for that :D

Which matches our calculated range perfectly.....

Crumpp says:

Now let's add back in our time for pilot reaction, instantaneous roll rate, and ADM 295 standard to see if it matches figure 6, RAE "Steady Roll Rate". 53 Degrees Per second + 13 Degrees Per second + 13 Degrees Per seconds = 79 Degrees Per second.

This is a best case scenario and too high an estimate. It gives us a good upper limit. Now let's add back the factored times for instantaneous roll rate: 53 Degrees Per seconds + 13 Degrees Per second + 6.5 Degrees Per second = 72.5 Degrees Per second. This represents a good lower boundary of our actual rate of roll from wings level.

That also does not make sense as the paper clearly states that it took in average 0.5 seconds for a pilot to achieve the required stick strength. I think we can agree that the steady roll rate at certain stick force can not be achieved before the force is achieved.

Which is why we are using the derivative. We take two points and account for the change in time and displacement.

Two things that must be accounted for and added back into the ADM standard in order to reach the Figure 6 "Steady Rate of Roll" numbers.

The ADM 295 criteria is actually the modern definition of steady state roll time to 45 degree angle of bank. If Figure 6 was the modern definition of steady state roll, we would not have to account pilot control input lag times or instantaneous rate of roll.

Our initial calculation converting out ADM standard to rate information in degrees per second would equal Figure 6's rates.

ADM standard for the FW-190 = .85 seconds to 45 degrees of bank at 400 MPH EAS @ 10,000 feet.

45 DEGREES/.85 seconds = 52.94 DEGREES PER second. That is far short of the ~67 degrees PER SECOND Figure 6 shows us as the roll rate!!

January 27, 2016

Thank you for replying. Unfortunately I'm not completely satisfied yet.

Not really. Steady rate of roll is defined constant acceleration rate of roll at a specific control surface deflection.

Interesting. Do you have any sources on this?

I can then imagine that comparing different results might be rather difficult and you would need at least the time to achieve the maximum roll rate and maximum roll rate (i.e. the constant roll rate for given stick force) to make any comparisons. Then again, with a very quick look it seems that the NACA paper is trying to answer exactly these kind of problems, so maybe good metrics for roll performance were still in search during writing of the RAE report...

Which matches our calculated range perfectly.....

Umm... nope. I still see some problems in your maths. I'll just use the section below as an example, although same applies to your result "6.5 deg / s":

Now lets account for the 1/2 second it takes for the pilot to go from zero aileron deflections to the 50lbs of input mark... 52.94 Degrees Per second * .5 = 26.47 Degrees Per second. Well that is not all rolling velocity because our aileron angle is changing moving our rate of roll from zero Degrees Per second to 26.47 Degrees per second at its maximum rate per second. Using the derivative we get ~13.235 Degree Per seconds added to the rate of change to account for the pilot input time.

If you take the roll rate of 52.94 deg / s and multiply it by time of 0.5 seconds it is a measure of angle. I.e. 26.47 degrees NOT degrees per second. Then you seem to do dw / dt (read the w as omega here ) and get 13.235 deg / s, implying that your dt = 2 seconds. So, the question remains: where do you get that arbitrary time of 2 seconds? If you take dt = 1/2 s, it of course becomes as 52.94 deg / s (DOH!). So I'm not confused with what a derivative is, but what you are actually taking a derivative of. Could you write this completely open?

Now let's add back in our time for pilot reaction, instantaneous roll rate, and ADM 295 standard to see if it matches figure 6, RAE "Steady Roll Rate". 53 Degrees Per second + 13 Degrees Per second + 13 Degrees Per seconds = 79 Degrees Per second.

This is a best case scenario and too high an estimate. It gives us a good upper limit. Now let's add back the factored times for instantaneous roll rate: 53 Degrees Per seconds + 13 Degrees Per second + 6.5 Degrees Per second = 72.5 Degrees Per second. This represents a good lower boundary of our actual rate of roll from wings level.

Taking into account what I wrote above, these calculations do not make any sense. You cannot add deg / s and deg or deg*s together. In order to add you need to have the same units. Please, try again writing the formulas in actual mathematical annotation (TeX code is good enough for me).

January 27, 2016

Thank you for replying. Unfortunately I'm not completely satisfied yet.

That is ok. I am

Then again, with a very quick look it seems that the NACA paper is trying to answer exactly these kind of problems, so maybe good metrics for roll performance were still in search during writing of the RAE report..

Yep, Lateral control research was not well developed.

f you take the roll rate of 52.94 deg / s and multiply it by time of 0.5 seconds it is a measure of angle. I.e. 26.47 degrees NOT degrees per second. Then you seem to do dw / dt (read the w as omega here ) and get 13.235 deg / s, implying that your dt = 2 seconds. So, the question remains: where do you get that arbitrary time of 2 seconds? If you take dt = 1/2 s, it of course becomes as 52.94 deg / s (DOH!). So I'm not confused with what a derivative is, but what you are actually taking a derivative of. Could you write this completely open?

A derivative is calculus. It is a way to measure to the change in distance over the change in time, getting it to one value we can use to estimate positional information.

I already did it for you in the very first post replying to your questions. Please go back and review it. I cannot teach a calculus class or aircraft performance class in this thread.

these calculations do not make any sense.

Actually they do very much make sense.....

Please review what has been written.

January 27, 2016

Here Pitbull....maybe this will help you.

What were are doing is finding the what our ailerons velocity is going to be when moving from Point A of zero deflections to Point B "steady rate of roll deflection. We want the rate it is traveling.

The main idea is the concept of velocity and speed. Indeed, assume you are traveling from point A to point B, what is the average velocity during the trip?

We are doing the physical.....

The definition of the derivative can be approached in two different ways. One is geometrical (as a slope of a curve) and the other one is physical (as a rate of change).

The Physical Concept of the Derivative

http://www.sosmath.com/calculus/diff/der00/der00.html

January 27, 2016

Interesting. Do you have any sources on this?

Yeah...

NACA Report 868 defines it for us.

In other words, if Figure 6 represented the maximized and steady state of Pb/2V then we would not have to account for the time our roll rate is changing and for the pilots reaction time to get it to align with ADM 295 standard.

It is really just that simple.

January 27, 2016

Still is thread about roll rate between La and FW in BoS game?

Edited January 27, 2016 by MK_RED13

January 27, 2016

Still is thread about roll rate between La and FW in BoS game?

Yes it is..

January 27, 2016

Sorry, i had to go when i opened the thread. In the test the FW 190 is in light configuration, 2 cannons only, and both planes at 60% fuel.

It matches the RAE 1231 data which is representative an FW-190 with full wing weapons.

January 27, 2016

To sum up Crump, are you basically saying the tests are recording the instaneous roll rate, not continous?

January 27, 2016

I believe have been friendly and courteous to you, Crump. Please allow me the same and answer to my questions directly, not in a condescending manner, thank you.

A derivative is calculus. It is a way to measure to the change in distance over the change in time, getting it to one value we can use to estimate positional information.

As I wrote, I do know what a derivative is. What is not clear is what YOU are doing. Proper mathematical formalism for your calculations, please.

You wrote:

As for the derivative it looks like 26.47 divided by two for reason. Try the change in deflection over the change in time. That is the basic definition of a derivative. Starting point = zero degrees deflection at the zero second Ending point = 26.47 degrees deflection at the 2nd second (26.47-0)/(2-0) Simply that and looks a lot like 26.47/2 because it is......lol.

(Just a quick edit: I'm using w here for the bank angle instead of omega)

Then 26.47 must be degrees, and what you are doing is dw/dt. Right or wrong?

So, according to that, your dt = 2 seconds. What is this time and where does it come from?

Or if the 26.47 is deg / s, you are doing d^2w/dt^2 which is the acceleration over 2 seconds. Again, what is this 2 seconds and why acceleration?

Which ever it is, either the numbers or the units don't add up.

Still is thread about roll rate between La and FW in BoS game?

Yes. In order to be able to compare the roll rates, we need to first figure out what the given roll rates for the FW in the report mean.

Edited January 27, 2016 by PitbullVicious

January 27, 2016

To sum up Crump, are you basically saying the tests are recording the instaneous roll rate, not continous?

Figure 6 includes both steady state roll and instantaneous. It is just the time to bank the RAE recorded in their investigation.

believe have been friendly and courteous to you, Crump.

I have been so with you. I am trying to answer your questions.

Or if the 26.47 is deg / s, you are doing d^2w/dt^2 which is the acceleration over 2 seconds. Again, what is this 2 seconds and why acceleration?

That is the acceleration.

S = Vi(t) + 1/2aT^2

S = 0(0) + 1/2 (13.235) 1^2

S = 13.235

Units would be degrees.....

It all does cancel. You can take the time to draw it out if you like.

.

January 27, 2016

That is the acceleration.

S = Vi(t) + 1/2aT^2

S = 0(0) + 1/2 (13.235) 1^2

S = 13.235

Units would be degrees.....

Sorry, but now I seriously feel like I'm being trolled. First your time was 2 seconds to get the acceleration (which does not make sense in the first place), then your time is 1 second to get the displacement from the acceleration. What you then is get an displacement in degrees and then add that up to a rate which is deg / s...

Oh, and on top of that 1/2 * 13.235 is not 13.235.

I'm sorry for being blunt, but looking at this it seems that you lack the grasp of even basic mathematics. I tried to give you a benefit of doubt, but as you are unwilling to answer my direct questions, I think we're through with this conversation.

I just hope that other people will see the faults in your calculations and understand what the implications are. Or jump in and show what I've done wrong. I'm always willing to learn and admit my mistakes, when clearly shown.

To everybody else:

Why am I being so pedantic?

If the aim is to convince the developers that something in their simulation is wrong, we need to make a strong case in order to get it heard from all the noise in the forums. That means good references (academic papers / technical reports) and good understanding that those references are in fact representative for the claimed problem (i.e. one most understand what exactly is stated in the paper, how the results were achieved and what they actually imply). These claims need then be written in clear, compact form, with proper mathematical formalism where appropriate.

Don't accept something just because it seems to support what you think is correct, but look into it a bit deeper and check the mathematics and ask questions if something seems weird. You are actually making a service to your cause as the more correct and clearly made the claims are, the more likely they are to be heard.

January 27, 2016

Measuring sustained roll rate would make no sense. Sustained roll rate is operationally pointless. You don't care how fast you go from 0-180 degrees after 5 rolls, you care how quick you go to that immediately.

January 27, 2016

I don't know if it is necessary to clear things up, but anyway:

If you look at the figure 6 you'll find the steady rate of roll. To estimate time to bank from that:

- you take the 400mph point (deg/s)

- you multiply this figure with 0.5 for an average, assuming a linear acceleration from initially 0 to that velocity (deg/s)

- you multiply this figure with 0.5s to arrive at the actual angle of bank after half a second, described in the British document as the time needed for acceleration (deg)

- you subtract this figure from 45°, which is the target angle of bank (deg)

- you divide this figure by steady rate of roll figure (s)

- you add to that the half second for the initial roll acceleration (s)

And you've arrived at the data given in the table, see attachment.

January 27, 2016

Thanks, JtD.

I don't know if it is necessary to clear things up, but anyway:

If you look at the figure 6 you'll find the steady rate of roll. To estimate time to bank from that:

- you take the 400mph point (deg/s)

- you multiply this figure with 0.5 for an average, assuming a linear acceleration from initially 0 to that velocity (deg/s)

- you multiply this figure with 0.5s to arrive at the actual angle of bank after half a second, described in the British document as the time needed for acceleration (deg)

- you subtract this figure from 45°, which is the target angle of bank (deg)

- you divide this figure by steady rate of roll figure (s)

- you add to that the half second for the initial roll acceleration (s)

And you've arrived at the data given in the table, see attachment.

And this is the same as estimating 0 deg / s for the first 0.25 secs and full rate after that. Assuming, as you say, linear acceleration for the first 0.5 secs.

Does this imply that the plane achieves it's maximum rate of roll (EDIT: within reasonable accuracy) as soon as the ailerons are deflected to their target angle (depending on the stick force), or is that 75 deg / s an average over some time / distance?

Edited January 27, 2016 by PitbullVicious

January 27, 2016

I don't know if it is necessary to clear things up, but anyway:

If you look at the figure 6 you'll find the steady rate of roll. To estimate time to bank from that:

- you take the 400mph point (deg/s)

- you multiply this figure with 0.5 for an average, assuming a linear acceleration from initially 0 to that velocity (deg/s)

- you multiply this figure with 0.5s to arrive at the actual angle of bank after half a second, described in the British document as the time needed for acceleration (deg)

- you subtract this figure from 45°, which is the target angle of bank (deg)

- you divide this figure by steady rate of roll figure (s)

- you add to that the half second for the initial roll acceleration (s)

And you've arrived at the data given in the table, see attachment.

Except for your assertion is backwards. Initial roll acceleration is always higher than steady state. Roll dampening forces build as the roll accelerates slowing our initial rate of roll down to the "steady state".

Therefore in the ADM if it represented initial roll rate + steady roll rate as defined by the NACA:

45 degrees of bank / .85 seconds = 52.94 degrees per second

If Figure 6 is steady state rolling as defined by the NACA, then we should be moving at a rate faster than the 75 degrees when Pb/2V is steady state.

In other words, if you are correct then the RAE was fundamentally flawed in their understanding of lateral control. I do not think that is the case.

January 27, 2016

And this is the same as estimating 0 deg / s for the first 0.25 secs and full rate after that. Assuming, as you say, linear acceleration for the first 0.5 secs.

Yes, it is.

Does this imply that the plane achieves it's maximum rate of roll (EDIT: within reasonable accuracy) as soon as the ailerons are deflected to their target angle (depending on the stick force), or is that 75 deg / s an average over some time / distance?

The British report says that it takes half a second to reach the desired aileron deflection/force of 50lb and the same time for the aircraft to reach steady rate of roll after the stick is first moved. This indeed implies the steady rate of roll is reached as soon as the ailerons are in the desired position, both happening after half a second.

The 75deg/s are the steady rate of roll of the Fw190 at 50lb stick force at 10.000ft at 400mph IAS. At these parameters, it is the maximum rate of roll.

January 27, 2016

The British report says that it takes half a second to reach the desired aileron deflection/force of 50lb and the same time for the aircraft to reach steady rate of roll after the stick is first moved. This indeed implies the steady rate of roll is reached as soon as the ailerons are in the desired position, both happening after half a second.

The 75deg/s are the steady rate of roll of the Fw190 at 50lb stick force at 10.000ft at 400mph IAS. At these parameters, it is the maximum rate of roll.

Thanks again. Then we are indeed reading the same paper and interpreting it the same way.

January 27, 2016

The British report says that it takes half a second to reach the desired aileron deflection/force of 50lb and the same time for the aircraft to reach steady rate of roll after the stick is first moved. This indeed implies the steady rate of roll is reached as soon as the ailerons are in the desired position, both happening after half a second. The 75deg/s are the steady rate of roll of the Fw190 at 50lb stick force at 10.000ft at 400mph IAS. At these parameters, it is the maximum rate of roll.

But is does not even say that....

According to the ADM 295 criteria, It takes a 1/2 second to move the ailerons to full deflections at 50 lbs of input. 1/4 of a second from the time the pilot moves the stick, the aircraft reaches steady state roll acceleration.

Your interpretation is just wrong JtD and have you have arranged the variables to fit the outcome you want instead of having them just fit the facts. You are more interested in proving me wrong than in what the facts are in this case.

Edited January 27, 2016 by Crump

January 27, 2016

And yet 75 * (0.85-0.25) = 45.

75 degrees/s being steady state roll

0.85s being JtD's calculated time to 45 degrees

0.25s being half the initial period

45 degrees being the target roll angle

As has already been pointed out, you can either calculate the effect of the average roll rate during the whole of the first 0.5 seconds, or just take the steady state roll speed for half of it.

JtD's calculated results are exactly the same as doing it the way underlined. No calculus required.

edit - as for "crickets chirping" - wrong season, right time of night. Pitbull lives in Tokyo - as is plainly stated under his avatar - where it is well past midnight, I expect he has to work tomorrow.

Edited January 27, 2016 by unreasonable

January 27, 2016

The British report says that it takes half a second to reach the desired aileron deflection/force of 50lb and the same time for the aircraft to reach steady rate of roll after the stick is first moved.

Is not what RAE 1231 says.

However, that is all minutia to the fact:

Initial roll acceleration is always higher than steady state. Roll dampening forces build as the roll accelerates slowing our initial rate of roll down to the "steady state". Therefore in the ADM if it represented initial roll rate + steady roll rate as defined by the NACA: 45 degrees of bank / .85 seconds = 52.94 degrees per second If Figure 6 is steady state rolling as defined by the NACA, then we should be moving at a rate faster than the 75 degrees when Pb/2V is steady state.

January 27, 2016

Thanks again. Then we are indeed reading the same paper and interpreting it the same way.

JtD's calculated results are exactly the same as doing it the way underlined. No calculus required.

Glad we're all on the same page, even with my linear acceleration thing.

Initial roll acceleration is always higher than steady state.

If initial acceleration was always higher than terminal velocity, you'd of course be right. Unfortunately, acceleration and velocity, be it linear or rotational, are not the same thing. Roll acceleration, at steady state roll, is zero. Which is the reason it is called 'steady state'.

Edited January 27, 2016 by JtD

January 28, 2016

At the request of a poster in this thread I have cleaned it up and reopened it.

Gentlemen... please do not let these discussions wax personal... It is fine to disagree.. and it is better to agree to disagree and move on where applicable.

January 28, 2016

[Edited]

In particular, he underlined part of RAE 1231 from this section:

"...the time to 45 degrees bank is then fairly accurately estimated by assuming the aircraft commences to roll (at the steady rate) 1/4 second after the pilot starts to move the stick." RAE 1231 p.3

He proposed that this invalidated JtD's method of estimation. You can see from my worked example that the result is the same as JtD's method.

If you want to think of it in geometric terms, the RAE formula is splitting a rectangle in half from top to bottom in the middle, JtD's formula is splitting the same rectangle in half from corner to corner. The area of the two is the same.

Please note the words "fairly accurately estimated". The degrees of bank achieved in an elapsed time equals the area under the curve (look at Crump's post 70 if it is still there for such a curve). RAE 1231 and JtD are using straight line approximations to estimate the area under the curve in the period before steady state roll is achieved. Obviously the RAE report writers thought that this was a small enough error to be ignored, given the other possible noise in the numbers.

If anyone wishes to prove them wrong by more accurately calculating the area under the curve, please share.

Edited February 3, 2016 by Bearcat

January 28, 2016

La-5 vs Fw-190...two videos,first with sequences from modded old Sturm,WT and BoS La-5 rolling around(what's the message?) and 2nd video with only BoS Fw-190.Then a chart with blue and red curves and no reference to real life La-5 data...really a comparison? What's the goal/target of this thread?

January 28, 2016

What's the goal/target of this thread?

To attract biased fanboys like you.

More seriously... there's a long time since this thread is not anymore about La-5 vs Fw 190 roll comparison, and rather focused on NACA #868 & RAE 1231.

If you want to compare ingame roll rate, if you have any data to share about roll rate, then come here: http://forum.il2sturmovik.com/topic/20679-bos-fighters-roll-comparison/

La-5 vs FW 190A-3 roll rate comparison

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