Rudder effectiveness

Curly,

interesting references, thank you for spending the time digging them up.

I find your calcs a bit hard to follow though, as values are introduced without explaining where they come from. I get different results when performing the same calculations.

Givens

Based on the AD0069271 report:

Rudder hinge moment H = C_Hr * q* s_r * c_r,

where s_r is the area (ft^2) and s_r * c_r is the area moment (ft^3).

s_r * c_r, per the report, is 9.72 ft^3.

q = rho*V^2/2, with air density rho in slugs/ft^3 or .00237 at SLS. V is in ft/s.

M.4

At M.4, .4*1126=450 ft/s, C_Hr as stated is .23 for -25° rudder deflection at 0° yaw angle

H = C_Hr * q* s_r * c_r = .23 * (.00237*450^2/2) * 9.72 = 540

Curiously, the unit for the hinge moment is never stated as far as I can tell, but it can probably be assumed to be ft lbs, as they haven't used sane units anywhere else. ;)

Assuming (and this is a big assumption) a completely linear relationship between pedal movement and rudder deflection, and assuming 6" of forward pedal travel, the pedal force at half a feet of travel should be twice the moment in ft lbs, or 1080 lbf.

That sounds a whole lot more plausible IMNSHO. My estimate is that 25° of deflection at 265 KTAS would rip the fin clean off - it's just not something you'd build aircraft for. Note that the data are from 20% scale wind tunnel tests. To get there with a mere 74 kgs of pressure on the pedal would mean you could essentially do it by accident while reaching for a map.

M.2

Repeat at M.2: C_Hr = .24, TAS = 225 ft/s.

H =.24 * (.00237*225^2/2) * 9.72 = 140 ft lbs, for 280 lbf of pedal force.

No comment on the M4.0 case presented... :pilotfly:

Pedal travel

Where did you get the 6" pedal travel figure from? DTI639028 gives 3.75 inches of travel, center to maximum deflection, based on two war-time reports. Sounds rather limited to me, but then again - I rarely look at my feet when flying. Three, six or eight inches has a profound effect on the final results, so this is an important figure to get right.

Crabbed approaches

I do not understand your comment regarding the crabbed approaches up to 25 knots [of xwind]. The rudder is effective for crabbed approaches up to direct xwinds equalling the true airspeed, at which point you will be hovering with the nose 90° off from the final approach course... :) Jesting aside, the rudder isn't used as long as you are crabbing. There is a recommendation in the FM not to land with xwind components in excess of 25 knots, but that's likely to be due to either structural limitations regarding the crab angle you can safely touch down with, or directional control issues on the rollout. It does indicate that you shouldn't touch down crabbed, so the rudder authority should be sufficient to remove most of the crab prior to touchdown with the 25 knots component. In that case, the air speed would be 120 + 2.5*5 = 132.5 KIAS (and KTAS, assuming SLS), so a 25 knot crosswind equates to 10.7° of crab. That's anecdotal evidence of the amount of rudder control available with normal control forces, FWIW.

(Pg. 6-8 of the F model manual is also interesting, regarding rudder control. Note the lack of warnings regarding overcontrolling at high airspeeds. Had pilots been able to deflect the rudder to angles posing structural hazards, there would have been such warnings.)

Rudder deflection in various flight conditions

Going backwards, the DTI639028 report figure for desirable rudder control force, from the 1945 US Army/Navy agreement, is 180 lbs. This gives a hinge moment of 90 lb ft. At 130 KTAS (220 ft/s),

C_Hr = H/(q * s_r * c_r) = 90/(.00237*220^2/2*9.72) = .16

220 ft/s is, coincidentally, almost exactly M.2 so we have a data point available without extrapolating. A .16 C_Hr will give you a little less than 20 degrees of deflection with no yaw.

Finally, let us move on to M.8, 900 ft/s. With no yaw, 180 lbs of control force would again give you 90 lb ft of hinge moment.

C_Hr = H/(q * s_r * c_r) = 90/(.00237*900^2/2*9.72) = 0.01. With no yaw, that's 1.5° or so of deflection.

Yes, it can be argued that the maximum rudder forces specified here are very conservative. Still, the fact remains - you're not going to produce much deflection at high airspeeds.

References

It would be easier if you stated what information you think is pertinent in the various references given. I'd stay clear of the DWC lecture slides, which the 12.5 figure stems from. I found a couple of 'interesting' things in them, and the book they state they are based on doesn't seem solid either, after a bit of googling on it.

Rudder efficiency vs flight dynamics

The rudder is efficient if it can generate a beta angle. This is what is important for a xwind landing.

The turn, or lack thereof, generated by that beta angle, is flight dynamics/aerodynamics.

This has amused me as this thread has run its course. :)

Rgds,

/Fred

I’m not sure what Belsimtik is attempting to model with the current rudder behavior. I could speculate..... it would be nice if the Developers would discuss their source material and how they arrived at the current behavior of the rudder.

Well done research & explanation Curly:thumbup:

Now that the devs have all the data they wanted, it's show-time! 7 days after your post & still no answer from the devs is troublesome though!

Curly,

interesting references, thank you for spending the time digging them up.

 

I find your calcs a bit hard to follow though, as values are introduced without explaining where they come from. I get different results when performing the same calculations.

 

 

2058.33/237.0163 * 8.12 * 4.65 does not equal .23

2058.33/(237.0163 * 8.12 * 4.65) does.

Where do you get the 4.65? Should be the area moment divided by the area as far as I can tell, 9.72/8.12 ft = 1.20 ft. That accounts for most of the differing results, then we probably have slightly different constants for the calculation of the dynamic pressure.

 

 

I’m using true dynamic pressure. Which at sea level, at a speed of mach .4 = 237.01 lbs/square foot. Where p= .076474 lbm/ft3 and v = 446.58 feet per second. And q = Dynamic pressure = 1/2pv^2. Thus q=237.01

Feel free to verify. http://www.engineeringtoolbox.com/dynamic-pressure-d_1037.html

http://www.aerospaceweb.org/design/scripts/atmosphere/

C_H = Hinge moment / q * Area of moveable surface aft of the hinge line * Mean aerodynamic chord of movable surface.

C_H = h/q*8.12ft* 55.99 inches or 4.665 feet

Thus, h= C_H * (q*8.12*4.66)

Givens

 

Based on the AD0069271 report:

 

Rudder hinge moment H = C_Hr * q* s_r * c_r,

 

where s_r is the area (ft^2) and s_r * c_r is the area moment (ft^3).

 

s_r * c_r, per the report, is 9.72 ft^3.

 

q = rho*V^2/2, with air density rho in slugs/ft^3 or .00237 at SLS. V is in ft/s.

 

 

M.4

 

At M.4, .4*1126=450 ft/s, C_Hr as stated is .23 for -25° rudder deflection at 0° yaw angle

 

H = C_Hr * q* s_r * c_r = .23 * (.00237*450^2/2) * 9.72 = 540

 

Curiously, the unit for the hinge moment is never stated as far as I can tell, but it can probably be assumed to be ft lbs, as they haven't used sane units anywhere else.

 

Assuming (and this is a big assumption) a completely linear relationship between pedal movement and rudder deflection, and assuming 6" of forward pedal travel, the pedal force at half a feet of travel should be twice the moment in ft lbs, or 1080 lbf.

Not sure what's up with your equation for dynamic pressure. It should be 1/2pV^2, Not 1/2pV^2/2., Your use of slugs is wrong too, they are not in agreement with the other measures. You need to multiply slugs by the force of gravity in feet per second. .0023*32.2 =.07631.

Also, your using the wrong term in your calculation of Ch. Where you're using 9.72, the area moment of the rudder,you should be using, the mean aerodynamic chord of the movable surface * The Area of the moveable surface aft of the the hinge line. It's written as Ch = Hinge moment / q * Saft * Caft on page 16 of the Sabre info pdf and on page 27 of the Cornell pdf. https://courses.cit.cornell.edu/mae5070/Caughey_2011_04.pdf .

Thus

Ch=.23 = h/ (237.0163*8.12*1.19)

h= (237.0163 * 8.12* 1.19) * .23

h= 526.775

The rudder on the Sabre is a simple wire configuration, it’s layout is viewable in the Maintenance manual linked at the bottom of the first post. You're estimate for pedal force also seems very high. Reversible systems basically provide a simple mechanical advantage. Which is calculated in my first post. If I use your definition for Mean aerodynamic chord of the rudder, it's is 1.19, (SrCr/Sr=Cr), (9.72/8.12=1.19). .The hinge moment and pedal forces goes down.

http://www.daerospace.com/FlightControlSystems/Reversible.php

(Gear Ratio 6/27.5) = .21 inch/ Degree =12.4962095 inch / radian

Where 12.49 represents the mechanical advantage of the pedals.

12.49*x=526.775

42.17 lbs of force at mach .4

force caluations at mach .4 with various rudder pedal travel below.

That sounds a whole lot more plausible IMNSHO. My estimate is that 25° of deflection at 265 KTAS would rip the fin clean off - it's just not something you'd build aircraft for.

What is your assessment based on? It’s not shear strength of rudder mounts. The ailerons have twice the area and those are fully deflect-able. So if you can build an aileron to deflect fully at said speed and above; it wouldn’t be a problem to do the same for a rudder. The hinge moment for the fully deflected rudder at 265 kts sea level is 528.364 lbs. It's reasonable amount of force for the mounts to withstand, or for the rudder it self to bare with out deformation or damage.

-Snip-

Pedal travel

Where did you get the 6" pedal travel figure from? DTI639028 gives 3.75 inches of travel, center to maximum deflection, based on two war-time reports. Sounds rather limited to me, but then again - I rarely look at my feet when flying. Three, six or eight inches has a profound effect on the final results, so this is an important figure to get right.

That 3.75 inches of pedal travel is for the mock up on which the force tests were conducted, not the actual Sabre.

Crabbed approaches

I do not understand your comment regarding the crabbed approaches up to 25 knots [of xwind]. The rudder is effective for crabbed approaches up to direct xwinds equalling the true airspeed, at which point you will be hovering with the nose 90° off from the final approach course... Jesting aside, the rudder isn't used as long as you are crabbing. There is a recommendation in the FM not to land with xwind components in excess of 25 knots, but that's likely to be due to either structural limitations regarding the crab angle you can safely touch down with, or directional control issues on the rollout. It does indicate that you shouldn't touch down crabbed, so the rudder authority should be sufficient to remove most of the crab prior to touchdown with the 25 knots component. In that case, the air speed would be 120 + 2.5*5 = 132.5 KIAS (and KTAS, assuming SLS), so a 25 knot crosswind equates to 10.7° of crab. That's anecdotal evidence of the amount of rudder control available with normal control forces, FWIW.

Please review the link for Rudder Design Chapter, 12 Design of Control Surfaces by Mohammad Sadraey. Pages 11-17. This provides a mathematical description of the crab approach.

“In order to keep the aircraft landing direction along the runway, the rudder is employed to counteract the yawing moment created by the wind. The rudder produces a verticaln tail lift along yaxis (L ) which consequently contributes the aircraft yawing moment and aerodynamic side force. The application or rudder is to create a crab angle (sigma) in order to prevent aircraft from yawing to the relative wind and avoid drifting awayfrom the runway. The rudder must be powerful enough to create the desired crab angle.

The crab angle is defined as the angle between fuselage centerline and the runway (i.e.heading direction). Figure 12.28 shows all the forces and moments affecting the final approach operation while the aircraft is in a crabbed landing.”

Approach speed is 140, touch down in 120. So working with a approach of 140kts,

Where B = tan^-1 (Vw/U_1)

10.124=tan^-1(25/140)

B=10.124

The relative wind or aircraft total speed is the vector summation of the aircraft forward speed and wind speed:

Vt= Sqrt U_1^2+Vw^2

142.21 kts= Vt

Using the side area, dynamic pressure and side drag we can calculate the force of the wind Fw.

Fw= ½ pv^2w * Ss * Cdy

Then using some of the known stability coefficients for the craft, we can work out the rudder deflection and crab angle angles needed.

1/2pvt^2* Sb(Cn +Cnb(B-sigma)+C_delta_r * Delta_r)+Fw*dc cos sigma =0

1/2Pv^2w*Ss*Cdy = 1/2pV^2t*s(Cyo + Cyb(B - sigma) +Cy_delta_r *Delta_r

If we solve for Delta_R and sigma we can tell if the rudder generates enough force to crab in those conditions.

Since we know that the craft is cleared to approach at 140 knots in a crab and the rudder deflects at most 27.5 degrees or .479 radians than the rudder is generating at least Fw force. Thus we can say Fay = Fw in 35 knt cross wind. Side force will alter the flight path. Thus we can say if rudder can generation enough side force in a X wind to crab, without a x wind the rudder will allow us to alter the flight path.

(Pg. 6-8 of the F model manual is also interesting, regarding rudder control. Note the lack of warnings regarding overcontrolling at high airspeeds. Had pilots been able to deflect the rudder to angles posing structural hazards, there would have been such warnings.)

Again based on your assumption that deploying large rudder deflection presents a structural danger.

Rudder deflection in various flight conditions

Going backwards, the DTI639028 report figure for desirable rudder control force, from the 1945 US Army/Navy agreement, is 180 lbs. This gives a hinge moment of 90 lb ft. At 130 KTAS (220 ft/s),

 

C_Hr = H/(q * s_r * c_r) = 90/(.00237*220^2/2*9.72) = .16

 

220 ft/s is, coincidentally, almost exactly M.2 so we have a data point available without extrapolating. A .16 C_Hr will give you a little less than 20 degrees of deflection with no yaw.

 

 

Finally, let us move on to M.8, 900 ft/s. With no yaw, 180 lbs of control force would again give you 90 lb ft of hinge moment.

 

C_Hr = H/(q * s_r * c_r) = 90/(.00237*900^2/2*9.72) = 0.01. With no yaw, that's 1.5° or so of deflection.

 

Yes, it can be argued that the maximum rudder forces specified here are very conservative. Still, the fact remains - you're not going to produce much deflection at high airspeeds.

The DTI doc states that ideal forces are 180 lbs. The report notes that all pilots even the weakest tested were capable of generating at least 300 lbs of force on the rudder pedals. Per 106019.pdf ,test pilots felt the rudder felt most like the stock Sabre when the max force was 300lbs. Thus, this would be a good benchmark for max force pedal forces our virtual pilot is capable of.

References

It would be easier if you stated what information you think is pertinent in the various references given. I'd stay clear of the DWC lecture slides, which the 12.5 figure stems from. I found a couple of 'interesting' things in them, and the book they state they are based on doesn't seem solid either, after a bit of googling on it.

Sadraey’ s books are taught all over the world in undergrad classes, what is your problem with it?

Rudder efficiency vs flight dynamics

The rudder is efficient if it can generate a beta angle. This is what is important for a xwind landing.

 

The turn, or lack thereof, generated by that beta angle, is flight dynamics/aerodynamics.

 

This has amused me as this thread has run its course.

 

Rgds,

 

/Fred

Beta angle is a function of (tan^-1 (Vw/U1) rudder efficiency is tied directly to flight dynamics. If Beta is altered, you’re inducing Fay which changes the flight path. You have to add Fay to the vector sum. And thus change the flight path. It’s basic Newtonian mechanics.

Curly,

it is always interesting to have an open discussion between adults. It is much better than arguing with kids who throw a hissy fit and negrep those who question their statements.

I was hoping for the former.

I am also relieved too see that I'm not the only one who finds your rather unorthodox take on algebra hard to follow. Somewhere in your flurry of edits you wrote that I was using the wrong formula when, in fact, it is the exact same formula you used for your calculations (as opposed to the one you posted, which I had to correct).

Getting the base facts straight

Rudder hinge moment C_Hr = H / (q*s_r * c_r) (1)

Note the parentheses.

Note the word "movable" in the above definitions.

And while we're at it,

Finally, for the vertical tail:

(All from AD069271/NA-50-1277).

Oh, and your second reference...

What's the magic word there?

The formula derived

From (1)

C_Hr = H / (q*s_aft * c_aft)

=>

H = C_Hr * q * s_aft * c_aft

I'll let you figure out the slugs and dynamic pressures on your own - you'll get there, just keep at it.

Where it all went wrong, and your plan just fell apart... (to paraphrase a good song!)

Where you went wrong was using the MAC for the entire vertical tail (sort of, you converted it correctly to 4.665 ft and then used 4.65 instead for your calculations) times the rudder area, instead of the area moment conveniently presented to you by the good people at North American.

Humble pie time! But just a small slice, thank you.

I take it you have no reference at all for the 6" of pedal travel? Well, let's use it anyway. And here, you are right - I miscalculated the mechanical advantage. That's what you get for rushing.

Hinge moment = force * moment arm.

For 27.5° of travel to result in 6" of movement, assuming a rigging arrangement converting the rotation to linear motion with a linear relationship between deflection of the rudder and the pedal, the moment arm r can be found through

27.5°/180°*pi*r=6"

r= 6"*180°/(27.5°*pi) = 12.5" = 1.04'

In other words, the pedal force is the hinge moment divided by 1.04, based on the specified assumption of ft-lbs. The formula google had found you used in-lbs. It is unfortunate that the AD069871 report does not specify the unit. There's another unit ambiguity in that they specify the average cord aft of the hinge line to be in inches, but specify the area moment in ft^3. If the moment is in in-lb, then your mechanical advantage calculations are correct - but then it stands to reason that the control surface average cord aft of the hinge line is indeed in inches, as specified, and we have to multiply the given area moment by 12. Net result: My figures.

Sanity check

The xwind landing case would likely be the design constraint used for the rudder system. For a sanity check let us use 132 KTAS and see if full deflection of the rudder will give us a pedal force which is in the ballpark and not a factor 12 off.

Thats, again, conveniently M.20 (225 ft/s) so C_Hr = .265 at 0° yaw.

H = C_Hr * q * s_aft * c_aft = .265 * (.00237*225^2/2) * 9.72 = 154 ft-lbs.

With mechanical advantage, 154/1.04 = 148 lfb at the pedal.

If we instead assume a hinge moment in in-lbs, we must assume that the rudder average chord aft of the hinge line is also to be specified in in rather than ft, giving

H = C_Hr * q * s_aft * c_aft = .265 * (.00237*225^2/2) * 9.72*12 = 1854 in-lbs.

Then, your mechanical advantage calculation is correct and we get a pedal force of

1854/12.5 = 148 lbf.

What a peculiar coincidence! The pedal force required for full rudder deflection at the recommended limiting crosswind is almost spot on the pedal force deemed manageable by the FAA.

Loose ends

As for the fin, we can discuss that when you stop using pounds of hinge moment as a force to make some kind of point. Or maybe not.

Your link on reversible control systems references rather interesting figures on rudder pedal forces as well. Whaddya know, 150 lbf. Now, how can this be? In the lab, 300-400 lbf from test subjects, yet the regs call for 150 lbf - written down from a previous figure of 180 lbf? The thing is: You cannot satisfactorily fly an aircraft while exerting all the force you can muster on a pedal. Test pilots would not sign off on a xwind capability requiring them to strain until they turn blue on short final, and the mapping between the maximum deflection of our Saiteks and the control force applied by our virtual alter egos should not mimic such a lab value - unrealistic in every day operations.

As for the book you're such a big fan of, read your quote. Assuming you are quoting it correctly, isn't there a thing or two which seems a bit peculiar right there? The lecture slides you linked had those "huh?" moments liberally sprinkled throughout the text, for anyone to find. Where is it used for undergrad training? If the slides based on it are representative, that's job security for the old hands in the industry right there.

Can you find anything peculiar in the image you posted?

More pie, and an exercise for the avid reader

I did make a mistake in calculating the crab angle, resulting in an error of almost -0.2°. Can you spot it?

Whoooosh

I'm afraid the high school physics weren't exactly the answer to the point I was making.

Summary

I don't think you present much of a case for anything to be changed.

Does this mean I believe everything is fine with the model? Nope, nor do I claim to have found anything wrong. Flat turns should be possible, and they weren't really a while ago. OTOH, they have very limited practical use. As long as I can de-crab and sideslip, I'm a happy camper. The latter would obviously be affected if the lateral forces due to beta angle are off (or missing), but I haven't done the testing to see if anything is really amiss.

Curly,

it is always interesting to have an open discussion between adults. It is much better than arguing with kids who throw a hissy fit and negrep those who question their statements.

 

Summary

I don't think you present much of a case for anything to be changed.

 

Does this mean I believe everything is fine with the model? Nope, nor do I claim to have found anything wrong. Flat turns should be possible, and they weren't really a while ago. OTOH, they have very limited practical use. As long as I can de-crab and sideslip, I'm a happy camper. The latter would obviously be affected if the lateral forces due to beta angle are off (or missing), but I haven't done the testing to see if anything is really amiss.

The equation for hinge moment coefficient is Ch= H / (q *Se * Ce)

The bottom term of the equation is a whole integer and treated that way the through the whole order of operations.

http://www.dept.aoe.vt.edu/~lutze/AOE3134/Stickfreecharacteristics.pdf

The area moment is in cubed feet it is a measure of volume. It should be apparent why you wouldn't use it in a force equation. Slugs are a measure of Mass not force, that's why you need to convert them to feet lbs for dynamic pressure calculations. This is why all your calculations are wrong.

I'll admit I made some errors in my entail post, but the pedal moments come out 505 ft lbs at mach .4 at sea level. Which means the rudder deflections are still to low. They also don't take into account dynamic pressures. Which as alt rises would mean more rudder throw for a given speed.

Sadraey is required course material at Boston University, TU Dresden and RMIT University.

http://www.bu.edu/me/files/2014/09/ME-408-F14-Syllabus-Geiger.docx

http://tu-dresden.de/die_tu_dresden/fakultaeten/fakultaet_maschinenwesen/ilr/aero/studium/lfa/LFT_LFA_E_2014.pdf

http://www1.rmit.edu.au/courses/c6131aero5952c1445

I’m using true dynamic pressure. Which at sea level, at a speed of mach .4 = 237.01 lbs/square foot. Where p= .076474 lbm/ft3 and v = 446.58 feet per second. And q = Dynamic pressure = 1/2pv^2. Thus q=237.01

Feel free to verify.

First off, it is the greek character 'rho', not 'p'. Getting that right will save you a lot of confusion later on. Lowercase 'rho' is the standard designator for density, in aeronautics and pretty much everywhere else in science.

Second, yes,

q=rho*V^2/2

But you didn't try calculating q yourself, using your suggested figures, did you?

Instead, you used the value you pulled off your lookup tables on the web - which, within the confines of numerical precision, is the same value I calculated and not what you would have arrived at.

Feel free to verify.

Now, didn't that quote suddenly get embarrassing? Yes, verification is a good thing, especially when relying on what google found you online.

This is why all your calculations are wrong.

Oh, indeed? And as you used the same value, even though you suggested the wrong method of calculating it, that would make your calculations.... ?

I also suggest a bit of reading. Start with your own references, conveniently repeated above.

The area moment is in cubed feet it is a measure of volume. It should be apparent why you wouldn't use it in a force equation.

What are you on about? No, I won't even bother outlining why that's wrong. Read your own references for a change. The important bit is, as mentioned, reproduced in this very thread for your convenience (even though it seems to have turned into a bit of an inconvenience for you).

I'll admit I made some errors in my entail post, but the pedal moments come out 505 ft lbs at mach .4 at sea level.

Pedal forces or pedal moments about a defined point and with a defined moment arm. Choose one.

And the rudder hinge moment for full deflection comes out to 540 lb-ft, as shown in my initial post. Glad we're in the same ballpark, finally. Show how you arrived at the figure of 505 lb-ft and we may be able to find the discrepancy.

That would be about 2.5 times the recommended maximum force for normal operation at the pedals, and still far in exceedance of the maximum achievable force in a lab setting.

The equation for hinge moment coefficient is Ch= H / (q *Se * Ce)

The bottom term of the equation is a whole integer and treated that way the through the whole order of operations.

What are you on about? No, you cannot skip the parentheses, as shown.

Youthful enthusiasm shall normally be applauded and supported. I would have been all pedagogical, but your attitude in the forums and out of public view just zapped my desire to go the extra mile for you.

I would leave you alone rather than spend my spare time administering physical violence to this particular deceased creature of equestrian nature, weren't it for the fact that some apparent traction had been gained by your attempt at providing seemingly credible 'evidence' for something being amiss.

Slugs are a measure of Mass not force, that's why you need to convert them to feet lbs for dynamic pressure calculations.

q = rho*V^2/2

Where rho is density kg/m^3

V is true airspeed m/s

[q]=[rho][V^2]=kg/m^3 * m^2/s^2 = kg / (m s^2)

F=ma

gives

[m]=[F/a]=N*s^2/m

so

[q] = N s^2 m/(m s^2) = N s^2 / (m^2 s^2) = N/m^2

That's for mass per volume. Go ahead and insert a moment per volume or force per volume, both of which you seem to be suggesting simultaneously, and see where it gets you.

Sadraey is required course material at Boston University, TU Dresden and RMIT University.

Did you actually read the links you provided? If you did, you're trying to pull a fast one and hoping no one else will read them. If you didn't, in your rush to google references to gain apparent credibility to your claims, well... I suggest you do so now.

No, research is not done by googling convenient search terms, browsing through the hits for formulas, plugging in numbers according to TSABR and then posting the results liberally sprinkled with links for faux credibility.

The important bit is reading and understanding the references you find and decide to use. Yes, it is an unfortunate fact of life that this will take a lot more effort.

I think I'm done here, so save yourself the trouble of replying.

It would at this point be very interesting to find data on the relationship between beta angle and rudder deflection. With that, we'd know exactly what is correct, as far as rudder efficiency goes. I suspect the devs have that information available. Perhaps a reference could be provided in order to settle that part of the discussion?

All well and good....

Might I recommend - professors - that we put down the chalk, and stroll from the Embry Riddle aeronautics lab to the A&P hangar?

In that location, we can hail the Belsimtek mechanics over to our F-86E that is waiting conveniently with all access panels and hatches removed.

If you can agree to "split the diff" between 505-540 lbs. - after much abuse of the Greek alphabet, maybe the Belsimtek A&P trainees-without-math-data-handbooks can re-rig the pulleys/cabling/fasteners/etc. to put this bird in a condition wherein we can slap an airworthiness certificate on it.

Sound like a plan? :thumbup:

Indeed.

First point of order would be to establish if the empirical rudder deflections achievable roughly match the calculated values.

To get further, we need the documentation on the beta/delta_r relationship.

I haven't got the time, but I put together a graph anyway assuming 6" pedal throw and 180 lbf available at the pedal. I don't know if I have much trust in the >=M1.0 values, and the available deflection at M.2 is an obvious extrapolation - force for full deflection calculated earlier to be less than 180 lbf.

I was looking for some F-86 stuff when I stumbled across an instructional\familiarization video for the Naval\Marine version of the F-86 E&F, the FJ2 Fury. In it pilots are warned to go easy on the rudder during firing to avoid skidding off the target and that the aircraft requires very little rudder. They start to talk about the rudder at about 8:09 in the video.

 

 

Apologies if this has already been posted.

Great find Moggster, I think many should focus on the practical aspect aka the video saying there was an effective rudder vice Monday Morning Engineering.

I think this discussion is going to simple dispute.

You (our customers) stated that there is some problem. We are doing our best to find the solution (whatever it is).

You have the video & calculations now. Any news ?

Looking forward to seeing the new rudder authority in the Sabre.

 

F-86F

• Rudder authority increased