Curly

National Stock Numbers confirm the the motor is model Ysr-113-tc-1 it weighs 387 lbs and contains 280 lbs of propellant. The propellant is listed as Tp-11-1159. https://nationalstocknumber.info/national-stock-number/1337-01-162-3421 This claim is backed by secondary sources in the NASA and DTIC databases. https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19870011601.pdf Page 3-194 in the NASA doc http://www.dtic.mil/dtic/tr/fulltext/u2/a342309.pdf Page 9: the fuel type and composition is listed. Specifications for this fuel can be found in Johns Hopkins Chemical Propulsion Information Agency “Solid Propellant Manual”. The manual may also be know as “Solid Propellant Mechanical Behavior Manual”. The agency since has be renamed and can be found here. https://www.erg.jhu.edu/ The work does not appear to be publicly available.

That would be a model with a perfect finish. Page 296-297 discusses the discrepancy between the results obtained with a model and full scale tests. It cites a .2 lower Cl in the real world vs a model. Owing to roughness, leakage, intakes and armament installations. Taking .2 off the Cl max of the model subsequently puts the model results in agreement with the wind tunnel and flight testing of the full scale f6f. Which provides a good analog for the 190's airfoil. Page 20 of NACA 829 goes into effect of the finish and seal on a service wing. Based on what's presented there, I think it's doubtful that there is anyway that a service condition 190 A or D has a clean configuration Cl max of 1.58. Not to mention Yo-Yo has presented plenty of both direct and anecdotal evidence that supports his claims of Cl max ~1.38.

Are you referring to CL max of the 2d airfoil sectionals in that report? As I have seen much lower CL max reported for full scale wind tunnel and flight tests of those aircraft. The F6F for example is shown to have CL max of 1.3-1.4 in the clean configuration and close to 1.7 with the flaps fully deployed. I find it plausible that in take off configuration, the CL max of the Dora is 1.58. The full flap of the F6F nets you a gain of .4 or .3 Cl max. It seems well within the realm of possibility that in the half flap configuration of the Dora you’re getting a increase of Cl max by .2 NACA reports 829 and 1044 both provide interesting result for wind tunnel and flight testing CL max numbers. Also are the illustrations you’re looking for in regards to aspect ratio and Cl located here http://history.nasa.gov/SP-367/chapt4.htm Figures 56 and 57? [url=http://history.nasa.gov/SP-367/chapt4.htm][/url]

I got board and did a little bit of research. 48 reels of micro film consisting of blueprints and technical drawings of the P-47, P-72, P-35, F-84, and other aircraft published by Republic Aviation Corp exists. According to the New York State Library it’s held in Hofstra University Library, Long Island Studies Institute, West Campus Library, Hempstead, N.Y. 11549. You can search for the catalog record here https://nysl.nysed.gov/uhtbin/cgisirsi.exe/x/x/0/49/ Search Technical AND drawings AND republic AND aviation. Syntax is important as searching for “Technical drawings Republic Aviation” will return no results. Though I could find no record of the item by querying the Hofsta Library database. Some more leg work or contacting the Library may be necessary to obtain said document. Interesting enough The Cradle Of Aviation Museum’s Archive keeps popping up in my searches. Their archive seems pretty rich and is located the town over from where the micro film is held. So if you're headed out there make two stops;). https://beta.worldcat.org/archivegrid/collection/data/155558167 This archive also seems to be present on this site http://aircorpslibrary.com/aircraft/p-47-republic-thunderbolt as they list 19,416 blueprint and drawings in their collection. Though I can’t say for sure as there is a access fee. Republic’s parent company Fairchild’s historical files are held at the Smithsonian Air and Space Museum. Though the collection doesn’t appear to hold a wealth of P-47 / F-47 data. The manifest of items indicates that it consists of drawings and manuals. Link to Smithsonian overview of the Fairchild collection: https://airandspace.si.edu/collection-objects/fairchild-industries-inc-collection-1919-1980 Link to manifest of items in the collection: http://sirismm.si.edu/EADpdfs/NASM.1989-0060.pdf Hope this helps a bit. Though if specifics as to what you’re looking for could be provided it might help us better serve you.

The original archives were destroyed when Republic Aviation went out of business. I would attempt to make contact with the Cradle of Aviation Museum, they're on Long Island where Republic was located. So the may have a few things. http://www.airspacemag.com/military-aviation/when-republic-aviation-folded-69197851/?no-ist

Both are correct. The difference is in the scale of the instruments. The first is formula to convert degrees to a 45 unit scale. The second is to convert degrees to a 30 unit scale. It's no surprise your reference is a US Navy document. The US Navy has historically used 30 degree scales in it's aircraft. With the face rotated so that the landing approach AOA is at 3 o'clock. The first formula is for a 45 unit indicator which is used in the F-15. The US Air Force has had less standardization in it's AOA displays than the US Navy historically. It was a very hot topic on what and how it should be presented during the time the F-15, F-16 and A-10 were designed. So it's not unusual that all the systems vary a bit. There is a nice NASA document about the history of AOA. While it doesn't say directly why the armed forces adopted the unit of AOA. It seems to imply that displaying units provides the pilot with pertinent information in the easiest to read fashion. Hence the Navy's 30 degree 3 o'clock AOA gauges. Here's a link to that NASA doc. http://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/20140011419.pdf

I devised a handy method for figuring a 3 degree glide slope for aircraft configured in metrics. Be advised it has some accuracy limitations depending on airspeed. This method is most accurate between 150 kph and 300 kph. My method is accurate to within .4 degrees between those speeds. Here is what you do. First, take your approach speed and multiply it by .01 then add 1. A simpler way to preform the function is to insert a decimal point after the first digit in your air speed and add 1 to that. For example lets say our approach speed is 250 KPH, to find our 3 degree glide slope we would just place a decimal point between the 2 and the 5 and add 1. So 250 becomes 3.5, or 250 * .01 + 1 = 3.5. So to fly a 3 degree glide slope at 250 kph we simple have to maintain a sink rate of 3.5 meters per second. This method works for a range of speed from ~150 to 300Kph with an accuracy of .4 degrees. This gives you a quick and easy method to fly a standard approach. They way this works is that if we convert our airspeed to meters per second and divide the sink rate by the airspeed and take the inverse tangent of that number, we’ll have our flight path angle (glide slope). Or the Inverse tangent of VVI / Air speed = flight path angle (alpha). There are some examples below Example 1 250 kph = 69.4444 mps The VVI from the above method is 3.5, 250 *.01 +1 = 3.5 3.5 / 69.4444 = 0.05040003225602 Arctan of .0504 = 2.8 degrees which is ~ 3 degrees Example 2 200 Kph = 55.5556 ms 3 mps VVI 3/55.55556 = 0.539 Arctan .053999 = 3.090 Example 3 280 Kph = 77.77 ms 3.8 Ms VVI 3.8 / 77.778 = .0488570032657 Arctan of .0488570032657 = 2.797 degrees 150 Kph @ 2.5 MPS VVI = 3.43 degree .4 degree deviation 300 Kph @ 4 MPS VVI = 2.74 degree .3 degree deviation.

I don’t want to put words in Yo-Yo’s mouth but, what he’s seems to be saying is that the linear equations Crumps charts are based do not accurately predict lift coefficients at high angles off attack. the NACA came to this conclusion around 1947 while testing swept wings. Even ignoring other factors not taken into account by Crumps model, propeller effects, etc, the margin of error in his methodology is so great that it renders the predictive and comparative power of the equation moot in high alpha conditions. Perkins and Hage is fine for most circumstances, but in the case being examined here a different approach is needed. The NACA, then NASA ran into similar issues as the performance of aircraft increased. Equations were no longer predicting experimental data. To reconcile this new methodologies were adapted, including Panel and Vortex Lattice methods. Even if such methodologies were adapted to predict CL, CD, etc in Crump’s model. The predictive power of the model would be doubtful because it needs to take in to account, the variability of propeller efficiency and propeller downwash effects. My interpretation of Yo Yo’s statements are, that there are limitations to the methodology used which say the flight model is wrong. The current test case exceeds the limitations of Crumps modeling and the more advanced modeling used in the sim would be a more accurate predictor of reality. Thus the 190 will stay the same until someone comes up with a proof or more accurate model which shows the current implementation is wrong. Section lift curves are not linear particularly at high angles of attack and analytical solutions are not feasible. Links to info on Vortex Lattice Method http://www.dept.aoe.vt.edu/~mason/Mason_f/CAtxtChap6.pdf http://www-mdp.eng.cam.ac.uk/web/library/enginfo/aerothermal_dvd_only/aero/vlm/vlm.html

im just going to delete my stupidity

What I'm asking is if the oleo strut and it's interaction are modeled. This would make two wheeled landings less bouncy. As the strut would absorb some the force of the impact. Thus lessening the monument / impulse which causes the pitch up encountered at touchdown.

Ballooning is caused by a force acting on the aircraft. That force is created by and equal to the force of the gear colliding with the runway. According to Yo Yo, all the gear is doing is prolonging the collision. The landing gear causes the momentum changes over a greater time period so there is less force acting on the airframe at any give time frame. Though because momentum is conserved the total force acting on the airframe is the same. Which makes sense if you consider landing gear springs that store the kinetic energy of the impact and release them over time. The momentum at landing with a positive sink rate causes the plane rotate about it’s CG. The reason why pushing the stick forward counters this ballooning behavior is that it creates a torque equal to but in the opposite direction of the hard landing torque. So a Mustang weighing 5000kg impacting the runway at with a sink rate of1 meter per second. Impacts with a Force = M A 5000kg *1^2 = 5000 Newtons of force on landing. So lets just assume that the moment arm from the elevator to the CG is the same distance as from the landing gear. When landing at 1ms, to counter the ballooning tendency of the aircraft the elevator has to generate 5000 Newtons of force. If you get slow and cannot generate this force at the elevator than you will likely see the ballooning tendency as the noted in the video in post . It’s not a CG or elevator effectiveness issue. What’s going is that the landing gear are acting as springs and simply displacing the force of the impact over a longer period of time. In realty the type of landing struts fitted to the P-51, oleo, reduces force at impact. During landing oil is forced through an orifice in the strut causing the temperature of the oil in the strut rise. Kinetic energy from the landing is transferred into thermal energy and dispersed thought the casing. Thereby dissipating part of the impact forces. Though they also function to prolong the collision and reduce forces on the aircraft ends up landing with a lower amount of Newtons of force. See page 13-8 to 13-12 for a description of the oleo strut. https://www.faa.gov/regulations_policies/handbooks_manuals/aircraft/amt_airframe_handbook/media/ama_Ch13.pdf

I have a question about this statement. I get that linear momentum is conserved in a collision, but don’t the Oleo type shock absorbers, which the p-51 is fitted with, convert some of the kinetic energy into thermal energy during their compression? Thus dissipating some of the kinetic energy at impact. I get that this would be true if the landing gear were simply springs, or dampened springs. But it seems that the landing struts have to absorb some of the energy of impact. Otherwise, carrier landings in a tricycle configurations, which you do not flair, would put tremendous strain on the nose gear as craft rotates nose down around the center of CG. It would also make an even more dubious proposition. It makes sense that shocks are absorbing some of the kinetic energy of impact when you consider that the maximum rate of descent at touchdown for the F/A-18 is around 24 feet per second, and the travel of the landing gear is 25 inches. Or is there something I’m missing in regards to the equations of motion.?

The thread size on the Warthog is M36 x 2. If you're going to a Home Depot in the US it is very unlikely that they will have any pipe stock or fittings with that threading. They maybe able to cut that thread size for you if they have metric tap and die set, though i get the feeling most stores do not. You'd want either 1 1/2 inch or 40mm pipe. This is why you see so many people making a Gardena extensions. Though you could 3'd print and adapter with metric threading on 1 end and us threads on the other. Or buy a tap and die set and cut the threads yourself.

Sent a Ian a PM about adding LoGet functions to DCS BIOS. He asked that I post the conversation as he thought it may help others. Hi Curly, thanks for your interest in DCS-BIOS! Take a look at the code for the CommonData export module. It exports generic position/altitude/heading data using LoGet* functions. Because DCS-BIOS only exports unsigned integers and strings, you will have to set some reasonable lower and upper bound and then map that to an integer range (I use 16-bit values, i.e. from 0 to 65535 for everything that involves floats). If you choose to modify the CommonData module, also make sure that your plane has an entry in AircraftList.lua to ensure that the CommonData module gets activated. Note that the CommonData module intentionally does not export everything that is available through LoGet* functions. The CommonData module will be activated in every aircraft DCS-BIOS knows about, so it has to work in all of them. I also have some concern about the volume of exported data. Exporting data that changes all the time but is not being used will still consume some processing power on connected microcontrollers. The current things it export are justified by the "moving map" use case. For example, the pitch and bank angles are not being exported. DCS-BIOS does not aim to be a TacView replacement, and physical or virtual panels won't use this data when the position of the actual in-cockpit ADI is available, which takes into account the pitch adjustment dial, failure modes (broken instrument, power off), etc. The correct place to export generic pitch and bank values would be a new "FC3" module that gets activated in all planes that don't have a clickable cockpit (or at least don't have their own DCS-BIOS export module yet). I think the same would apply to the acceleration value, as e.g. the A-10C and Ka-50 have separate G-meters. PS: It occurs to me that I have just spent a few minutes writing an explanation that could also be useful to other people. Can you create a thread on the public boards to continue this so that this information, any follow-up questions and their answers are available to others?

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. 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) 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. 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- That 3.75 inches of pedal travel is for the mock up on which the force tests were conducted, not the actual Sabre. 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. Again based on your assumption that deploying large rudder deflection presents a structural danger. 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. Sadraey’ s books are taught all over the world in undergrad classes, what is your problem with it? 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.

The issue with the F-86’s rudder comes down to 2 points, can the rudder be effectively displaced by the pilot throughout the flight envelope, and does the rudder generate enough force to create yaw and sideforce. First, lets look at rudder displacement. From Fox’s video and testing it appears that the full range of the rudder deflection is not available thorough the flight envelope. I was able to track down the hinge moment coefficient and calculate the rudder pedal forces on the actual F-86. http://www.dtic.mil/get-tr-doc/pdf?AD=AD0069271 At the highest hinge moment coefficients and large dynamic pressures, the pilot is still able to exert enough force to deflect the rudder and keep the rudder deflected much further than in game. And at much greater yaw angles than is possible in the current flight model. While forces may be high, they are within the realm of forces that a pilot is capable of generating. As defined by the NACA and Airforce reports. IE forces up 300-410 lbs http://www.dtic.mil/dtic/tr/fulltext/u2/639028.pdf A later modification program of Sabers, which provided a hydraulically boosted rudder, found that the craft behaved and felt similar to the original when the equivalent max force of 300lbs of was transmitted through the pedals via the boost system. Thus in the sim we should be able to hold rudder deflections and yaw angles that require up to and above 300 lbs of pedal actuation force. As was the case by the men who piloted the Sabre. http://www.dtic.mil/dtic/tr/fulltext/u2/106019.pdf Below are the hinge moment calculations. On the issue of rudder effectiveness, The F-86 flight manual states that the rudder is effective for a crabbed approach up to 25 knots. If the rudder has enough power to crab in a 25 kt crosswind, then those forces will be enough to cause changes of it's flight path vector without a crosswind component. Also, as velocity increases, for a given rudder displacement, the rudder actually generates more lift and greater side forces and yaw moments. Furthermore the rudder is well sized for the vertical tail. With a rudder area to tail area ratio of .242 it’s similar in proportions to large passenger craft. http://faculty.dwc.edu/sadraey/Rudder%20Design.pdf http://www.dept.aoe.vt.edu/~lutze/AOE3134/YawMoment.pdf https://courses.cit.cornell.edu/mae5070/Caughey_2011_04.pdf http://www.dept.aoe.vt.edu/~mason/Mason_f/LDstabdoc.pdf Personal accounts also indicate that it was possible to turn the Sabre 180 degrees using the rudder only. https://books.google.com/books?id=aDNVY4lgOAQC&lpg=PA15&ots=aSy4pWoLqK&dq=f%2086%20sabre%20hydraulics%20out%20%20rudder%20cable%20turn&pg=PA15#v=onepage&q&f=false While not Yaw moment or force reports for the f-86, it seems that the modeling of the rudder forces, either due to lack of rudder displacement or incorrect modeling of forces prevents the DCS exhibiting accurate rudder forces. If access to reports listed below is made available. It would be possible to more accurately make an assessment of the rudder performance of the F-86: December 1950: Johnston, E. W. F-86E Characteristics for Lateral Directional Dynamic Stability Studies. North American Aviation, Inc. Report NA-52-286 March 1952 NA 50 928 Aerodynamic Dimensional Data F86E. NA 50 1277 NA 47 1043, NAAL 99 NA 49 467, NAAL 99 From observing the behavior of the craft In game I’m not sure what Belsimtik is attempting to model with the current rudder behavior. I could speculate and say they took the hinge moment coefficients, and interpreted these as the ranges for the rudder at various speeds and normal forces displaced by the rudder. Which would be easy to do, because the report is poorly formatted and labeled. Thus making it seems like the entire range of motion isn’t available, because the report’s author was trying to fit a lot of data onto 2 pages. Though this is just speculation, thus it would be nice if the Developers would discuss their source material and how they arrived at the current behavior of the rudder. F-86 maintenance manual, shows rudder system layout: http://www.avialogs.com/viewer/avialogs-documentviewer.php?id=15729

I'd like to enter for the 109, thanks.

I see that the net effect of the PTC is modeled. What I meant was, is the system as a whole modeled? You could cook up some tabular data and LERP between points based on air speed and acceleration to determine Stab position to achieve a constant 1 g trim. When I ask is PTC really modeled, I mean PTC as an integer which is computed along with other factors to determine the position of flight surfaces. The position of the PTC arm is fundamentally an integer (it's part of hydro-mechanical computer) which is summed with the position of the Pitch Ratio Changer (PRC) to drive a PRCA hydraulic boost actuator which moves all the downstream linkage and also drives a linkage input to the PRCA roll axis and to the Aileron Rudder Interconnect (ARI). With both the ARI and PRCA feeding data across to each other. With PRCA feed into Lateral Longitudinal Mixer and the ARI directly into the rudders. Resulting in a complex dynamic in which both the ARI and PRCA are determining the position of the flight control surfaces along with CAS compensating. If you use the Mechanical lateral control authority schedule alone to determine aileron, stabilator and rudder position of a given input you would get results which tend to be on the oversteer side. The missing piece is the roll and pitch rate changers. Which, modify the amount of control surface deflection based on stick input and air speed data. The scheduling issues seem to be a result of the lack of the the pitch and roll rate ratio controller and changer. The reason I asked if it was really modeled was because A. the the Pitch ratio gauge in the F-15 is static and B. disengaging the pitch CAS results in unexpected behavior. The hydraulic Pitch system should still keep the PTC functionality, unless turning it off put's it in emergency mode. Where even there it should lock to more a stick neutral position. The force to G ratio should stay fairly consistent through most of the flight envelope. Except at a high angles and high and low speeds. When the pitch ratio controller is near zero and you can't get enough elevator to pull the desired load factor. Other wise CAS should be driving the stabilator surfaces to match the load factor based on stick position. It's one of the prime functions of that system.

Is the PTC really modeled then? It seems likes it's more of a conventional trim system. Ie hitting the trim causes the elevator pitch to change. In the F 15, the trim signals alter zero force position reference. For example: If the pilot trims into a 2g turn at zero stick pitch force. The 2g becomes the reference for zero stick force instead of the normal 1g. To me it seems like a very basic trim model is implented. I'm not even sure how well the CAS is modeled. Maybe it's because I fly with some non linearity, but it seems like G per stick defection stays does not stay constant. Which is what CAS does.

CAS matches performance to for stick force. The stick force is set at 4.25 lb/g, if the pilot pulls 8.5 lbs of for force on the stick the and the craft does not pull 2 g's the CAS system will power servo motor to drive the stabilators up to +-10 degrees to match the required input. The CAS servos connect to the mixer linkage which will drive the stabilator hydraulics directly. The mechanical inputs are transmitted to flight control surface actuators by way of the control stick boost pitch compensator (CSBPC). The CSBPC is a hydromechanical analog computer made up of two units: a pitch and roll channel assembly (PRCA) and an aileron rudder interconnect. The pitch trim controller (PTC) is part of the PRCA, it's travel is limited by the pitch ratio changer.The PTC is driven by a control valve which operates as a function of normal acceleration (load factor) and control stick position. The PTC is damped to prevent disturbances from affecting the mechanical control system. PTC travel is limited as the PRC nears minimum ratio. When pitch CAS is engaged, the CAS interconnect (CASI) servo controls the PTC. Limited pitch commands are applied to the PRCA CAS interconnect (CASI) servo. The CASI servo drives the PRCA pitch trim compensator and forces the mechanical system to track the pitch CAS. The tracking function minimizes differences between CAS and mechanical commands so that if pitch CAS fails, the mechanical system can take control at the point of failure. The function prevents CAS from back driving the control stick and keeps the neutral point the same when CAS engaguges. When they talk of the PTC driving CAS It's descbing the hydro-mechanical analog computer taking input and compenstaing for CAS. http://www.f15sim.com/images/F-15_Longitudinal_Control-2.jpg Perhaps the trim probelm at 40,000 ft has to do with with PTC only having +-10 degrees of stabilator deflection? As they figured CAS and PTC were the same. PTC really should have almost unlimited stabolitor deflection unless the mechanical advantage (number of degrees of collective stabilator deflection per degree of longitudinal stick deflection) via the PRC is low. http://www.f15sim.com/images/F-15_Pitch_Control-1.jpg http://books.google.com/books?id=NUnlAwAAQBAJ&dq=f+15+cas+servo&source=gbs_navlinks_s

Maybe the wind isn't applying greater force to the tail of the aircraft, and instead provides only a singular moment to body the of the aircraft. So that the weathervaning behavior is not exhibited. That or, the wind over comes the lateral static friction of the wheels, and you're basically skidding, down the runway . I also think alot of people's troubles with landing the Dora has to do with the approach setup Wags uses in the landing video. He advises using an approach speed of 220 and VVI of 2.5 to 5. That's a glideslope of 4 to 8 degrees. Both are on the steep side for an approach. For a 3 degree glide slope at 220kph, you'd want a VVI of 1.8 (call it 2). This should make the flare and touchdown a bit easier to manage. You'll have to spend less time hovering over the runway at 200kph, or flaring out. Plus it should be easier to see your touch down area with a shallower approach.

How is power derived from Alpha, what is the relation of alpha to power? That is a calculation of the mass of the required airflow needed to achieve the mixture strength at a given a fuel flow. It is not a calculation of the actual airflow inside the manifold. Mixture setting (Air mass/Fuel Mass) is the determinate of power output of combustion. Power output of combustion is the determinate of engine power output. A mixture of setting of 12.5 produces more power than a mixture setting of 14.9. Therefore, to determine the needed airflow of an engine at a given power setting and mixture strength; One observes fuel flows needed to derive power at the given fuel air mixture settings. From there you can calculate airflow needed. Gasoline also has limits on where where it is combustible. At sea level gasoline will not ignite if it is less than 1.7 % of the mixture. It also won't ignite if it's more than 7.5% of the mixture. As altitude increases and the air gets thinner and cooler those numbers increase. At alpha of .0011 there should no longer be combustion in the engine. .0011=(.017/1)/14.9 .0011=(.034/2)/14.9 And before alpha=.0011 power should be dropping rapidly. Back to issue of BSFC, which is where all this started. If you alter fuel flows you have to recalculate BSFC. Assuming a base fuel flow of 1055 and constant power output of 1770 hp at 3250, 2136 inch Displacement. Doubling the fuel flow takes the BSFC to 1.192, from .59. Thus resulting in a rise of the VE of the engine need to achieve the same power. 2.86= ( 9411 x 1770 x 1.192) / (2136 x 3250) If the mixture ratio stays the same and you double fuel flow, VE goes up more than double, 1.41 was base. VE is an expression of how much air compared to the volume of the cylinders, is needed for the power output of the engine. It allows us to reasonably evaluate the performance of an engine. At those fuel flows and that mix setting, for that power, you have to put a volume of air that is 2.86 times the volume of the cylinders in. We can then compute the sea level manifold pressure necessary to archive this. VE * 29.92 = Manifold absolute pressure in inches of mercury, HG. As 2.86*29.92= 85.5712 inHG Convert HG to ATA 85.5712/28.95 = 2.95 ATA It doesn't look reasonable, knowing the limits of forced induction the engine's capacity to sustain those pressures. VE and TE are great tools for evaluating the performance of an engine. Which is where this whole thing started. The debate was, does cooling provide a reasonable means to increase the horsepower of the Jumo by 100. By showing that only marginals gains in thermal efficiency we're need. I showed it was reasonable to say additional cooling is an efficient means to derive 100 hp.

The mixture is leaning out, it should be outputting less power. .7= (10.4/1)/14.9 .7= (20.8/2)/14.9 .8 = (10.4/.087)/14.9 .8= (20.8/1.74)/14.9 You're taking, roughly, 13 % of the fuel out from the mix; With every decline of .1 Alpha. You should see a drop in power. There is a lot less gas in the cylinders, combustion will be weaker. Unless you're starting out overly rich, than leaning would actually increase power. Or does your model derive power out based on the difference of Alpha from 14.9? Stoichiometry, 14.9, fuel air mixture, isn't the point of peak power in the fuel air mixture. It's just the point where all of the air input with the gas is also consumed. Peak power typically occurs somewhat richer than stoichiometry, down in the 12's (12.5 is typically assumed as best power mixture). While peak efficiency tends to be closer to stoichiometry but in the 15's. Mixture controls and a throttle that controls airflow are useful to a piston engine pilot because they allow one to control the power band. So that they are always maximizing the VE of their craft. A throttle which controls fuel flow is more appropriate for a jet, because the dynamics of the engine fuel flow has a very direct relation with power output. "Alpha = (Air mass flow/Fuel mass flow)/14.9" This is an equation to calculate the difference between current mix and stoichiometry. It's not a good equation to derive power out put of a piston engine from. As you're not taking into account the varying power output achieved by mixture settings and what's realistically achievable by the engine and the induction system. If you keep fuel flow constant in the equation, then rises in mass flow reach such high levels, that one is not operating in a realistic band of manifold pressures. Again, a 13 percent increase in mass airflow is need to keep the power constant. If the volume of cylinders stays constant, at some point your either likely to exceed the engine's capacity to sustain those MP, or be at such a place where the equipment need to compress the air to those level, would no longer be efficient. IE running a 4 horsepower supercharger to gain 2 horsepower. .7= (10.4/1)/14.9 .8= (11.92/1)/14.9 Further more, we can calculate the mass of the air based on the fuel flow. At a fuel flow of 1055lbs per hour (from the Jumo) assuming the same mixture here of 14.9 Air flow = 14.9 * fuel flow 10,972= 14.9 * 1055 (both sides in Lbs per Hour) 7= (10.4/1)/14.9 10,972/1055 Convert to Standard cubic feet per min Mass flow = Fuel flow(pph) * 3.246 3418 Scfm =1055 * 3.246 Air mass has rise with alpha, you hit a series of limiting returns at this point. If you wanted peak power, you'd be better off enriching the mixture. As your Mass air flow would be come realistically achievable. Before auto mix systems pilots tended to run to rich, because they thought it would give them more power. Though they easily could end up with less power than their planes we're capable of producing, trying to dump gas into a craft who's beyond the VE to use it. This also lead to maintenance nightmares for ground crews. As running rich would tend to cause issues with plug fouling, ect.

It's not the increased temperature of the cylinder heads (heat) that creates the motion of the piston. The actual physical force, that moves the piston, is the expansion of fuel air mixture, as it expands when it combusts. A pressure wave smacking into the piston head. You can heat the pistons all you want and you will not get more power. Unless that heat is accompanied by either a great volume or a more rapid expansion of the fuel air mixture. Put a piston on a stove in a vacuum, turn it to high, let me know when it moves. Heat is a form of work and units of measure of both can be converted into each other. A British Termal Unit is defined as the heat energy required to raise the temperature of one pound of pure water by one degree F, and is equivalent to 778 foot-pounds of work / energy. One horsepower (33,000 ft-lbs per minute) is the equivalent of 42.4 BTU's per minute or 2545 BTU's per hour 33,000 / 778 = 42.4165 42.4165 × 60 = 2544.98 The thermal efficiency of a heat engine is the percentage of heat energy that is transformed into work. TE= work / heat(energy). If you do less work for more heat(energy), your have a lower TE. Look at the Jumo 213-A. It only captures .22 of the potential work that fuel is capable of outputting, 1770 HP = TE * 1054.58 * 19,000, If it was capable of converting all the BTU's into work it would output 7867.166 HP. Since we know the energy output of the fuel is constant, we can then say the engine is only 22 percent efficient in capturing the potential energy of the fuel. Thus we say the TE of 213-a .225 or TE=.225. The block of the engine is physically cooler when MW 50 is introduced to the cylinders. As it's being sprayed with a water coolant mixture. As show by the previous calculations, it's likely a large percentage of the increase power output is due to the better thermal efficiency. Don't get me worng, I'm an not saying the majority of the increased power output isn't due to VE enchantments provided by the MW 50. I'm saying that the 100 hp quoted in the manual due to cooling (TE gains) is reasonable. The portion of the sentence that is somewhat misleading is; "due to the fact that a cooler engine can pull in more air." For the sentence to be more succinct it should read; "Turning the system on immediately increases engine power by almost 100 HP due to increases in the thermal efficiency of the engine." It should then say "The system provides as much as 440 additional horsepower, over max output, due to gains in airflow and thermal efficiency. More power through cooler air is a result of Charles' Law. Which states Volume is proportional to temperature. A cooler manifold environment means you can get more air in the same area for the same amount pressure, ATA in the case of Fw190 measurement. This means your restoring the parts of oxygen per unit of air back to sea level conditions, or greater, depending on how much you cool the air. The anti detonation effects derived from mw50 are due to the increased air in the mixture. Which prevents early detonation. You're essentially diluting pure gas with air so it doesn't ignite before top dead center. The increased horsepower benefits of the anti detonation effects are reaped in the form of more aggressive the ignition timings, closer to top dead center. Which results in a higher compression ratio and hence a higher VE. The brawn in the MW50 is really nothing with out the brains of the MBG control unit.