RV-7 Landing Gear Weight Distribution in 3 point stance

Hello,

I'm working on a college aerodynamics project on the Van's RV-7 (one of my partners is building one, so we chose it). It's nothing crazy, the class is for pilots not engineers. Anyway, for this week I need to calculate the landing roll distance for an RV-7, and part of the calculation depends on the weight distribution for the aircraft (since only the mains provide braking).

I have looked at the weight distribution from the weight and balance thread but those are taken at 0 pitch attitude (understandable to make calculating the CG easier) but in the 3 point the aircraft appears to be in about a 12 degree pitch attitude. In the absence of better data I have been estimated a 700/700/100 lb loading for a 1500 lb RV-7, which is a few % more than the numbers I saw on the W&B thread at 0 degrees pitch. I realize this will vary with CG too. For every 50 pounds more on the tailwheel, the landing roll seems to increase by about 3%. So far at 1500 lb gross weight, sea level, ISA conditions, we calculated a 341 foot ground roll with no flaps, assuming touchdown at stall speed of 48.6 KTAS (1500 lb, flaps up).

Anyway, if there is any info about this I'd appreciate it. Here is a link to the table of all the other work we've done on the RV-7 so far if anyone cares. I know it's not the most accurate since I had to make a lot of assumptions about the airfoil (extrapolated a 23013.5 from a NACA 23012 and NACA 23015 at 1,000,000 reynolds number, estimated oswald efficiency at 0.8, estimated form drag at a 0.02 additive to coefficient of parasite drag, questionable propeller efficiency model):
RV-7 Performance Spreadsheet
Developer options in excel need to be turned on for it to work right.

Thanks to anyone who responds.

Empty

Empty, my 7 weight distribution is 500/501/67 in a level attitude.

The typical arms for payload and fuel for the RV7 are listed on Vansaircraft.com. Shouldn?t be much of an effort to calculate the loaded weight distribution.

All the best with your study!

Simon

From a non-engineer, not sure where the conflict lies. Perhaps it's an issue on grass, but on pavement, if you brake hard enough, the tail comes off the ground. At that point, all the weight is on the mains. I would have expected a bigger braking change in a nose wheel a/c, with the weight transferring to the non-braking nose wheel causing loss of braking ability.

Now, loss of braking ability due to wing lift can be a real thing (at least early in the landing roll).

Thanks for the replies. I had seen reports on the W&B thread that if the tailwheel was not elevated (to 0 pitch attitude) that the tailwheel weight reading on scales would be significantly higher than the 60-80 lb values I typically saw on there.

However, if tailwheel loading can be largely countered by pilot technique (nose forward stick and judicious braking to keep it unloaded), maybe I won't need to factor it in any more than I already had. At what point in the landing roll (assume calm wind) would you estimate that you lose elevator effectiveness? 20 knots?

I was reading on some piper super cub forums that their tailwheel loading was close to 10% of the aircrfaft's gross weight in the 3 point stance, which got me a little worried about just using 100 lb for a 1500 lbs RV-7.

rv7charlie said:

From a non-engineer, not sure where the conflict lies. Perhaps it's an issue on grass, but on pavement, if you brake hard enough, the tail comes off the ground. At that point, all the weight is on the mains. I would have expected a bigger braking change in a nose wheel a/c, with the weight transferring to the non-braking nose wheel causing loss of braking ability.

Now, loss of braking ability due to wing lift can be a real thing (at least early in the landing roll).

Click to expand...

The formulas that were given to us to use are:

Stopping force = Residual Thrust - Average Drag - Friction Force

Friction Force = .75 x (weight on main wheels) + .02 x (weight on tailwheel)

Residual Thrust = 10% of brake horsepower, converted to thrust using "Thrust = (325 * HP)/ Vs", times a 1.4 factor to compensate for increasing thrust with decreasing velocity

Average Drag derived from experimental data

Stopping force = (180 HP x .10)*1.4 - 102.3 lb - ((1400 * .75)+(100 * .02))


Stopping force = ((325*18)/Vs)*1.4 - 102.3 lb - 526 lb

Stopping force = (5850 / 48.5)*1.4 lb - 628.3 lb = -459.4 lb

F = ma

-459.4 lb = (1500 lb/ 32.2 fps^2) * a

-459.4 lb = 46.6 lbs^2ft^-1 * a

a = - -9.85837 fps^2

Stopping distance = Vs^2 / 2a

Stopping distance = (48.5 knots)^2 / (2 *9.85837 fps^2)

48.5 knots * 1.69 = 81.965 fps

Stopping Distance = (81.965 fps)^2 / 19.71674 fps^2

Stopping Distance = 6718.261225 ft^2s^-2 / 19.71674 fts^-2

Stopping Distance = 340.738947 ft


I think we are disregarding residual lift because we assume a 3 point landing touchdown at stall, even though I'm guessing there will still be some.

Narwhal said:

I think we are disregarding residual lift because we assume a 3 point landing touchdown at stall, even though I'm guessing there will still be some.

Click to expand...

You?re correct - lift doesn?t instantly drop to zero as you cross critical AOA, even in a three point landing in a taildragger, and depending on the airfoil it may be a rapid or relatively gentle drop. Either way you?ll initially have a potentially significant amount of lift (and thus, reduced braking), gradually falling to zero as the airspeed decreases.

Interesting topic, nice to read.

rv7charlie said:

... loss of braking ability due to wing lift can be a real thing (at least early in the landing roll).

Click to expand...

On my Lancair, I can't even apply the brakes at all until I'm below 45 KIAS, lest I flat spot the tires.