Heinkel 222 Reference in Ladislao Pazmany Landing Gear Book

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Anybody ever heard of the Heinkel He 222 - referenced in Pazmany's "Landing Gear Design for Light Aircraft", page 186.

I quote -

"The landing gear illustrated in Figure 10-121 employs a unique torsion spring arrangement. Four torsion bars are restrained from rotation at the top by means of knurled bosses pressure fitted to a plate. At the bottom end, the torsion bars pass through bronze bushings and extend below the cap which closes the leg. Four bellcranks are attached to the torsion bars. Each pair of bellcranks is interconnected by a short link. A tie rod at each side connects a bellcrank to the wheel fork."

"The gear was invented by Wilhelm Van Ness in West Germany and patented in Austria in 1952. The gear was used in a light airplane made by the Heinkel Company (He 222)."

LandingGear.jpg

I'm interested because this particular design seems to be more within the home builders' fabrication range than a pneumatic oleo strut. Any information on this aircraft/landing gear would be gratefully received - I can't find anything useful on the Web.

Regards
Tim
 
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hosscara

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Heineken also makes beer; you might've heard of it.
Heinkel, on the other hand, did make all sorts of interesting things including bubble cars & scooters.
They also made some really interesting aircraft.

Was the He222 not an experimental derivative of He219 Uhu, the night fighter?

I also searched in my book: German Aircraft Landing Gear by Günther Sengfelder; it doesn't get a mention...hmmm strange.

It would be easier to make shock-absorber than a proper oleo, but what about rubber donuts?
 

cluttonfred

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Note the 1952 patent date mentioned above, which suggests that this may be a post-WWII development. Heinkel did return to the aircraft business briefly in the 1950s before merging into VFW in the 1960s. I have put out some feelers, will let you know if I come up with anything.
 

fly2kads

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The VW Beetle has torsion bar suspension: the front has a stack of steel strips; the rear has a solid bar. Possible inspiration for this landing gear would have been all over the roads of post-war Germany.
 

Himat

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Anybody ever heard of the Heinkel He 222 - referenced in Pazmany's "Landing Gear Design for Light Aircraft", page 186.

... The gear was used in a light airplane made by the Heinkel Company (He 222)."

View attachment 50860

I'm interested because this particular design seems to be more within the home builders' fabrication range than a pneumatic oleo strut. Any information on this aircraft/landing gear would be gratefully received - I can't find anything useful on the Web.

Regards
Tim
Could it be that the airplane was sold and mostly known under a different name?
Now, what obscure German post war airplane are there?
 

cluttonfred

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I asked the question over at secretprojects.co.uk and the only response that might work is the Siebel SIAT 222, which was a single prototype developed into the Siebel SIAT 223 Flamingo that was produced in limited numbers. The 223 definitely used rubber-in-compression trailing link suspension all around and the earlier 222 looks like it used that for the main gear, but your torsion bar setup might have been used for nosewheel. The pic is just not clear enough to tell, though I have seen no reference to Heinkel or Van Ness in connection to the SIAT designs, so I don't think we're there yet.

SIAT 222.jpg SIAT_223_Flamingo_D-ECRO_Le_Bourget_06.67.jpg 1024px-MBB_223_Flamingo_PFM.jpg
 
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Thanks for everybody's input here. I'm always impressed with the depth of knowledge that shows up in this forum.

In the absence of any sources of information, I tried to assess the applicability of this landing gear design for my project (3600 lb taildragger).

From the web I found the following equations for torsion rods:

Energy U from applied moment T, shear modulus G, polar moment of inertia J, and rod length L

U = (T^2*L) / (2*G*J)

Theta = (T*L) / (G*J) (the shaft twist in radians)

For a steel rod .8 inches radius, 48 inches long we can calculate some of our numbers...

L = 48
R = 0.8
E (steel) = 2.9E+07 (or 29000000)
v (steel) = 0.3

So G = E / 2(1+v) = (1.12E+07) (or 11153846)

J = PI * R^4 / 2 = 0.6434

If we start with a guess for Theta of 15 degrees (0.2618 radians), we can calculate the moment T required to twist the rod 15 degrees.

T = (G*J)*Theta / L = 39141

U = (T^2*L) / (2*G*J) = 5124 [This is per rod]

So for 8 rods total, 15 degrees torsion will suck up a total U of 40988.

(If we don't provide damping somehow this energy will come out again as one very big bounce).

----------------------

Making some guesses about the trailing link geometry

If the link pivots 20 degrees and the arm is 11 inchs long we have a vertical stroke of 3.76 inches.

Assuming the tyre deforms to give us another 4 inches of vertical travel, we have a 7.76 total inches travel.

Assume a gross weight of 3600 lb (W),
Design sink speed of 8 fps (V)
g = 32.2 fps^2

E kinetic = 1/2 * W/g * V^2

Remembering to convert feet per second to inches to per second...

E kinetic = 1/2 * [3600 / (32.2 * 12)] * (8*12)^2 = 42931

E potential = (W/3)*(vertical travel) = (3600/3)*(3.76+4) = 9312

I came up with a rough guess for the amount of energy absorbed by the each tire by using the value Hiscock uses in his design example on page 182 of "Design of Light Aircraft".

E tire = 9000

Total E that needs to be absorbed by the torsion springs would be

E Required = E kinetic + E potential - E tire = 43243

So we're in the ballpark here.

(There's no damping whatsoever on in the torsion springs, so an external shock absorber will have to be added to the trailing link, and it will absorb some energy itself).

------------------

Other considerations:

Weight of 8 rods = 30 lb

Width of the gear leg - at least 3.5 inches, probably closer to 4 inches (it has to hold 2 rods side by side).

------------------

Final thought - max shear stresses in the rod.

S = T * R / J = 48667 psi

Assuming the ultimate shear strength of steel is half of the ultimate tensile strength, and using a safety factor of 1.5

Tensile Strength ult = 48,667 * 2 * 1.5 = 146000 psi

So we would require a (non brittle) steel rod with a tensile strength of approx 150ksi

So mild steel is definitely out, 4130 is a non starter too (97ksi tensile strength).

Anybody know of a reasonable grade of steel that might fit the bill?

Regards
Tim

PS - If my math is correct it will be a miracle... :)
 
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