Gear Loading Factor Question

Why does an increase in gear load factor decrease shock stroke?
Using the equation to determine shock stroke I input a gear load factor. For the aircraft landing gear that I am designing, I am assuming a gear load factor of 5 which is what is generally used for Navy fighter aircraft. This is because I am designing landing gear for a STOL aircraft and I want it to be able to drop right onto the landing spot more like a carrier landing. The question I have is why does an increase in gear load factor decrease the shock stroke? Perhaps I don't understand the concept of gear load factor very well, but I would assume that a harder landing would demand a greater shock stroke.

I don't know what equation you're using, but for a given vertical speed a longer shock stroke absorbs the energy over a greater distance, so the load is less.

Do you have a copy of "Landing Gear Design for Light Aircraft" by Ladislao Pazmany? I've used this book when designing the landing gear on a couple of home-builts. Lot's of good information. My most popular design was a conventional gear for the Moni Motorglider. It has been accepted in several countries.

Richtdow: It depends on your gear design. With bungees, are you looking for total vertical displacement? With an air/hydraulic cylinder at higher landing loads, the piston/cylinder will act as though it is solid. In heavier airplanes, there is a metering pin in the oriface so that as the load gets higher (more vertical force on the gear) the pin shutsdown the oriface so the gear doesn't "bottom out" (and then drive up through the wing).

PS. You're definitely correct in that 5G is very high (yes, carrier style … 3G/10fps is Part 23). You might want to look at the certification regulations for drop testing landing gear to get a valid height number for dropping it in. A lot of your shock absorption will be in the tire itself, though.