What causes lower unit failure?

OP wants to quickly drive the lower unit, he didn't write how long it should last.

 

I think his pitch to gear leg ratio math went the wrong way. Unless this is some really balsy motor being hooked up to the 2.5 ratio leg.

 

Yeah. The duty cycle will only be about 10 minutes per hour, and I really don't care if I have to change the oil every 2 hours or something.

I'm aiming for 4o kn, but I would be satisfied with 30. If that has to be done by spinning the input shaft at 10k rpm -- we will see

Actually I'm hoping the vessel is just really low drag. Definitely possible that there won't be enough torque -- I haven't finished the calcs yet.

I can't really find any hard figures on what the torque curve for such a small outboard looks like. Using some spitballing, I figure that the 15 hp (which is the same leg I have) puts out about 21 newton-meter (15.5 lb-ft) peak. My motor has the power output drop precipitously at 21 N-m, and does about 12 N-m at maximum power. That is assuming I go direct drive. If there is an overdrive, the torque values will be even lower.

So, I don't think I'll blow up the lower by exceeding torque limits, but possibly from overspeeding the bearings, or maybe vibrational/harmonic issues.

 

"Thinking" is a vague indication of reality.
Knowing is physics and data.

You're going to have to give up a lot more information before this thread can move from speculation to quantifying what you have and need.
Until then, good luck with your mystery project.
Post pictures and feedback.

 

9.9/15hp OMC they have different lower units.

 

This is an unnecessarily combative reply. I did a calculation right there in the post and then qualified it with an "I think" because I acknowledge the weakness of the calculation in the absence of more precise data.

 

Mathematically, horsepower equals torque multiplied by rpm. H = T x rpm/5252, where H is horsepower, T is pound-feet, rpm is how fast the engine is spinning, and 5252 is a constant. If you have the HP at what RPM, then torque can be easily calculated. That will let you know exactly.

 

Yes, indeed. But it is my understanding that peak power is usually rated at or near max max RPM (5500 or so) while peak torque is usually quite a bit lower (4000-4500). Since I can't find a dynamometer graph for any small outboards, I just did what you described to find torque at peak power and guessed that peak torque was ~20% higher than that based on the proportions of the dynamometer data that I could find for larger engines.