I could use some input on a problem I have. We have a boat house in a very remote location where we keep a small ski boat weighing about 2,400 lbs. It sits on a small cart on rails we salvaged from an old mine shaft. The total run is about 150 feet. It is flat for most of the run but it rises about 2 or 3 feet over a 6 or 8 foot run to climb out of the shallow water. I can push the boat easily once on the concrete boat house floor.

My dad, who has passed, installed to handle a smaller boat 10 years ago and when he got a larger boat of around 2,000 pounds the torque broke one of the motors mounting 'feet'. So now when pressure is applied, the rubber love-joy gear between the motor shaft and spool slips.

Dad also did not have a brake installed with the setup, so once it starts down the steep part, you have no control at all. This is a lite use situation where we really only need to pull the boat into the boat house once or twice each summer.

I am a financial planner, and in exchange for helping me solve myengineering problem, I will be happy to answer a few financial question or steer you in the right direction.

So, do I need a new motor? Size? Type?

I can email a photo, but I can't seem to paste here.

Thanks,

Terry

With a 3 foot rise on an 8 foot slope, you are pulling 3/8 of the boat's weight with the winch, which would by about 900 pounds. To find the required motor torque you need to know the diameter of the winch spool. They are usually around 6 to 8 inches in diameter, but yours may be different. Since torque is the product of spool radius and tension, a larger spool will need larger torque. Let's say that (for a safe assumption) the spool is 1 foot diameter. This will need (900lb)(0.5ft) = 450 ft*lb of torque. If your radius is different from what I assumed, you can insert it (in units of feet) into the formula to get your torque.

Since you are talking about a remote location, I don't know whether you have electricity there or not. Is this an electric or a gasoline motor? Either way, it must produce at least 45 ft*lb of torque for this application. If the existing motor is strong enough, you may only need to replace the mounting foot. A new one can probably be welded on for you, but it may not be as strong as the original.

My recommendations:

1) cheap short term solution -- weld a new mounting foot on the motor and reduce the size of the spool. This will put substantially less torque on the motor and it is unlikely to break again. The motor will raise the boat more slowly than before, but at least it will work, and since you only need to pull it about 8 feet, it shouldn't take too long for the motor to lift the boat.

2) long term solution -- keep the existing spool and replace the motor with one rated to have a torque 50% larger than what your torque calculation says. If this is a gasoline motor, they often don't tell you the torque, but they will usually tell you the maximum horsepower and the speed in rpm at which the maximum occurs. You can find the torque by dividing the HP rating by the RPM rating, and then multiplying by 550*60/(2pi), roughly 6000. (Because 1HP = 550 ft*lb/sec, and 60RPM = 1 rotation per second, and 1 rotation = 2pi radians, in case you wanted to know.) The stronger motor will of course come with stronger mounting feet to withstand the greater torque.

In either case, I have a further recommendation about lowering the boat. The best kind of break for this job is a capstan. All you need is a wooden cylinder about 6 inches in diameter, fixed so it cannot turn, and long enough to get several wraps of line around it without overlapping them. You just wrap the line from the boat several times around the capstan and then stand there and hold your end of the line taut. Each wrap reduces the tension by a factor of exp(2pi*u) where u is the friction coefficient between the line and the wood. This works best with a flexible line like nylon rope, but can be done with steel cable too, you just need a larger capstan because it is not so flexible.

The coefficient of friction is usually about .5, which reduces the tension by a factor of 20 for each wrap. Even with a small coefficient (as in wet conditions) of 0.1, the tension will reduce by a factor of almost 2 for each wrap. Since you are a financial planner, you will understand that this works like compound interest, so the change is cumulative and drastic with just a few wraps. For a 900 lb tension, just 2 full wraps in dry conditions should reduce the tension to a comfortable 2 lb or so. You just play the line out hand over hand, and the boat neatly goes down the ramp at a controlled speed.

Hope this helps!

--Stuart Anderson