How much does the jack plate weight? It weighs approximately 70lbs.

To correctly determine the jack plate size, measure from the transom (thick black line) to the red line in front of the trim rod. IMPORTANT: The measurement you provide will determine the L Bracket and hardware needed for your Jack Plate.

For a few examples, consider the general guidance below:

1. For Boats 20' and Longer: Use a 10” jack plate.
2. For Boats Just Under 20' Long: Use a 6” or 8” jack plate.
3. * For Boats Under 18' Long: Use a 4” jack plate.

And that's the pressure at p1. And we can remember that p2. Is also equal to p1 so that pressure under the first piston is the same as the pressure under the second piston.

A hydraulic jack plate provides the boater with the greatest flexibility to adjust the vertical position of the engine while running. The engine can be placed at the optimal height for everything from rough water (lower vertical position) to shallow water (high vertical position).

For a few examples, consider the general guidance below:

1. For Boats 20' and Longer: Use a 10” jack plate.
2. For Boats Just Under 20' Long: Use a 6” or 8” jack plate.
3. * For Boats Under 18' Long: Use a 4” jack plate.

Area of the cylinder where a 4 is going to be applied hydraulic. Pressure is going to be applied to extend the cylinder. And you'll be able to lift or move. So many pounds on the cylinder.

The capacity of a pump depends on the least of safe pressure that the pumping and lifting cyclinders and their seals and piston rings can take assuming that the lever and joints are strong enough. Area of jack piston is pi x r x r, where r is half the inner diameter of the jack's cyclinder.

Then when I get back on the trailer. I lower it back down. So that it really takes the pressure off the transom. Put it on a transom saver.

So p1 is going to be equal to F. 1 over a 1. We can then ask what is the force coming off the second piston f2. And we know that force is pressure times area.

Hydraulic Formulas

1. Force = psi x Area of Piston.
2. Effective Area = Diameter x . 785.
3. Pressure = Force / Area of Piston (answer in psi)
4. Displacement = Area of Piston x Stroke = (answer in Cubic Inches)

Hydraulic Jack, Capacity: 2-3 Ton