# Planing Hull Transom Breakaway Angle

Discussion in 'Hydrodynamics and Aerodynamics' started by tkdchris, Apr 19, 2017.

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### tkdchrisEngineer

I'm looking at the mounting height for an outboard that is set back from the transom on a box, the cavitation plate needs to be just below the surface of the water as it would be if the outboard was mounted on the transom. Is there a formula for working out the the height adjustment required.

The theory as I see it is that when the craft is planing the water breaks away from the transom cleanly, which is lower than the dynamic waterline. This lower water will then of course raise up the further aft from the boat you go.
This will be a function of speed, deadrise and distance aft of the transom. Potentially running draft could also play a part.

Any help would be greatly appreciated.

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### Mr EfficiencySenior Member

A rough rule of thumb would be at least an inch per foot of setback. But really you need to have the ability to go up or down a hole or two after on-water testing, to fine-tune matters. And make sure the boat can handle the centre of gravity going rearward.

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### cmckessonNaval Architect

This states that, right at the transom, the slope of the surface is:

dz/dx = root-g root-2 root-T20 / v

where:

root-g means the square root of the gravitational acceleration
root-2 means the square root of 2.
root T-20 means the square root of the draft of the transom
v means the speed of the boat.

The units must be consistent, e.g. m/s2, metres, metres/second

The formulation is only exact right at the transom. The slope diminishes (according to this model) as you move aft of the transom.

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### jarmo.hakkinenJunior Member

Savitsky, Morabito et al have been researching this, and their formulas are as attached. They used those to study stepped hull performance.

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### Mr EfficiencySenior Member

So the trim angle is of negligible importance ?

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### cmckessonNaval Architect

Mr. Efficiency: In the ballistic analogy the trim has negligible effect because even with trim there is very little initial vertical velocity imparted to the water. In effect the water particles "fall upward" from at rest, driven by gravity (i.e. hydrostatic pressure.)

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### tkdchrisEngineer

jarmo.hakkinen: thanks, that's exactly what I've been looking for. When you add in the extra distance the outboard will be lowered into the water due to the vessel trim angle by moving it aft it looks similar to Mr Efficiency's rule of thumb.

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