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Cutting Transom Angles

On September 12, 2014, in News From Gayle, Outboard Powered, Plywood Construction, by Gayle Brantuk
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Figuring out how to cut the angle on a transom can be a little challenging. Joshua Burks is currently building our Zip which is a 14′ runabout that has a 12 degree angle on the transom as do many of our designs. Joshua is documenting his build with video. He’s done a really good job at walking us through his build and explaining what he does. You can see all his videos on his You Tube Chanel.

The second video pertains to today’s article about cutting the transom angles. One of the considerations is how much material to add to the transom to allow for the angle. We have explained how this is done in a formula and photo below, but for those of you who are visual, Joshua’s video is excellent and really shows how much more simple it is than you might think. I hope you enjoy the video:

And for those of you who prefer the “formula” method–enjoy the article below!

 

The transom pattern perimeter is generally given to the outside back
surface, see sketch above. However many boats have transoms angled (A)
aft, in the case of outboards about 12° to 15°. Thus the angle
required on the bottom, and usually the sides, requires more material
(X) than the pattern outline. Cutting the angle away from or greater
than the given pattern contour solves the problem. Or, the extra
material required for the angle at any given point can be obtained.
The sketch shows the total thickness (T) of the transom plus the motor
board or frame. The sketch above can be duplicated using the angle
given on the plans and the total thickness of the transom. Distance
“X”, the amount of additional material required larger than
the pattern, can be measured. Optionally X can be obtained by simple
arithmetic.

AS A FORMULA: X = Angle factor (A, as listed in chart) multiplied by T
(thickness)

X = A x T

As an example solve X for a 1 1/2″ thick transom and motorboard
or frame with a 12° angle.

X = .213 x 1.5 = .3195″

To find the answer in sixteenths of an inch divide by .0625.

X in 1/16ths =.3195 ÷ .0625 = 5.04 or 5/16″

CHART
Angle in degrees “A”Angle Factor
10 .176
12 .213
13 .231
14 .249
15 .268
16 .287

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