# What will the angle of a boat with an outboard motor be?

• I
HRG
I want to calculate the angle of a wooden boat at rest in salt water, with a 55 lb outboard motor mounted on the transom + 3 gallons of gasoline (18 lbs) + 2 adults weighing about 150 lbs each. So just figure a total weight at the back end of the boat being 55 + 18 + 300 = 373 lbs.

The boat will be a jon boat that has a flat bottom. It will be 4 feet wide x 12 feet long. Assume that the part of the boat that is "in the water" is a rectangular box that is 4 feet wide x 8 feet long. The 4 feet to the bow plate is curved upward so will not be in the water. So for the calculation, just figure a 4' x 8' rectangular box with 373 lbs on one end.

For info, so you don't have to look it up, one cubic feet of sea water weighs approximately 64 lbs.

What will be the angle of the top of the boat relative to the surface of the water, when the boat is at rest in the water? (The top and bottom of the boat are parallel)

EDIT: I came up with an angle of 2.6 degrees.

Last edited:

How much does the boat weigh ?

Arjan82
depends on where these people and the 3 gallons of gasoline sit.

HRG
How much does the boat weigh ?
Since the boat will be perfectly level with nothing in it, please disregard the weight of the boat. I did add the weight of the gas and 2 adults in my original post. Just assume that all of that weight is concentrated about 1 foot from the back end of the boat. I just need the "approximate" angle the top of the boat will assume.

HRG
depends on where these people and the 3 gallons of gasoline sit.
Please just assume that the 373 lbs is 1 foot from the back of the boat. I just want to know the "approximate" angle the top of the boat will be relative to the surface of the water. I'll post my answer when I recalculate it and see if others come up with approximately the same angle. Thanks.

Assume stern sinks displacing a wedge of water 4' wide and 8' long.
55 lb outboard; 64 lb seawater per cu ft;
Additional displacement will be 55 / 64 = 0.859 cu ft = volume of wedge.
Displaces 0.859 = ½ * 4' * 8' * depth
Wedge (thick end) depth = 0.859 / 16 = 0.0537 ft over 8'
Atan( 0.0537 / 8 ) = 0.38°
Now I see that you have changed the problem.

HRG
Assume stern sinks displacing a wedge of water 4' wide and 8' long.
55 lb outboard; 64 lb seawater per cu ft;
Additional displacement will be 55 / 64 = 0.859 cu ft = volume of wedge.
Displaces 0.859 = ½ * 4' * 8' * depth
Wedge (thick end) depth = 0.859 / 16 = 0.0537 ft over 8'
Atan( 0.0537 / 8 ) = 0.38°
My calculation with just the 55lb motor was .996 inches for the right triangle's short side. Ended up with 0.6 degrees tilt. I'm recalculating with 373 lbs on the back of the boat. Thanks.

Arjan82
Since the boat will be perfectly level with nothing in it, please disregard the weight of the boat. I did add the weight of the gas and 2 adults in my original post. Just assume that all of that weight is concentrated about 1 foot from the back end of the boat. I just need the "approximate" angle the top of the boat will assume.
Have you ever stood in such a boat? It can change many degrees in angle when moving around the boat. Also, the weight of the boat is important if it is significant enough compared to this 373 lbs.

Assume stern sinks displacing a wedge of water 4' wide and 8' long.
373 lb load; 64 lb seawater per cu ft;
Additional displacement will be 373 / 64 = 5.828 cu ft = volume of wedge.
Displaces 5.828 = ½ * 4' * 8' * depth
Wedge (thick end) depth = 5.828 / 16 = 0.365 ft over 8'
Atan( 0.365 / 8 ) = 2.6°

HRG
Assume stern sinks displacing a wedge of water 4' wide and 8' long.
373 lb load; 64 lb seawater per cu ft;
Additional displacement will be 373 / 64 = 5.828 cu ft = volume of wedge.
Displaces 5.828 = ½ * 4' * 8' * depth
Wedge (thick end) depth = 5.828 / 16 = 0.365 ft over 8'
Atan( 0.365 / 8 ) = 2.6°
Bingo. That is exactly the angle that I calculated using 373 lbs. 2.6 degrees.
Thanks!

HRG
I want to calculate the angle of a wooden boat at rest in salt water, with a 55 lb outboard motor mounted on the transom + 3 gallons of gasoline (18 lbs) + 2 adults weighing about 150 lbs each. So just figure a total weight at the back end of the boat being 55 + 18 + 300 = 373 lbs.

The boat will be a jon boat that has a flat bottom. It will be 4 feet wide x 12 feet long. Assume that the part of the boat that is "in the water" is a rectangular box that is 4 feet wide x 8 feet long. The 4 feet to the bow plate is curved upward so will not be in the water. So for the calculation, just figure a 4' x 8' rectangular box with 373 lbs on one end.

For info, so you don't have to look it up, one cubic feet of sea water weighs approximately 64 lbs.

What will be the angle of the top of the boat relative to the surface of the water, when the boat is at rest in the water? (The top and bottom of the boat are parallel)

EDIT: I came up with an angle of 2.6 degrees.
Here's my calculations:

1 cu ft = 1728 cu inches, 1 cu ft of sea water weighs 64 lbs.

x cu in/373 lbs = 1728 cu in/64 lbs.
x = 10071 cu inches.

volume of a right angle wedge = 10071 cu inches
(area of a right triangle) width of boat = 10071
(short leg of right triangle x long leg / 2) 48" = 10071
(short leg x 96" / 2) 48" = 10071
(short leg x 4608) / 2 = 10071
short leg x 4608 = 20142
short leg = 4.37 inches

Using a right triangle calculator on the web:
short leg = 4.37 inches ... long leg = 96 inches.
The angle = 2.6 degrees.

*** not as simple and elegant as Baluncore's answer but the result is the same angle, 2.6 degrees ***