radius of a circle sought (relates to a boat I am designing.)

[peterAustralia]

This problem relates to a boat I am designing.

I want the boat hull to have a gunnel (deck- hull junction) to have an arc

of a circle shape when viewed from above.
Viewed side of the gunnel is flat (thus all in the horizontal plane)

Now at the widest point, the main hull is 1.00m wide, and is thus 0.50m from the centerline.
Now the widest point is 4.00m from the bow.

If I make the bow and widest point fixed, I need to know the radius of the curve of arc (the circle thingy) so I can define the offsets for all other sections along the hull.

I am pretty sure it is solvable,,, just that I am having a hard go at it.

Please find attached a diagram that might explain it better.

When I try to solve it I get into complicated equations
If you imagine a line between the bow and the widest point of the boat, and then go into the center of the arc, you end up with an isosceles triangle. The two large angles are found by tan (theta) = (0.5/4.0) thus theta = approx 85.426 degrees

Now all angles of triangle add up to 180 degrees
so the angle of curve that the arc scribes = 180 - (85.426 + 85.426) = 9.1478 degrees

anyway,,, thats kinda where I am up to
not sure what I have done is right or not

another question. If the topview to the boat has a constant circular arc when viewed from above, and the gunnel is level, and if each section of hull has the same elliptical curve when viewed end on (cross section), does that mean each part of the hull has exactly the same compound curve in all 3 dimensions. Thus could a mold/mould be built for a small area that has compound curve (curved in both axes) and that would fit all areas of the hull?

 

[markmain]

This problem relates to a boat I am designing.

I want the boat hull to have a gunnel (deck- hull junction) to have an arc

of a circle shape when viewed from above.
Viewed side of the gunnel is flat (thus all in the horizontal plane)

Now at the widest point, the main hull is 1.00m wide, and is thus 0.50m from the centerline.
Now the widest point is 4.00m from the bow.

If I make the bow and widest point fixed, I need to know the radius of the curve of arc (the circle thingy) so I can define the offsets for all other sections along the hull.

I am pretty sure it is solvable,,, just that I am having a hard go at it.

Please find attached a diagram that might explain it better.

When I try to solve it I get into complicated equations
If you imagine a line between the bow and the widest point of the boat, and then go into the center of the arc, you end up with an isosceles triangle. The two large angles are found by tan (theta) = (0.5/4.0) thus theta = approx 85.426 degrees

Now all angles of triangle add up to 180 degrees
so the angle of curve that the arc scribes = 180 - (85.426 + 85.426) = 9.1478 degrees

anyway,,, thats kinda where I am up to
not sure what I have done is right or not

another question. If the topview to the boat has a constant circular arc when viewed from above, and the gunnel is level, and if each section of hull has the same elliptical curve when viewed end on (cross section), does that mean each part of the hull has exactly the same compound curve in all 3 dimensions. Thus could a mold/mould be built for a small area that has compound curve (curved in both axes) and that would fit all areas of the hull?

Have you researched at boatdesign.net? They have a ton of boat related formulas there for you and they have many boat builders there who are experts with the math that you're needing; and helpful documents there as well. There may be some boatbuilders here, but that site is dedicated to your subject matter and may offer assistance as well.

 

[peterAustralia]

Have you researched at boatdesign.net? They have a ton of boat related formulas there for you and they have many boat builders there who are experts with the math that you're needing; and helpful documents there as well. There may be some boatbuilders here, but that site is dedicated to your subject matter and may offer assistance as well.

Found the answer eventually. Had to google and found a formula on the web. Answer radius = 25.222m

I had an isosceles triangle. I knew the distance of the base, and I knew the angle, and using the formula I was able to find the radius of the curve.

In respect to the formula used, the radius is the same as b, which is the length of two identical triangle side
the distance 'a' refers to the hypotenuse of a right and triangle of 4m x 0.5m
a becomes the base of the isosceles triangle

Aside, there are no specific formulas for this specific problem on boatdesign.net

Formula that i used to solve it can be found here

http://math.stackexchange.com/quest...osceles-triangle-how-to-find-length-of-the-tw