Robotic Mechanics and Modeling/Vehicles

Vehicles

Considerations for Boats (Cardboard Canoes)

Buoyant Forced Based on the Volume of a Block

Calculated deph of submersion of a block with a length of 4.5 ft., a width of 17 in. for two individuals of 60 lbs and 150 lbs, along with a safety factor of 20%.

When designing a box-like structure such as a cardboard canoe, we might start by getting a rough idea of how deep the canoe will sit in the water based on the displaced volume $V$ and lateral dimensions $L$ and $W$. As shown in the figure, this type of calculation is something that can be done with an elementary school-aged child using the equation for buoyant force $B_{f}$:

$B_{f}=\rho gV$

where $\rho$ is the density of the liquid, $g$ is acceleration due to gravity, and $V=HWL$.

Code

%reset -f
import numpy as np

m_child=27.2 # kg or 60 lbs
m_adult=68 # kg or 150 lbs

W=0.4318 # m Width of Boat or 17 in.
L=1.372 # m Length of Boat or 4.5 ft.

rho=1000 # kg/m^3 Density of water
g=9.8 # m/s^2 Gravity

H=(m_child+m_adult)*g/(rho*g*W*L)

print("Height of ",np.round(H,3)," m or ",np.round(H/0.0254,1)," in.")
print("Height with 20% safety factor = ", np.round(H/0.0254*1.2,1)," in.")

For the figure shown, the depth of submersion is 6.3 in. without including the safety factor of 20% and 7.6 in. While simple and determining whether or not your payload will sink or float, this calculation is important when determining the stability of a canoe.

Tilt Based on Two Masses Placed in the Floating Shell of a Block

Calculation of the depth and tilt of a buoyant beam with two applied masses.

When building a canoe for two people of significantly different masses, it may make sense to check to see how much tilt and submerged depth to expect. We can envision two forces $F_{1}$ and $F_{2}$ being applied at distances $a$ and $b$ from the left end of a beam of length $L$ in the X-direction, along with two superposed distributed buoyant forces acting upward in the Y-direction. One distributed buoyant force is uniform with a depth of $h_{1}$, while the other increases proportionally to the depth or tilt $(h_{2}/L)$. Thus, we can write equations of equilibrium for vertical forces in the Y-direction and moments about the left end of the beam in the Z-direction as follows:

$F_{y}=\rho gWLh_{1}+{\frac {1}{2}}\rho gWLh_{2}-F_{1}-F_{2}=0$

$M_{z}={\frac {1}{2}}\rho gWL^{2}h_{1}+{\frac {1}{3}}\rho gWL^{2}h_{2}-aF_{1}-bF_{2}=0$

Code

%reset -f
import numpy as np

rho=1000 # kg/m^3 Density of water
g=9.8 # m/s^2 Gravity

L=4.5 # ft
L=L*12*0.0254 # m
W=17 # in. Width of Boat or 20 in.
W=0.0254*W # m

m_child=27.2 # kg or 60 lbs (F_1 for this problem)
F1=m_child*g
a=1.5 # ft Distance of m_child from left
a=a*12*0.0254 # m

m_adult=68 # kg or 150 lbs
F2=m_adult*g;
b=3.5 # ft Distance of m_adult from left
b=b*12*0.0254 # m

h1=2*((L*F1+L*F2)-a*F1-b*F2)/(rho*g*W*L**2)
h2=2*((F1+F2)/(rho*g*W*L)-h1)

print("Depth of uniform distributed buoyant force h1: ",np.round(h1/0.0254,1)," in.")
print("Max depth of linearly distributed buoyant force h2: ",np.round(h2/0.0254,1)," in.")

For the length of 4.5 ft. and width of 17 in., the height $h1$ is 4.4 in. and the height $h2$ is 3.8 in. We would expect the left end submerged 4.4 in. and the right end submerged 8.2 in.

Stability and Metacenter (Portion Copied from Link)

Ship stability diagram showing center of gravity (G), center of buoyancy (B), and metacenter (M) with ship upright and heeled over to one side.
As long as the load of a ship remains stable, G is fixed. For small angles M can also be considered to be fixed, while B moves as the ship heels.

The metacentric height (GM) is a measurement of the initial static stability of a floating body. It is calculated as the distance between the centre of gravity of a ship and its metacenter. A larger metacentric height implies greater initial stability against overturning. The metacentric height also influences the natural period of rolling of a hull, with very large metacentric heights being associated with shorter periods of roll which are uncomfortable for passengers. Hence, a sufficiently, but not excessively, high metacentric height is considered ideal for passenger ships.

If the metacenter is above the center of gravity of the ship, then there will be a restoring torque to keep the ship upright. If the metacenter is below the center of gravity of the ship, then the torque coupled between the weight acting at the center of gravity and the center of buoyancy will cause the ship to overturn. Positive metacentric heights refer to stable configurations, while negative metacentric heights refer to an instability.

For a smooth shape, the condition for stability is that of having a positive metacentric height given by the following:

$I_{o}/V-{\overline {GB}}>0$

where $I_{o}$ is the area moment of inertia about the tilt axis, ${\overline {GB}}$ is the distance between the center of gravity and the center of buoyancy, and $V$ is the submerged volume. For our example with a width of 17 in., a length of 45 in., and submerged depth of 6.3 in., we would have issues with stability as calculated in the code.

Code

%reset -f
import numpy as np

W=17 # in. Width of Boat
W=0.0254*W # m
L=45 # in. Length of Boat
L=0.0254*L # m
h=6.3 # in.
h=h*0.0254 # m

Io=L*(W**3)/12
V=h*W*L
gamma=10 #in
gamma=gamma*0.0254 #m Distance above bottom of boat
b=0.5*h #m Distance above bottom of boat

MG=Io/V-(gamma-b)
print("MG (meters):", np.round(MG,2))

print("MG (inches):", np.round(MG/0.0254,2))

Practical Issues with a Cardboard Canoe

• Make sure you have enough plastic to wrap around the wetted regions.
• Add stiffeners to make sure long sections do not buckle. One method of stiffening the structure might involve attaching rolled cardboard to the sides of the canoe. Another involves partial slitting and folding of cardboard into triangular extrusions to attach to the sides of the canoe.