displacement of water I just read the following puzzle on puzzling SE and it made me wonder, is there a proof which can be formulated to see if an object of any size and weight would cause more displacement inside the boat or outside.
When in the boat the weight of the object will push the boat down displacing water. 
When not in the boat the volume of the object will displace water.
We can make an infinitely small object with an enormous mass and place it in the boat (lets say the boat doesn't sink) the water would be displaced by an enormous amount. But the volume of it (when submerged) would hardly do anything to the water.
but now for the opposite, can you create an enormous object which does submerge but displaces more water when submerged than in the boat. obviously creating an enormous object is not the problem but it needs to be denser than water.
If something needs to be denser than water it needs to be heavier than it whilst having the same volume right? So my logic would say an object which submerges cannot displace more water than when inside the boat. Am I right? and how would one prove that?
 A: Let's name our variables so we get a better idea of what is happening.
Say the object you throw off the boat has a volume of $V_o$ and a mass of $m_o$. The mass of the boat is $m_b$. The water has a density of $\rho_w$ and the object has a density of $\rho_o = \frac{m_o}{V_o}$
When the boat is in the water and it is not sinking, the force pushing the boat down is equal to $$F_g=(m_o + m_b)g$$ because that is the gravitational force.
The force pushing the boat up is equal to the weight of the displaced water, which has a volume of $V_1$ in this case:
$$F_{up} = \rho_w\cdot V_1\cdot g$$
Since the boat is not moving up or down, you can see that $F_g=F_{up}$, meaning that $$V_1 = \frac{m_o + m_b}{\rho_w} = \frac{m_o}{\rho_w} + \frac{m_b}{\rho_w} = \frac{\rho_o}{\rho_w}\cdot V_o + \frac{m_b}{\rho_w}$$
When you throw the object in the water, two things can happen:
The object sinks
In this case, the object dispaces a total of $V_o$ of water (since it all sinks). On the other hand, the boat now has a downward force of $m_b\cdot g$ and an upward force of $$F_{up}=\rho_w\cdot V_{bw}\cdot g,$$ where $V_{bw}$ is the volume of the water displaced by the boat which can be calculated to be $$V_{bw} = \frac{m_b}{\rho_w}.$$
The total amount of water dispaced after throwing the object in the water is therefore $$V_2 = V_o + \frac{m_b}{\rho_w}$$ which, because $\rho_o>\rho_w$ (we know this because the object sinks) is MORE than the amount of water displaced when we were carrying our object in the boat.
Your idea was indeed correct: If I have an object with a fixed volume in the boat, then the boat displaces more water if the object is heavier. However, once i throw it overboard, it displaces the same amount of water no matter its weight (so long as it sinks).
The object floats
This time, the boat still displaces $V_{bw}$ of water, but the object, now floating, displaces less. The upforce on the object is $$F_{up} = V_{ow} \cdot \rho_w\cdot g,$$ where $V_{ow}$ is the volume of the water, displaced by the object. Since the downforce on the object is $m_o\cdot g$, you can see that $$V_{ow} = \frac{m_o}{\rho_w} = \frac{V_o\cdot \rho_o}{\rho_w} =\frac{\rho_o}{\rho_w}\cdot V_o.$$
The total amount of water displaced in this case, $V_3$, is then equal to $$V_3-V_{ow} + V_{bw}=\frac{\rho_o}{\rho_w}V_o + \frac{m_b}{\rho_w} = V_1$$
So now, the amount of displaced water does not change if you throw the object overboard.
