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Granted that knowns are easier to deal with than unknowns, but let's let's deal with some "knowns" here. Let's see...one knot is...Oh yeah ! One nautical mile....per hour. Ten minutes of coasting after the "speedometer" reads one knot equals about one sixth of one nautical mile or +/- 1000 feet.
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Not really, because the boat is decelerating during this time, the distance will be shorter. If the deceleration is a linear rate, distance would be easier to calculate, but the deceleration observed from higher speeds in the game shows deceleration is not linear. This is how it should be as the drag forces change in a non linear manner as the boat slows down. Besides, break this down even further and you are talking an object at one knot slowing to a stop with a forward velocity of only 1.7 ft per second.
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Originally Posted by pythos
Have you ever watched a train come to a stop? I have and this auto train going no more than 5 MPH took for ever and a day to stop.
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Yes, we have coal drags consisting of 100+ cars running through the town all the time. One of them with 110 loaded hoppers came to a stop from ~10 mph in less than a mile after it struck a car that was stalled at the crossing (their speed limit is actually higher through town, but there are 3 street crossings, so they go slower). The car was knocked off the track, so the car was not increasing the drag.
You are talking about a controlled brake, and this situation was an emergency brake.
Let me go into a little more detail as to why this analogy doesn't apply. Trains and cars ride on axles designed to reduce friction to a minimum. In addition, they are moving through a much less dense and compressible medium- air. The friction created by the axles is totally different than the drag and friction created by water flowing around an object. How long do you thing it would take a train to stop that was pushing against WATER?
One is a fairly simple and straightforward problem easy to compensate for. The other involves getting into fluid dynamics and is a completely different animal. First, you are pushing a hull through a liquid, not a gas. So instead of the object moving through a compressible medium, it moves through an incompressible one
(
http://www.fas.harvard.edu/~scidemos/NewtonianMechanics/IncompressibilityofWater/IncompressibilityofWater.html ).
Kinetic energy is spent through the change in the fluid as it's pushed aside and around the hull. This is considered a deformation of the medium and energy is transferred from the boat to the water in the process. This is only one component taken into consideration.
Drag over a hull has several parameters, go here to read more on this:
http://www.sailinganarchy.com/YD/2003/whatadrag.htm
So you have drag acting over the entire surface of the hull, plus all exposed parts that are sticking out from the hull and all hull penetrations.
In addition, once the propellers are stopped they have a huge drag coefficient. One book equates a stationay small prop as having the equivalent drag components of a flat hand placed into the water with the palm facing the direction of movement. They are designed to grab and push the water to create propulsion. So when stationary they are a huge component of the total drag. In otherwords, they are similar to applying a brake. How much depends on the configuration of the propellers (blades, pitch, hub, etc) and their location in relation to the hull. (ironically they create more local drag when rotating, but this is outweighed by the thrust created).
So once again, comnparing a train and a boat are not compatible analogies. So I'm not going to argue that point anymore- it just doesn't apply.