'Impossible' rocket drive works and could get to Moon in four hours

gravityblock · August 07, 2015, 04:31:11 AM

Quote from: MarkE on August 07, 2015, 03:29:18 AM
Integral calculus absolutely applies. Doctor heal thyself.

Integral calculus absolutely applies if you want to form an improper visualization. You're the one in need of a healing, not me.

Gravock

gravityblock · August 07, 2015, 05:14:35 AM

Quote from: MarkE on August 07, 2015, 03:29:18 AM
IOW: The position is the integral of the velocity, just as I described.There you are absolutely wrong. Let us use two examples:

Mass of the boat plus occupants: 100kg
Mass of each rock: 2kg
Ejection speed of the first rock with respect to the frame of reference: 10m/s
Case 1: Ejection speeed of the second rock with respect to the frame of reference: -10m/s
Case 2: Ejection speed of the second rock with respect to the boat: -10m/s
Initial condition: Boat is at rest.

Eject first rock to the right, velocity change to the boat is: 10m/s * 2kg + V_{change_boat} * (100kg + 2kg) = 0 V_{change_boat} = -10m/s*2kg/(102kg)
Boat position now integrates at -0.19608 m/s.
Some time later the second rock is ejected.

How does ejecting a 2kg rock from the boat increase the boat's mass from 100kg to 102kg?

Also, why are you using -10m/s before the 2 second rock is ejected to integrate the boat's position at -0.19608m/s? Is this how you do math MarkE? Come on, get it right........then we can see how it draws an improper visualization.

Gravock

MarkE · August 07, 2015, 05:54:10 AM

Quote from: gravityblock on August 07, 2015, 05:14:35 AM
How does ejecting a 2kg rock from the boat increase the boat's mass from 100kg to 102kg?

Also, why are you using -10m/s before the 2 second rock is ejected to integrate the boat's position at -0.19608m/s? Is this how you do math MarkE? Come on, get it right........then we can see how it draws an improper visualization.

Gravock

I have shown my work. Once again: The boat and its occupants are 100kg of mass. Each rock is 2kg. That's a total of 104kg. When the first rock is ejected its change in momentum is identically matched by an opposing change in momentum of the boat plus the second rock mass of: 102kg. In order to satisfy CoM, the change of 2kg moving to the right relative to the frame of reference at 10m/s must be identically matched by the remaining 102kg moving to the left at 2kg*10m/s / 102kg = ~0.19608m/s.

If the boat had started out with a much bigger pile of propellant rocks, say 900kg, then when the first rock is ejected, the boat would move to the left at: 2kg*10m/s / 998kg = ~ 0.020m/s, roughly one tenth the speed as in the first case. As the pile of propellant goes down, the change in velocity of the boat with each uniform toss gets larger and larger.

If you think I have made any mistakes, then you are free to show your own work.

gravityblock · August 07, 2015, 06:26:20 AM

Quote from: MarkE on August 07, 2015, 05:54:10 AM
If the boat had started out with a much bigger pile of propellant rocks, say 900kg, then when the first rock is ejected, the boat would move to the left at: 2kg*10m/s / 998kg = ~ 0.020m/s, roughly one tenth the speed as in the first case. As the pile of propellant goes down, the change in velocity of the boat with each uniform toss gets larger and larger.

If you think I have made any mistakes, then you are free to show your own work.

According to you, the boat is oscillating back and forth away from the starting point after both rocks are ejected, and the distance the boat travels between the oscillations grows larger and larger with each additional cycle of rocks being thrown. This is not in-line with the original hypothetical as shown by the video demonstration. Your words have already proven it gives the wrong visualization. Thank you!

Gravock

gravityblock · August 07, 2015, 07:13:32 AM

@all,

In the hypothetical, the process proceeded without the loss of the boat's mass after the ejection of the rocks. MarkE however wants to ignore this part of the process in the hypothetical so he can give a false perception that the use of integral calculus absolutely applies. However, by doing this, he'll form a wrong visualization. Velocity, mass, and time has no relevancy in the hypothetical, since the second rock is thrown after the boat travels a distance of 100m. The second rock thrown will then stop the boat. The only thing that has relevancy here is the distance the boat travels. So, there's absolutely no reason to use the position to integrate the velocity in regards to the original hypothetical when distance is the only thing relevant in the hypothetical.

In the hypothetical, the mass is not given and the velocity is unknown without the value of the mass. Since the velocity and mass is unknown, then so is the time it took the boat to travel 100m. The only thing given in the hypothetical is the distance the boat travels before the next rock is thrown in the opposite direction. This is because the distance is the only thing relevant in the hypothetical.

Gravock