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

[MarkE]

@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.  So, there is no reason to use the position to integrate the velocity in regards to the original hypothetical.

Gravock

You have not shown any error in my analysis or math.  You cannot eject mass from something and still have the same mass as when you started.  Position is the integral of the velocity.  The velocity is in turn the integral of the acceleration.  The boat accelerates due to reaction force from accelerating each propellant rock as it is ejected.  Each rock that is ejected reduces the remaining mass on the boat.  As I have shown: the net change in velocity that results from each ejection gets bigger and bigger as the boat plus remaining propellant mass shrinks.  Where the mass of each rock is small with respect to the mass of the boat the magnitude of adjacent velocity steps is small but real.

[gravityblock]

You have not shown any error in my analysis or math.  You cannot eject mass from something and still have the same mass as when you started.  Position is the integral of the velocity.  The velocity is in turn the integral of the acceleration.  The boat accelerates due to reaction force from accelerating each propellant rock as it is ejected.  Each rock that is ejected reduces the remaining mass on the boat.  As I have shown: the net change in velocity that results from each ejection gets bigger and bigger as the boat plus remaining propellant mass shrinks.  Where the mass of each rock is small with respect to the mass of the boat the magnitude of adjacent velocity steps is small but real.

MarkE,

In a hypothetical, you can eject mass from something and still have the same mass as when you started, thus the reason it's called a hypothetical (are you really this disconnected?).  In the real world, a person in a second boat could be placing rocks in the boat as they're being ejected in order to maintain the same mass of the boat.  As I have shown in my previous post, the velocity, mass, and time are irrelevant in the hypothetical since the mass of the boat is unknown and remains unchanged during the ejection of the rocks.  The only value given that has relevancy in the hypothetical is the distance the boat travels before the second rock is thrown.

Gravock

[MarkE]

MarkE,

In a hypothetical, you can eject mass from something and still have the same mass as when you started, thus the reason it's called a hypothetical (are you really this disconnected?).

LOL, you eject the mass and keep it do you?

In the real world, a person in a second boat could be placing rocks in the boat as they're being ejected in order to maintain the same mass of the boat.

There are multiple problems with that proposal.  The first is: Some energy must be expended in order for them to track the movements of the first boat so that they are in place to transfer new rocks.  The second is that the act of transferring rocks from one boat to the other effects a momentum exchange.

As I have shown in my previous post, the velocity, mass, and time are irrelevant in the hypothetical since the mass of the boat is unknown and remains unchanged during the ejection of the rocks.

You have done no such thing.  The mass of the boat includes the propellant that remains at any given moment.  If the boat starts out with a single 2kg rock or a million 2kg rocks, the momentum of the laden boat changes equally and oppositely to the momentum change of the ejected rock.  The smaller the mass is of the ejected rock to the remaining mass of the boat plus all of its contents, the smaller the proportional velocity change of the boat versus the ejected rock.

The only thing given in the hypothetical is the distance the boat travels before the second rock is thrown.  How are you going to apply integral calculus when the only values given in the hypothetical is the distance?

You are so completely lost.

Gravock

[gravityblock]

Alright, moving on.....

Since we're now talking about mass and momentum, here's a device which demonstrates action without reaction made in the form of a beam balance.  The balls of steel having a different weight placed in jets exert no influence on the beam balance (first image below).  Here's a video demonstration of a similar beam balance device with no counter reaction (second image below). So, if a boat had 20 jets holding 20 rocks, then the mass of the boat would remain the same throughout the process of ejecting all 20 rocks from the boat.  There's also no issues with momentum in this example.

Gravock

[MarkE]

Miss Information is being joined by Miss Direction today.