Author Topic: 'Impossible' rocket drive works and could get to Moon in four hours (Read 95993 times)

TinselKoala · « **Reply #225 on:** August 13, 2015, 02:08:04 AM »

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It is all dependent upon that one singular moment in time when a mass reaches the peak of it's upward motion then stops and becomes weightless. For a brief moment in time it acts like it's in outer space with little or no gravity rather than here on Earth. Think about that... no matter it's weight for a brief moment in time it is weightless.

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For an instant, the velocity is zero. The acceleration is in fact at a maximum at that same instant.

No, that is not right either, neither one of you. Neglecting air friction, in a gravitational field like that of the Earth, if a mass is launched upward from the surface by, say, a spring, or out of a gun muzzle, then the mass is in _free fall_ and only under the acceleration of gravity, from the instant it departs the spring or gun muzzle until the instant it hits the ground again... where it is _still_ under the constant acceleration of gravity even though its velocity is once again zero. The gravitational acceleration is always constant and always downward. During the upward _coasting_ travel of the object, the downward gravitational acceleration acts to reduce the velocity of the object until it comes to an "instantaneous" stop at the top of its arc. The gravitational acceleration is still there, is still constant, and now acts to accelerate the object downward.

MarkE · « **Reply #226 on:** August 13, 2015, 02:29:44 AM »

Quote from: TinselKoala on August 13, 2015, 02:08:04 AM

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No, that is not right either, neither one of you. Neglecting air friction, in a gravitational field like that of the Earth, if a mass is launched upward from the surface by, say, a spring, or out of a gun muzzle, then the mass is in _free fall_ and only under the acceleration of gravity, from the instant it departs the spring or gun muzzle until the instant it hits the ground again... where it is _still_ under the constant acceleration of gravity even though its velocity is once again zero. The gravitational acceleration is always constant and always downward. During the upward _coasting_ travel of the object, the downward gravitational acceleration acts to reduce the velocity of the object until it comes to an "instantaneous" stop at the top of its arc. The gravitational acceleration is still there, is still constant, and now acts to accelerate the object downward.

I agree that is absolutely true for all projectiles. My understanding was that AC was talking about building a harmonic oscillator.

gravityblock · « **Reply #227 on:** August 13, 2015, 05:47:43 AM »

Both TK and MarkE have conveniently left out the acceleration equation that is inside F=mA. It's a common practice for MarkE and his minions to conveniently leave things out. If the initial and final speed are both the same, then there is no acceleration, regardless of the value for the time interval. The acceleration equation and the result of that equation speaks for itself, as we see below.

Gravock

gravityblock · « **Reply #228 on:** August 13, 2015, 06:08:05 AM »

If a force is being applied to an object (such as a gravitational force), and the object is resisting this force with the same intensity (such as being on the ground), then there is no net force and there is no acceleration, thus F=mA is false and Net F=mA is true.

Gravock

MarkE · « **Reply #229 on:** August 13, 2015, 06:20:41 AM »

Quote from: gravityblock on August 13, 2015, 05:47:43 AM

Both TK and MarkE have conveniently left out the acceleration equation that is inside F=mA. It's a common practice for MarkE and his minions to conveniently leave things out. If the initial and final speed are both the same, then there is no acceleration, regardless of the value for the time interval. The acceleration equation and the result of that equation speaks for itself, as we see below.

Gravock

Once again you are worse than wrong. Things can and do accelerate in one direction and later accelerate in the opposite direction with the effect that the starting and ending velocities are zero. Yet, the acceleration and velocity values during the movement are both far from zero. The simplest motion profile that a typical digital servo can execute is a triangular velocity ramp and double half parabolic relative position move: A_MAX from X_START to ( X_START + X_END )/2 and -A_MAX from ( X_START + X_END )/2 to X_END.

MarkE · « **Reply #230 on:** August 13, 2015, 06:51:06 AM »

Quote from: gravityblock on August 13, 2015, 06:08:05 AM

If a force is being applied to an object (such as a gravitational force), and the object is resisting this force with the same intensity (such as being on the ground), then there is no net force and there is no acceleration, thus F=mA is false and Net F=mA is true.

Gravock

A = (F₁ + F₂ ... F_N)/m is indistinguishable mathematically and physically from:

A = F₁/m + F₂/m ... F_n/m.

Consequently, since there is no distinction between F=mA taking all F into account, and "Net F=mA" again taking all F into account one statement cannot be true and the other false.

gravityblock · « **Reply #231 on:** August 13, 2015, 06:56:25 AM »

Quote from: MarkE on August 13, 2015, 06:20:41 AM

Once again you are worse than wrong. Things can and do accelerate in one direction and later accelerate in the opposite direction with the effect that the starting and ending velocities are zero.

You have now conveniently left out dV = v1 - v0.

dV = (5 minus -5) = 10 <-------the starting and ending velocities are not the same and does not equal 0, thus Net F=mA is True and F=mA is False

Gravock

MarkE · « **Reply #232 on:** August 13, 2015, 07:00:50 AM »

Quote from: gravityblock on August 13, 2015, 06:56:25 AM

You have now conveniently left out dV = v1 - v0.

dV = (5 minus - 5) = 10 <-------the starting and ending velocities are not the same and does not equal 0, thus Net F=mA is True and F=mA is False

Gravock

LOL.

gravityblock · « **Reply #233 on:** August 13, 2015, 07:13:53 AM »

Quote from: MarkE on August 13, 2015, 06:51:06 AM

A = (F₁ + F₂ ... F_N)/m is indistinguishable mathematically and physically from:

A = F₁/m + F₂/m ... F_n/m.

Consequently, since there is no distinction between F=mA taking all F into account, and "Net F=mA" again taking all F into account one statement cannot be true and the other false.

There is a distinction between a Net Force and No Net Force. To think otherwise is absurd. Net F=mA does not take all F into account as you have wrongly and falsely asserted. Net F=mA does not take into account a force that is being equally resisted with the same intensity of said force. Thus, one statement is true and the other is false.

Gravock

raburgeson · « **Reply #234 on:** August 13, 2015, 08:04:22 AM »

Well get it working and on the shelf, we have a large group ready to buy. I can imagine smuggling this in. That is the way that we would have to bring in a 300 MPG VW.

MarkE · « **Reply #235 on:** August 13, 2015, 08:18:57 AM »

Quote from: gravityblock on August 13, 2015, 07:13:53 AM

There is a distinction between a Net Force and No Net Force. To think otherwise is absurd. Net F=mA does not take all F into account as you have wrongly and falsely asserted. Net F=mA does not take into account a force that is being equally resisted with the same intensity of said force. Thus, one statement is true and the other is false.

Gravock

LOL, forces obey linear superposition. There is no mathematical and no physical difference between applying each force individually, or taking only the sum of all forces. If one neglects to count all the forces, then one can apply for an accounting job at Enron.

gravityblock · « **Reply #236 on:** August 13, 2015, 08:48:48 AM »

Quote from: MarkE on August 13, 2015, 08:18:57 AM

LOL, forces obey linear superposition. There is no mathematical and no physical difference between applying each force individually, or taking only the sum of all forces. If one neglects to count all the forces, then one can apply for an accounting job at Enron.

You're neglecting to account for an object that is resisting a force with the same intensity of said force, thus there is no net force and no acceleration of this object. Resistance is a force, and it is you is neglecting to count all the forces, not me. This is another psychological projection by you. You're definitely more than qualified for an accounting job an Enron.

There is a mathematical and physical difference between applying each force individually or taking only the sum of all forces. There is a difference in throwing rocks from a boat simultaneously in opposite directions (taking only the sum of all forces), as compared to throwing rocks at different times in opposite directions (applying each force individually). One results in a Net Force and a motion, and the other results in no net force and no motion.

Gravock

TinselKoala · « **Reply #237 on:** August 13, 2015, 10:08:11 AM »

Quote from: gravityblock on August 13, 2015, 05:47:43 AM

Both TK and MarkE have conveniently left out the acceleration equation that is inside F=mA. It's a common practice for MarkE and his minions to conveniently leave things out. If the initial and final speed are both the same, then there is no acceleration, regardless of the value for the time interval. The acceleration equation and the result of that equation speaks for itself, as we see below.

Gravock

I most certainly did NOT leave anything out. I showed, with checkable valid outside references, how the acceleration due to gravity IS the "A" in F=mA and produces the measurable and _calculatable_ weight that a stationary object exerts on a scale. The equation F=mA is a definition, just like Ohm's Law is a definition. Both define each of three quantities in terms of the other two. Mass is that which responds to acceleration by a force. Acceleration is what happens when you apply a force to a mass. Force is the result of accelerating a mass. The force that we call "WEIGHT" is the result of a mass being _stationary_ in an accelerating field: that of gravity.
You can do the math, and see that it gives the correct answer. Under your mistaken assertions, the math yields incorrect and/or inconsistent answers.

MarkE · « **Reply #238 on:** August 13, 2015, 10:08:34 AM »

Quote from: gravityblock on August 13, 2015, 08:48:48 AM

You're neglecting to account for an object that is resisting a force with the same intensity of said force, thus there is no net force and no acceleration of this object. Resistance is a force, and it is you is neglecting to count all the forces, not me. This is another psychological projection by you. You're definitely more than qualified for an accounting job an Enron.

There is a mathematical and physical difference between applying each force individually or taking only the sum of all forces. There is a difference in throwing rocks from a boat simultaneously in opposite directions (taking only the sum of all forces), as compared to throwing rocks at different times in opposite directions (applying each force individually). One results in a Net Force and a motion, and the other results in no net force and no motion.

Gravock

The equations I wrote tell all. If you are intent on continuing to resist reality, flail away to no avail.

gravityblock · « **Reply #239 on:** August 13, 2015, 10:24:17 AM »

MarkE and TK has once again violated the principal of cause and effect! They have violated the principle of sequence of an analysis of the phenomenon or the process being described. Newton's first law of dynamics states, “Every body continues in its state of rest, or of uniform motion in a straight line, unless it is compelled to change this state by forces impressed upon it”. In this statement, we see at once a violation of the principle of the cause and effect relationships.

Motion is a result of a net force, but it is missing in Newton’s first law; there is no mathematical model of this law, which describes its constant movement in space, but a body ignores it and moves with constant velocity V. The discrepancies being described are a cause of a violation of the principle of sequence of an analysis of the phenomenon or the process being described. This principle requires a description of the process or the phenomenon from its very beginning, not from the middle.

In order to return the principle of the cause and effect relationships into the former Newtonian dynamics, it is necessary to put the law of the accelerated motion of a body to the first place. As a result, we’ll get a new dynamics. In order to differentiate it from the old dynamics, Kanarev calls it “Mechanodynamics”.

Gravock

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