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NASA's suicide missions
21 years 8 months ago #5101
by Mac
Replied by Mac on topic Reply from Dan McCoin
LB,
Now that is interesting. I've done it, others have done it (not necessarily on a pendulum but effectively the same claim to net thrust), the government is doing it but you want to keep my nickle?
I think I am going to charge "Indian Giver" on this one.<img src=icon_smile_wink.gif border=0 align=middle>
Don't guess you want to expound a bit on your unit?
Now that is interesting. I've done it, others have done it (not necessarily on a pendulum but effectively the same claim to net thrust), the government is doing it but you want to keep my nickle?
I think I am going to charge "Indian Giver" on this one.<img src=icon_smile_wink.gif border=0 align=middle>
Don't guess you want to expound a bit on your unit?
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- Larry Burford
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21 years 8 months ago #4974
by Larry Burford
Replied by Larry Burford on topic Reply from Larry Burford
???
Am I being too subtile here?
Does anyone else see what Mac is overlooking?
Regards,
LB
Am I being too subtile here?
Does anyone else see what Mac is overlooking?
Regards,
LB
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21 years 8 months ago #4870
by Mac
Replied by Mac on topic Reply from Dan McCoin
LB,
Maybe I am missing your point or vice-versa. I agree that it does require net thrust to hold an object off perpendicular on a pendulum.
But you seem to express that that is the only test. I disagree with that. The key is "Net Force". If a device lifts itself in a vacuum chamber it has net force. TTB's units have supposedly done that and indeed they are in affect pendulums in that they were hanging on a lead wire before they were fired. Then they don't merely push off perpendicular they actually lift and fly carrying more than teir own weight.
I don't see why you disqualify that as an inertial drive.
Maybe I am missing your point or vice-versa. I agree that it does require net thrust to hold an object off perpendicular on a pendulum.
But you seem to express that that is the only test. I disagree with that. The key is "Net Force". If a device lifts itself in a vacuum chamber it has net force. TTB's units have supposedly done that and indeed they are in affect pendulums in that they were hanging on a lead wire before they were fired. Then they don't merely push off perpendicular they actually lift and fly carrying more than teir own weight.
I don't see why you disqualify that as an inertial drive.
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21 years 8 months ago #4874
by Larry Burford
Replied by Larry Burford on topic Reply from Larry Burford
A Lift Test is the equivalent of a Pendulum Test where the angle-of-the-dangle goes to 90 degrees or more.
Quite acceptable to me ...
... if ...
the <b>claim</b> can be substantiated. ANYONE can make a claim about ANYTHING. It means nothing.
Were you suitably impressed by the test results I provided to back up my claim?
You really don't know how badly I WANT there to reactionless thrusters in the Universe. And I always take a look when someone claims to have done it.
Earlier you asked me to comment about the Clark device. I designed a gizmo <b>exactly</b> like the one in his 1978 patent. And I did it in the late 70s. Its a very reasonable way to design that particular device. But just before I started to build a prototype I figured out what was wrong with the idea. And that is, when you transfer the mass from arm A to arm B you generate an impulse that exactly balances the impulse created by swinging the mass through half a circle on arm A. It will vibrate like hell, and probably move all over the place if you put it on a pendulum. But the sum of all that movement will be zero in a fairly short time (tens of seconds or less).
NO ONE has done it. I even have deep suspicions about the possibility that some super secret military outfit has this kind of tech. I've worked on some very black, very cool projects both as a member of the American armed forces and as a civilian. There were a lot of other geeks working right beside me. This particular kind of thing would be so hard to keep bottled up that an unreasonable number of people would literally have to be shot.
And that could not be kept quiet, even if the reason for it could. I've never blurted a word about anything I actually worked on. And I'm proud of that. But if I had ever worked on a thruster I really don't think I could have kept quiet.
Did I mention that this would be big news?
Regards,
LB
Quite acceptable to me ...
... if ...
the <b>claim</b> can be substantiated. ANYONE can make a claim about ANYTHING. It means nothing.
Were you suitably impressed by the test results I provided to back up my claim?
You really don't know how badly I WANT there to reactionless thrusters in the Universe. And I always take a look when someone claims to have done it.
Earlier you asked me to comment about the Clark device. I designed a gizmo <b>exactly</b> like the one in his 1978 patent. And I did it in the late 70s. Its a very reasonable way to design that particular device. But just before I started to build a prototype I figured out what was wrong with the idea. And that is, when you transfer the mass from arm A to arm B you generate an impulse that exactly balances the impulse created by swinging the mass through half a circle on arm A. It will vibrate like hell, and probably move all over the place if you put it on a pendulum. But the sum of all that movement will be zero in a fairly short time (tens of seconds or less).
NO ONE has done it. I even have deep suspicions about the possibility that some super secret military outfit has this kind of tech. I've worked on some very black, very cool projects both as a member of the American armed forces and as a civilian. There were a lot of other geeks working right beside me. This particular kind of thing would be so hard to keep bottled up that an unreasonable number of people would literally have to be shot.
And that could not be kept quiet, even if the reason for it could. I've never blurted a word about anything I actually worked on. And I'm proud of that. But if I had ever worked on a thruster I really don't think I could have kept quiet.
Did I mention that this would be big news?
Regards,
LB
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21 years 8 months ago #4978
by Mac
Replied by Mac on topic Reply from Dan McCoin
LB,
You have indeed designed, if not built. The problem that most lose sight of is that the acceleration/deceleration forces do tend to counter cenfrifugal imbalance.
Got to run but later tonight I'll provide you a bit more insight on how to get around that fact.
You have indeed designed, if not built. The problem that most lose sight of is that the acceleration/deceleration forces do tend to counter cenfrifugal imbalance.
Got to run but later tonight I'll provide you a bit more insight on how to get around that fact.
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21 years 8 months ago #4879
by Mac
Replied by Mac on topic Reply from Dan McCoin
LB,
Sharpen your pencil.
Start with a compound platform with a stationary frame and a rotating frame attached.
Now start with a moveable mass at the center. All is balanced.
As the rotating platform turns move the movable mass out radially at a linear rate 1% per 3.6 degrees rotation.
During that rotation the mass undergoes acceleration and that force produces a resistive torque on the rotating platform. To make things easy here use two counter rotating platforms so that torque to the stationary frame is cancelled.
During the rotation there is a linear buildup of centrifugal force which tries to pull the inertial frame in a circular motion but is partially restrained by the compound (dual rotating platforms being diametrically opposed timing wise.
Once they have completed 360 degrees rotation you will find that the force of acceleration has cancelled any net force on the frame and that peak centrifugal force is now 90 degrees from the mass motion vector.
STOP.
You just created a large impulse or momentum transfer to the frame in a given vector. That vector and the peak centrifugal force vector are 90 degrees apart. The integration of the centrifugal force opens the angle between the peak centrifugal force vector and momentum force vector but it is substantially less than 180 degrees and they are not equal force.
The deceleration force is dispersed in a linear vector. The acceleeration force is applied throughout 360 degrees and cancels itself out.
DO THE NUMBERS - YOU HAVE A NET FORCE. It is created by power driving the torque input to accelerate the mass and the mass meomentum transfers that energy into a linear vector.!!!!
Now you move the mass back to the center at a linear rate along a straight line without any acceleration or decelleration forces on the mass since it is no longer rotating.
When back in the center you repeat the cycle.
I'm only telling you this because we now have something much better, this is still a high stress, high mechanical shock design. We have overcome that and are able to produce a continuous output.
<img src=icon_smile_big.gif border=0 align=middle>
Sharpen your pencil.
Start with a compound platform with a stationary frame and a rotating frame attached.
Now start with a moveable mass at the center. All is balanced.
As the rotating platform turns move the movable mass out radially at a linear rate 1% per 3.6 degrees rotation.
During that rotation the mass undergoes acceleration and that force produces a resistive torque on the rotating platform. To make things easy here use two counter rotating platforms so that torque to the stationary frame is cancelled.
During the rotation there is a linear buildup of centrifugal force which tries to pull the inertial frame in a circular motion but is partially restrained by the compound (dual rotating platforms being diametrically opposed timing wise.
Once they have completed 360 degrees rotation you will find that the force of acceleration has cancelled any net force on the frame and that peak centrifugal force is now 90 degrees from the mass motion vector.
STOP.
You just created a large impulse or momentum transfer to the frame in a given vector. That vector and the peak centrifugal force vector are 90 degrees apart. The integration of the centrifugal force opens the angle between the peak centrifugal force vector and momentum force vector but it is substantially less than 180 degrees and they are not equal force.
The deceleration force is dispersed in a linear vector. The acceleeration force is applied throughout 360 degrees and cancels itself out.
DO THE NUMBERS - YOU HAVE A NET FORCE. It is created by power driving the torque input to accelerate the mass and the mass meomentum transfers that energy into a linear vector.!!!!
Now you move the mass back to the center at a linear rate along a straight line without any acceleration or decelleration forces on the mass since it is no longer rotating.
When back in the center you repeat the cycle.
I'm only telling you this because we now have something much better, this is still a high stress, high mechanical shock design. We have overcome that and are able to produce a continuous output.
<img src=icon_smile_big.gif border=0 align=middle>
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