There is no "pull" of gravity, only the PUSH of flowing ether!
in [Physics]
|
Prev: "The Einstein Hoax"
Next: ALL DIZEAZZEZ ARE DEZERVED ! ESPECIALLY THE CANCER GOODY, BACKBONE OF THE JUICY DIZEAZZEZ INDUSTRY
From: BURT on 1 Jun 2010 15:42 On Jun 1, 12:25 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > On May 25, 6:46 pm, PD <thedraperfam...(a)gmail.com> wrote: > > Dear PD: Your comment, below, about equal and opposite thrust force > and resistance force would be astute, if not for this fact: Objects at > rest, waiting to be dropped, are already experiencing a downward force > equal to their static weight. As soon as the retaining force is > released, there is a dynamic imbalance that starts the object > accelerating at 'g'. The downward force can never exceed the object's > inertia. If 'a' is substituted for 'g', the acceleration could > (perhaps) be greater, and the force would not be limited to the > inertia. The latter is true ONLY for objects dropped near the Earth. > My correct kinetic energy equation, KE = a/g (m) + v / 32.174 (m) > works anywhere in the Universe, as long as the "Pound" of reference is > that for 'g', or on Earth. NE > > > > > > > On May 23, 4:08 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > On May 20, 9:42 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > Dear PD: If you had ever taken a course in structural engineering, > > > there are TWO distinct types of force interactions. The STATIC ones > > > always have the opposing forces being equal. > > > Yes, indeed, so the net force is zero, so the acceleration is zero. > > Hence the word "static". > > > > But the DYNAMIC ones > > > only have a FORCE equal to the LESSER resistance of the two. > > > What??? > > > Let's take a simple example. There is a wooden block on a wooden > > inclined plane, and the plane is tipped up until the block starts to > > slide, accelerating. List the forces acting on the block, what their > > origin is, what direction the force is pointing. > > > > You've > > > never realized that DYNAMICS limits the downward force on the falling > > > object to be whatever the INERTIA of the object is. Since the inertia > > > of a one pound mass will always be just one pound, there can never be > > > an exponential KE increase, because the force and the resistance are > > > equal. > > > WHAT???? > > If the force and resistance were equal, then there would be two equal > > and opposite forces acting on the object, and you'd be back to a > > static situation. Try again. > > > See the block example. > > > > Since your precious WORK would have to increase exponentially, > > > that violates simple dynamics, because the distances traveled by > > > falling objects differ so markedly from second to second. I hope that > > > you are sitting down, PD, because the latter statement means that > > > there can neither be an increasing Work nor and increasing KE, from > > > just the accruing COASTING components, which you are so reluctant to > > > acknowledge. The COASTING components of the distance of fall for all > > > dropped objects KILLS your made-up science! NoEinstein > > > > > On May 20, 7:40 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > On May 18, 9:18 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > On May 17, 6:04 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > > On May 17, 2:59 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > Dear PD, the Parasite Dunce: You have changed the subject AWAY from > > > > > > > KE, > > > > > > > No, I didn't. I responded directly to your post about ether flow on > > > > > > muons. > > > > > > > > because you can't find any place in any text that states: "Work is > > > > > > > being done even if there is no resistance. (sic) The only requirement > > > > > > > to have work is that there be a displacement. (sic) Thus, if a hockey > > > > > > > puck slides twice as far across the ice, twice as much work was done, > > > > > > > and there is twice as much KE in the puck, even if the ice is > > > > > > > frictionless. (sic)." > > > > > > > You are not paying attention. > > > > > > Remember, PD: I am King of the Hill in science. It isn't "my job" to > > > > > pay attention to you. > > > > > :>) > > > > Just make sure you tell yourself that each day in the mirror. If you > > > > like, please add the line, "And I am the heir to the throne of the > > > > kingdom, by birthright." You may also consider adding, if you are > > > > feeling confident, "And I am irresistible to women." > > > > > > > There is no work if there is no force present, even in the absence of resistance. > > > > > > I correct you: 'There is no work if there is no force present', AND > > > > > there is no corresponding resistance > > > > > There is no work done if there is no net force, regardless whether > > > > there is resistance or not. > > > > > > > There is work if there is a force present, even in the absence of resistance. > > > > > > Dear PD: Newton's Laws of Motion require: "For every action there > > > > > must be an equal and opposite reaction." It is IMPOSSIBLE to apply a > > > > > force... UNLESS there is a corresponding resisting force! > > > > > Newton's 3rd law is not a statement about a resisting force. That is a > > > > common mistake by high school students that is corrected in virtually > > > > every textbook on the subject. For example, from the high school text > > > > that I've been quoting from, page 133: > > > > "One important thing to remember about action-reaction pairs is that > > > > each force acts on a different object. Consider the task of driving a > > > > nail into wood.... To accelerate the nail and drive it into the wood, > > > > the hammer exerts a force on the nail. According to Newton's third > > > > law, the nail exerts a force on the hammer that is equal to the > > > > magnitude of the force that the hammer exerts on the nail. > > > > "The concept of action-reaction pairs is a common source of confusion > > > > because some people assume incorrectly that equal and opposite forces > > > > balance one another and make any change in the state of motion > > > > impossible. If the force that the nail exerts is equal to the force > > > > the hammer exerts on the nail, why doesn't the nail remain at rest? > > > > "The motion of the nail is affected only by the forces acting on the > > > > nail. To determine whether the nail will accelerate, draw a free-body > > > > diagram to isolate the forces acting on the nail.... The force of the > > > > nail on the hammer is not included in the diagram because it does not > > > > act on the nail. According to the diagram, the nail will be driven > > > > into the wood because there is a net force acting on the nail. Thus, > > > > action-reaction pairs do not imply that the net force on either object > > > > is zero. The action-reaction forces are equal and opposite, but either > > > > object may still have a net force acting on it." > > > > > > That is > > > > > like applying the "force" of a skyscraper in a marsh. The maximum > > > > > static force that can ever be applied is determined by the supporting > > > > > capacity of the marsheffectively ZERO. My thesis in Architecture > > > > > was: "Float Foundations for Poor Soils". Essentially, I created > > > > > structural boats under buildings to support those by the bouyancy of > > > > > the marsh. For the record, I made an 'A' on that thesis. > > > > > NoEinstein > > > > > > > > The resistance on electrons imposed by the ether IS the force being > > > > > > > measured in those early Lorentz experiments. > > > > > > > Sorry, what "Lorentz experiments"? > > > > > > Lorentz experimented, extensively, with trying to measure the velocity > > > > > and the mass of electrons. Those were inside vacuum tubes, and were > > > > > speeded up by electromagnets. But, strangely, the electrons > > > > > encountered exponentially more resistance the closer the velocity came > > > > > to 'c'. > > > > > Reference, please. > > > > > > The equation beta = 1 / [1 - v^2/c^2]^1/2 was written years > > > > > before the M-M experiment. Lorentz shoehorned such to also explain > > > > > (sic) the nil results of M-M. Lorentz was a mathematition who DABBLED > > > > > in science. Expect bad results whenever that happens. > > > > > > > > So, the very experiments > > > > > > > you inquire about, only need the correct CAUSE, not a new set of > > > > > > > experiments! > > > > > > > A correct cause would be accompanied by calculations, indicating the > > > > > > size of the effect expected due to this cause. Without those > > > > > > calculations, you've got nothing. > > > > > > PD, the drag on electrons due to the ether that is clumping in front > > > > > is very close to Lorentz's Beta. > > > > > Prove that. > > > > > > The MATH is close to correct, but > > > > > the cause is ether drag. > > > > > Prove that. Show the derivation. Your bluff is called. > > > > > > Note: Ether can drag electrons and massive > > > > > objects, but it never drags photons! NoEinstein - Hide quoted text - > > > > > - Show quoted text -- Hide quoted text - > > > > > - Show quoted text -- Hide quoted text - > > > > > - Show quoted text -- Hide quoted text - > > > - Show quoted text -- Hide quoted text - > > - Show quoted text - What is the strength of gravity?
From: NoEinstein on 2 Jun 2010 19:46 On May 25, 6:46 pm, PD <thedraperfam...(a)gmail.com> wrote: > Dear PD: Most structural engineering problems involve the use of statics, rather than dynamics, to solve the equations. In statics, a one pound force, matched head-on by a one pound resisting force, wont allow the point of reaction to go accelerating away. On more than one occasion, when Ive said that the force must be equal to the resistancefor falling objects, that involves dynamics, not statics. Ill explain the difference. Suppose that there is a one pound object that is the payload of a rocket with a one pound net thrust; i.e., the rocket thrust is one pound MORE than the weight of the rocket minus the payload. Under those conditions, Newtons Second Law says that the rocket will be accelerating at 32.174 feet per second, each secondbetter known as the acceleration of gravity, g. As is the case for one pound objects on the surface of the Earth, the uniform force of gravity will cause any such object to press against the ground with a force of one pound. The ground, in turn, must offer an upward resistance of one pound to order to maintain static equilibrium (zero motion). Einstein himself noted that a person in a windowless box thats accelerating at g cant tell whether the box is resting, stationary, on the Earth, or flying through space. The box will be experiencing the weight of the person, and the person will experience the support (force) offered by the box. As required by Newtons Third Law, the action (the persons weight) is exactly matched by the reaction the supporting force provided by the box. Those two are equal and opposite, exactly as in the solving of problems in statics. The design equations for the box would be identical to the ones used on Earth for static loading. The INERTIA of falling objects isnt a constant force thats equal to the static weight, but can have any value *from zero up to the static weight. If a one pound block is resting on frictionless ice, that block can be caused to slide at, say, one ounce of force. In order to experience the maximum inertia of one pound, it will require applying a force, via some point of reaction, that is capable of moving 32.174 feet per second in the first second. So, the inertial resistance of any mass is VELOCITY dependent, within the limits, *above. Consider that payload being accelerated by a one pound net thrust rocket. If the payload weight is one pound, the acceleration will be g. If the payload weight were somehow reduced to .5 pounds, while the net rocket thrust remains one pound, the acceleration would jump to 2g. If the payload weight, somehow, could be increased to two pounds, the acceleration would be cut to .5g. It is very well known that all compact, near-Earth falling objects are accelerating at g'. Since the uniform force of gravity, acting on the one pound object, always causes such to accelerate at g, then, the force must be matched by the inertia of the objectas required by Newtons Third Law of Motion. PD, you have repeatedly claimed that there can be a downward one-pound force of gravity, WITHOUT there being a resistance. But heres why that cant be: Gravity requires an object to act upon. An object that has no resistance, also has no inertia, and will, therefore, have zero mass. You, erroneously, claim that a force can remain at one pound without there being a resistance. Heres why that is never the case: A big black house fly is circling your den, and aggravating you. So, you roll a newspaper into a one pound bat and swing away at the fly. After several misses, you finally hit the pest. Your hand and the newspaper were traveling exactly 32.174 feet per second. You figure you just smashed the fly with a one pound force on its tiny body. But much to your surprise, the fly hits the floor, readjusts its wings, and flies away, apparently unharmed. The physics of the above is simple: The impact of the fly and the newspaper will never cause a force thats greater than the inertia of the FLYquite small indeed! Since the flys body is always functioning under its full weight, accelerating it to 'g isnt a killing force. The fly could be stunned, allowing you to smash it flat, but you can never do that with a bat of ANY weight that isnt traveling faster than 32.174 feet per second. If the fly had ZERO resistance or zero inertia, the newspaper wouldnt experience any force at allexactly as if you had missed hitting it at all. Learn this, PD: The available force of gravity on a one pound dropped object, is 100% expended in causing the linear increase in the kinetic energy, as is given by my equation, KE = a/g (m) + v / 32.174 (m). Each second of fall, there is a COASTING carryover velocity from the end of each of the previous seconds. Coasting is what causes the free- drop curve to be parabolic in shape, rather than linear. Since 100% of the force of gravity produces only a linear increase in KE, then, there is no net force available to act through the remaining distance of fall, to cause the work that you imagine is being done. You err, PD, by assuming that a one pound force is available for both the coasting portion of the curve, and the straight line portion. As for a hockey puck already sliding on frictionless ice, there is no force needed to keep the puck moving, or coasting. If you still suppose that there is then you are ignoring the Law of the Conservation of Energy. Ive taken this long amount of time to walk-you-through the physics, not because I expect you to ever get it, but because I want 97.5% of the readers to know, that I know, that I know! NoEinstein > > On May 23, 4:08 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > On May 20, 9:42 am, PD <thedraperfam...(a)gmail.com> wrote: > > > Dear PD: If you had ever taken a course in structural engineering, > > there are TWO distinct types of force interactions. The STATIC ones > > always have the opposing forces being equal. > > Yes, indeed, so the net force is zero, so the acceleration is zero. > Hence the word "static". > > > But the DYNAMIC ones > > only have a FORCE equal to the LESSER resistance of the two. > > What??? > > Let's take a simple example. There is a wooden block on a wooden > inclined plane, and the plane is tipped up until the block starts to > slide, accelerating. List the forces acting on the block, what their > origin is, what direction the force is pointing. > > > You've > > never realized that DYNAMICS limits the downward force on the falling > > object to be whatever the INERTIA of the object is. Since the inertia > > of a one pound mass will always be just one pound, there can never be > > an exponential KE increase, because the force and the resistance are > > equal. > > WHAT???? > If the force and resistance were equal, then there would be two equal > and opposite forces acting on the object, and you'd be back to a > static situation. Try again. > > See the block example. > > > > > Since your precious WORK would have to increase exponentially, > > that violates simple dynamics, because the distances traveled by > > falling objects differ so markedly from second to second. I hope that > > you are sitting down, PD, because the latter statement means that > > there can neither be an increasing Work nor and increasing KE, from > > just the accruing COASTING components, which you are so reluctant to > > acknowledge. The COASTING components of the distance of fall for all > > dropped objects KILLS your made-up science! NoEinstein > > > > On May 20, 7:40 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > On May 18, 9:18 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > On May 17, 6:04 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > On May 17, 2:59 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > Dear PD, the Parasite Dunce: You have changed the subject AWAY from > > > > > > KE, > > > > > > No, I didn't. I responded directly to your post about ether flow on > > > > > muons. > > > > > > > because you can't find any place in any text that states: "Work is > > > > > > being done even if there is no resistance. (sic) The only requirement > > > > > > to have work is that there be a displacement. (sic) Thus, if a hockey > > > > > > puck slides twice as far across the ice, twice as much work was done, > > > > > > and there is twice as much KE in the puck, even if the ice is > > > > > > frictionless. (sic)." > > > > > > You are not paying attention. > > > > > Remember, PD: I am King of the Hill in science. It isn't "my job" to > > > > pay attention to you. > > > > :>) > > > Just make sure you tell yourself that each day in the mirror. If you > > > like, please add the line, "And I am the heir to the throne of the > > > kingdom, by birthright." You may also consider adding, if you are > > > feeling confident, "And I am irresistible to women." > > > > > > There is no work if there is no force present, even in the absence of resistance. > > > > > I correct you: 'There is no work if there is no force present', AND > > > > there is no corresponding resistance > > > > There is no work done if there is no net force, regardless whether > > > there is resistance or not. > > > > > > There is work if there is a force present, even in the absence of resistance. > > > > > Dear PD: Newton's Laws of Motion require: "For every action there > > > > must be an equal and opposite reaction." It is IMPOSSIBLE to apply a > > > > force... UNLESS there is a corresponding resisting force! > > > > Newton's 3rd law is not a statement about a resisting force. That is a > > > common mistake by high school students that is corrected in virtually > > > every textbook on the subject. For example, from the high school text > > > that I've been quoting from, page 133: > > > "One important thing to remember about action-reaction pairs is that > > > each force acts on a different object. Consider the task of driving a > > > nail into wood.... To accelerate the nail and drive it into the wood, > > > the hammer exerts a force on the nail. According to Newton's third > > > law, the nail exerts a force on the hammer that is equal to the > > > magnitude of the force that the hammer exerts on the nail. > > > "The concept of action-reaction pairs is a common source of confusion > > > because some people assume incorrectly that equal and opposite forces > > > balance one another and make any change in the state of motion > > > impossible. If the force that the nail exerts is equal to the force > > > the hammer exerts on the nail, why doesn't the nail remain at rest? > > > "The motion of the nail is affected only by the forces acting on the > > > nail. To determine whether the nail will accelerate, draw a free-body > > > diagram to isolate the forces acting on the nail.... The force of the > > > nail on the hammer is not included in the diagram because it does not > > > act on the nail. According to the diagram, the nail will be driven > > > into the wood because there is a net force acting on the nail. Thus, > > > action-reaction pairs do not imply that the net force on either object > > > is zero. The action-reaction forces are equal and opposite, but either > > > object may still have a net force acting on it." > > > > > That is > > > > like applying the "force" of a skyscraper in a marsh. The maximum > > > > static force that can ever be applied is determined by the supporting > > > > capacity of the marsheffectively ZERO. My thesis in Architecture > > > > was: "Float Foundations for Poor Soils". Essentially, I created > > > > structural boats under buildings to support those by the bouyancy of > > > > the marsh. For the record, I made an 'A' on that thesis. > > > > NoEinstein > > > > > > > The resistance on electrons imposed by the ether IS the force being > > > > > > measured in those early Lorentz experiments. > > > > > > Sorry, what "Lorentz experiments"? > > > > > Lorentz experimented, extensively, with trying to measure the velocity > > > > and the mass of electrons. Those were inside vacuum tubes, and were > > > > speeded up by electromagnets. But, strangely, the electrons > > > > encountered exponentially more resistance the closer the velocity came > > > > to 'c'. > > > > Reference, please. > > > > > The equation beta = 1 / [1 - v^2/c^2]^1/2 was written years > > > > before the M-M experiment. Lorentz shoehorned such to also explain > > > > (sic) the nil results of M-M. Lorentz was a mathematition who DABBLED > > > > in science. Expect bad results whenever that happens. > > > > > > > So, the very experiments > > > > > > you inquire about, only need the correct CAUSE, not a new set of > > > > > > experiments! > > > > > > A correct cause would be accompanied by calculations, indicating the > > > > > size of the effect expected due to this cause. Without those > > > > > calculations, you've got nothing. > > > > > PD, the drag on electrons due to the ether that is clumping in front > > > > is very close to Lorentz's Beta. > > > > Prove that. > > > > > The MATH is close to correct, but > > > > the cause is ether drag. > > > > Prove that. Show the derivation. Your bluff is called. > > > > > Note: Ether can drag electrons and massive > > > > objects, but it never drags photons! NoEinstein - Hide quoted text - > > > > - Show quoted text -- Hide quoted text - > > > > - Show quoted text -- Hide quoted text - > > > > - Show quoted text -- Hide quoted text - > > - Show quoted text -
From: NoEinstein on 2 Jun 2010 19:53
On Jun 1, 3:42 pm, BURT <macromi...(a)yahoo.com> wrote: > Dear Burt: Gravity from flowing ether can produce downward FORCES that match the mass of the objects. The 'strength' of natural gravity on Earth never exceeds one pound for a one pound mass, or always matches the static weight of the object in question. NoEinstein > > On Jun 1, 12:25 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > On May 25, 6:46 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > Dear PD: Your comment, below, about equal and opposite thrust force > > and resistance force would be astute, if not for this fact: Objects at > > rest, waiting to be dropped, are already experiencing a downward force > > equal to their static weight. As soon as the retaining force is > > released, there is a dynamic imbalance that starts the object > > accelerating at 'g'. The downward force can never exceed the object's > > inertia. If 'a' is substituted for 'g', the acceleration could > > (perhaps) be greater, and the force would not be limited to the > > inertia. The latter is true ONLY for objects dropped near the Earth. > > My correct kinetic energy equation, KE = a/g (m) + v / 32.174 (m) > > works anywhere in the Universe, as long as the "Pound" of reference is > > that for 'g', or on Earth. NE > > > > On May 23, 4:08 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > On May 20, 9:42 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > Dear PD: If you had ever taken a course in structural engineering, > > > > there are TWO distinct types of force interactions. The STATIC ones > > > > always have the opposing forces being equal. > > > > Yes, indeed, so the net force is zero, so the acceleration is zero. > > > Hence the word "static". > > > > > But the DYNAMIC ones > > > > only have a FORCE equal to the LESSER resistance of the two. > > > > What??? > > > > Let's take a simple example. There is a wooden block on a wooden > > > inclined plane, and the plane is tipped up until the block starts to > > > slide, accelerating. List the forces acting on the block, what their > > > origin is, what direction the force is pointing. > > > > > You've > > > > never realized that DYNAMICS limits the downward force on the falling > > > > object to be whatever the INERTIA of the object is. Since the inertia > > > > of a one pound mass will always be just one pound, there can never be > > > > an exponential KE increase, because the force and the resistance are > > > > equal. > > > > WHAT???? > > > If the force and resistance were equal, then there would be two equal > > > and opposite forces acting on the object, and you'd be back to a > > > static situation. Try again. > > > > See the block example. > > > > > Since your precious WORK would have to increase exponentially, > > > > that violates simple dynamics, because the distances traveled by > > > > falling objects differ so markedly from second to second. I hope that > > > > you are sitting down, PD, because the latter statement means that > > > > there can neither be an increasing Work nor and increasing KE, from > > > > just the accruing COASTING components, which you are so reluctant to > > > > acknowledge. The COASTING components of the distance of fall for all > > > > dropped objects KILLS your made-up science! NoEinstein > > > > > > On May 20, 7:40 am, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > On May 18, 9:18 am, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > On May 17, 6:04 pm, NoEinstein <noeinst...(a)bellsouth.net> wrote: > > > > > > > > > On May 17, 2:59 pm, PD <thedraperfam...(a)gmail.com> wrote: > > > > > > > > > Dear PD, the Parasite Dunce: You have changed the subject AWAY from > > > > > > > > KE, > > > > > > > > No, I didn't. I responded directly to your post about ether flow on > > > > > > > muons. > > > > > > > > > because you can't find any place in any text that states: "Work is > > > > > > > > being done even if there is no resistance. (sic) The only requirement > > > > > > > > to have work is that there be a displacement. (sic) Thus, if a hockey > > > > > > > > puck slides twice as far across the ice, twice as much work was done, > > > > > > > > and there is twice as much KE in the puck, even if the ice is > > > > > > > > frictionless. (sic)." > > > > > > > > You are not paying attention. > > > > > > > Remember, PD: I am King of the Hill in science. It isn't "my job" to > > > > > > pay attention to you. > > > > > > :>) > > > > > Just make sure you tell yourself that each day in the mirror. If you > > > > > like, please add the line, "And I am the heir to the throne of the > > > > > kingdom, by birthright." You may also consider adding, if you are > > > > > feeling confident, "And I am irresistible to women." > > > > > > > > There is no work if there is no force present, even in the absence of resistance. > > > > > > > I correct you: 'There is no work if there is no force present', AND > > > > > > there is no corresponding resistance > > > > > > There is no work done if there is no net force, regardless whether > > > > > there is resistance or not. > > > > > > > > There is work if there is a force present, even in the absence of resistance. > > > > > > > Dear PD: Newton's Laws of Motion require: "For every action there > > > > > > must be an equal and opposite reaction." It is IMPOSSIBLE to apply a > > > > > > force... UNLESS there is a corresponding resisting force! > > > > > > Newton's 3rd law is not a statement about a resisting force. That is a > > > > > common mistake by high school students that is corrected in virtually > > > > > every textbook on the subject. For example, from the high school text > > > > > that I've been quoting from, page 133: > > > > > "One important thing to remember about action-reaction pairs is that > > > > > each force acts on a different object. Consider the task of driving a > > > > > nail into wood.... To accelerate the nail and drive it into the wood, > > > > > the hammer exerts a force on the nail. According to Newton's third > > > > > law, the nail exerts a force on the hammer that is equal to the > > > > > magnitude of the force that the hammer exerts on the nail. > > > > > "The concept of action-reaction pairs is a common source of confusion > > > > > because some people assume incorrectly that equal and opposite forces > > > > > balance one another and make any change in the state of motion > > > > > impossible. If the force that the nail exerts is equal to the force > > > > > the hammer exerts on the nail, why doesn't the nail remain at rest? > > > > > "The motion of the nail is affected only by the forces acting on the > > > > > nail. To determine whether the nail will accelerate, draw a free-body > > > > > diagram to isolate the forces acting on the nail.... The force of the > > > > > nail on the hammer is not included in the diagram because it does not > > > > > act on the nail. According to the diagram, the nail will be driven > > > > > into the wood because there is a net force acting on the nail. Thus, > > > > > action-reaction pairs do not imply that the net force on either object > > > > > is zero. The action-reaction forces are equal and opposite, but either > > > > > object may still have a net force acting on it." > > > > > > > That is > > > > > > like applying the "force" of a skyscraper in a marsh. The maximum > > > > > > static force that can ever be applied is determined by the supporting > > > > > > capacity of the marsheffectively ZERO. My thesis in Architecture > > > > > > was: "Float Foundations for Poor Soils". Essentially, I created > > > > > > structural boats under buildings to support those by the bouyancy of > > > > > > the marsh. For the record, I made an 'A' on that thesis. > > > > > > NoEinstein > > > > > > > > > The resistance on electrons imposed by the ether IS the force being > > > > > > > > measured in those early Lorentz experiments. > > > > > > > > Sorry, what "Lorentz experiments"? > > > > > > > Lorentz experimented, extensively, with trying to measure the velocity > > > > > > and the mass of electrons. Those were inside vacuum tubes, and were > > > > > > speeded up by electromagnets. But, strangely, the electrons > > > > > > encountered exponentially more resistance the closer the velocity came > > > > > > to 'c'. > > > > > > Reference, please. > > > > > > > The equation beta = 1 / [1 - v^2/c^2]^1/2 was written years > > > > > > before the M-M experiment. Lorentz shoehorned such to also explain > > > > > > (sic) the nil results of M-M. Lorentz was a mathematition who DABBLED > > > > > > in science. Expect bad results whenever that happens. > > > > > > > > > So, the very experiments > > > > > > > > you inquire about, only need the correct CAUSE, not a new set of > > > > > > > > experiments! > > > > > > > > A correct cause would be accompanied by calculations, indicating the > > > > > > > size of the effect expected due to this cause. Without those > > > > > > > calculations, you've got nothing. > > > > > > > PD, the drag on electrons due to the ether that is clumping in front > > > > > > is very close to Lorentz's Beta. > > > > > > Prove that. > > > > > > > The MATH is close to correct, but > > > > > > the cause is ether drag. > > > > > > Prove that. Show the derivation. Your bluff is called. > > > > > > > Note: Ether can drag electrons and massive > > > > > > objects, but it never drags photons! NoEinstein - Hide quoted text - > > > > > > - Show quoted text -- Hide quoted text - > > > > > > - Show quoted text -- Hide quoted text - > > > > > > - Show quoted text -- Hide quoted text - > > > > - Show quoted text -- Hide quoted text - > > > - Show quoted text - > > What is the strength of gravity?- Hide quoted text - > > - Show quoted text - |