CHAPTER 27          INDEX TO OTHER PAGES

                 April 1998                  


  1. The first question we should come to - is to ask “what is inertia”, and “is it, or can it be construed as a force”? Most fundamentally inertia is motion, the displacement of matter, simply that.  Except to reiterate that the motion of or upon matter endows it with the quality known as rigidity, to keep it in place, or into its direction of movement, as to be self sustaining, to have character, and independence.

  2. As then force is motion, motion comes in two directions of, and inertia being motion, it comes in these by the terms of linear and angular.  All substance then being made up of atoms, and atoms being systems of parts in planetary motion, it accordingly is defined as having a momentum of angular force or inertia.  [Reference illustration figure 27-1).

  3. Whether or not therefore mass be at rest, or in rectilinear motion, it has inertia.  Inertia then if not drawn upon, may be said to have a potential of inertia.  This potential (the total of the atoms) can be utilized in full or in any part thereof.  And the same of course is true of linear inertia.


  1. Whenever we see anything that has power, or energy, anything at all that may be construed as force or action, or reaction, whatever can we possibly conclude as being the factual power - is so for that which is called “Motion?”

  2.  Let’s be realistic and not give ourselves to what the ancient may have told us.  We must reason for ourselves, and be wiser than a dozen Einstein’s.  What may be the power of a tornado, or the waves pounding on the shores, or a vehicle to crash over a barrier - if not for the motion thereof?  

  3. For it is motion that drives an airplane, and motion that sends a bullet out of its barrel by the expanding molecules seeking room.  And it is the motion of the sub-atomic particles that make each and every atom the likes of a gyro at full rpm.

  4. And let us not shortchange ourselves to conclude that saying; Yes motion, but there is substance, and without the substance there is no motion, wherefore substance also is energy.  For here in part, I beg to differ with you.  

  5. Yes for your sake I concur that without substance there is no motion, but - substance the material itself as such, is not energy, nor will it ever be, or come to that, regardless of what fantasies Einstein may have dulled man with.

  6.  Look again at motion and at matter.  Can matter be without motion?  Yes a stone can lay upon the ground and not be in motion, and as such it can not hurt, or cause a dent in your car.  

  7. But when motion is brought upon it, it can hurt and make a dent, which then is not because it is matter, but matter in motion, the motion being the force or energy of and/or upon that stone.  

  8. For motion in elementary terms does not subsist without matter. Wherefore “Motion” is the prime ingredients for all that may be called force or energy. (And where is man to know me for a difference?)

  9. We have this notion that gravity is as gravitational waves that extend everywhere and just draws things together, and even called it a force.  If then we can call it “a force” we should likewise have understood “motion” as the “prime ingredients” thereof.  

  10. For again when we look at a wheel that is without motion, it cannot be called a gyro, or of force, but with angular movement the momentum is “the” motion”, its motion.


  1. Next, let us ask what gravity appears unto us?  And to asses the same by illustration figure 27-2 - let us visualize the ring as were it a space station way out in space where for our assumption there is no gravity, and begin by assuming the ring at rest (not rotating).

  2. In that case both ladies, the one at point A and at point B will feel themselves weightless, and not moving.  If then the ring were caused to move into direction Z, both ladies would feel a movement into a direction opposite to the direction of the ring’s movement for as long as the acceleration of that linear movement lasts.  

  3. After which at the instant the acceleration stops for a constant velocity, both ladies would again come to feel that they are no longer in motion, and still as weightless as they had been before. From this we assess that movement cannot be felt by us except by a change of, which is to say, by and of acceleration and/or deceleration only.

  4. Next, let us from a rest (idle) position of the ring cause it to rotate as indicated.  The lady at point A would then begin to feel weight (direction Y1), as if a force of gravity pushed or pulled her down.  We said “as if gravity” - knowing there is no gravity, thus we conclude it must be a force of centrifugal that she feels.  

  5. But then we come to question what a force of centrifugal is, or if there is such a force?  The only force that we instigated was an angular movement of the ring, and how can that be construed as a force of centrifugal - which in turn is always noted as linear and - in fleeing away?  

  6. If it really were a centrifugal force - we should have instigated a force pushing down on top of her to cause her movement to flee from (as in from center). Thus we resort again to movement and ask ourselves; what in turn we instigated to the lady by rotating that ring?  

  1. The answer to which is twofold, #1; we caused her to move from her fixed position, which must be noted as an acceleration to go from a speed of zero to the speed at which the ring will continue to rotate at a constant angular velocity.  

  2. The lady thus will feel herself tilted to and/or driven to direction S for the duration of the acceleration, after which she would feel nothing more than her weight upon the ring.

  3. “Her weight upon the ring" then is an interesting question.  For why would we say “her weight” if we did not point to her mass, or more accurately, to the inertia of her mass?  If then we have a good understanding of inertia, we may call her weight as inertia, or a specific measure of her inertia.

  4. Thus we conclude that the force of the atoms of which the lady is constructed that tend to keep her in a fixed position, is her own gravity, her own weight.  Which she will come to feel upon herself the instant any action is produced to force her from her position by a movement in a change of direction.  

  5. This is correct since movement without change of direction that likewise will exhibit our force of inertia, is by a change of momentum (speed, velocity) [acceleration].

  6. This change in momentum then, in which a person other than its weight will feel a movement upon herself, is true for the lady at point B.  Assuming she was standing upon the ring, but not firmly secured thereto. Here the start of the rotation of the ring would have given her a nudge into the direction X causing her to float free from the ring.  

  7. At this point, once the lady has been set in motion, she will continue to maintain that constant movement, and feel nothing in the way of movement and/or weight or gravity.  All because her inertia being her weight, her affixed force in the nature of things, maintains whatever constant of inertia may be upon her.

  8. If then we tie her shoes to the ring so that she will be forced to rotate along with the ring, she will come to feel her weight in a direction that unto her appears opposite to gravity, in a direction away from the center of the ring.  Or as we are wont to say in a centrifugal direction.  

  9. But is her direction really centrifugal or not rather a force at right angle to her direction of so called gravity?  Centrifugal means a direction straight upwards from her, while in fact that is but a sight deception or mispronounced notion according to what she feels.  

  10. The actual impression towards movement is her resistance to the ever present directional change, which is into the direction noted X.

  11. The lady at point B, tied with a rope to point C, and swung around like a potato on a string, is called as being under the duress of a centrifugal force, just as the lady at point A is.  Yet in neither case is the force one of centrifugal, but of inertia, the force of one’s own weight in the mass of which he or she is constructed.

  12. If however instead of the ring we assume it as the sphere of the earth with point C the center of its gravitational force, then we need not tie off the shoes of the lady at point B.  The force then to present her with her weight would be a force of gravity, (which is a force different from any dynamic action, such as a string, or rotating space station) wherein she would come to feel her true weight.

  13. But how is this to say “her true weight”?  The statement is of course optional, since the true weight of anything is what one might call true.  The full exhibition of the potential inertia of the human body can easily come to crush it under the power thereof.  

  14. Our weight in the force of gravity is thus however much force gravity in itself will impose upon our inertia, to say - our in-based, or fundamental inertia, the sum of the atoms in their angular momentum.

  15. For if we added the inertia that may be placed upon us by a change in directional movement, to add “uniform angular movement", or acceleration in a straight path, these seem to be distinct.  The factors that point to this distinction will become apparent as we proceed further in our investigation of gravitational force and how to determine the same.

  16. Reiterating once that; the weight of either lady on the ring is directly proportional to the angular momentum, velocity, rpm, of the ring.  They might feel themselves to weigh 75-lb of the 150-lb that a force of gravity would place on them, or by an increase of the ring’s angular momentum they can come to weight any amount to the point whereby they collapse under the strain thereof.  

  17. Gravity may thus be called “the gentle lady", to draw on us according to the angular moment of our inertia, and not express itself so dearly upon our velocity of movement - as the factors relevant thereto seem to indicate.


  1. Gravitational force is accumulative.  A book laid on a table may be drawn by the force of 1-lb.  Placing a second book on top thereof will bring the force to 2-lb.  If instead we placed the second book next to the first each will show only 1-lb of force again.  Adding mass one on top of the other thus adds gravitational force, and/or affixing masses one to the other will register with their combined mass or weight in gravity.

  2. And so in speaking of gravitational force being accumulative, should we add the weight of the atmosphere that presses upon us, or ignore it when measuring the weight of the car for an actual force of gravity - excluding all else?  We know that a column of water (Figure 27-3) one-foot square and one mile high or deep will come to an accumulative weight of 329,789-lb (one cu/ft of water recorded at 62.46-lb).  

  3. While then the top cubic foot of water registers a g force of 62.46-lb, the bottom cubic foot of water must have a g force 5280 times as great, since there are 5280 cu/ft of water in a column one mile deep.  And the fact that the force is indeed that great is obvious from the tremendous pressure one mile down into the ocean.

  4. Assuming then the ocean at that point to be one mile in depth and a human person were to stand upon the ocean floor, would his weight upon the ocean floor also be in the thousands of pounds - like the water?  Aside of the fact that he would be crushed to death by the pressure, I do not believe his weight to be any more than at sea level - give or take a little.  

  5. Yet if the column of water were replaced by a column of steel and the same person would stand under it at the ocean floor, he would be crushed vertically and driven into the floor of the ocean.

  6. Thus we conclude that mass which can flow around us (liquids, gasses) do not accumulate to our individual mass, the pressure thereof being from all directions.  Since therefore the air of the atmosphere can get under the car as much as above it, we shall discount it when measuring the mass of the vehicle for a gravitational force upon it.


  1.  If we are to look for a constant of gravity it would have to be based upon the rate of acceleration which appears to be a constant to work for more than just the acceleration of gravity.  When therefore we wish to measure gravity for a force by way of a constant, then we must restrict ourselves to that measure which at all times is that force which moves or otherwise draws mass of any kind directly to the center of the gravitational object.

  2. The Cavendish’s experiment by which presumably the constant of gravity was found is however totally none existent and in error.  The attraction of the balls to one another has nothing whatsoever to do with gravity.  If there is any attraction between them worth mentioning the cause thereto may be electro-static, or in the movement of the precession of the earth or in the converging lines of magnetic force that pass upon them, or by combination of these.  

  3. But nothing that does not move directly in line with the direction of gravitational force by means thereof can be utilized for a measure of gravity.  As therefore these balls were allowed a movement perpendicular to the direction of gravity it instantly disqualified them.

  4. I can understand why Newton was wrong in his idea regarding gravity, he assumed correctly that bodies were attracted to one another.  But he also beheld the apple to fall down.  And so came the error as if bodies act by their whole mass rather than by a specific force that is upon them.   

  5. For while it was said; “that bodies act as if their whole mass, or even force, is located at their center”, that said center could very well be the central point from which the force radiates or comes and goes.  

  6. For we have seen magnetic force how by its formation it is the one and only force that draws to a central point.  Since then magnetic force was not unknown at that time, how did man come to omit, or ignore it?

  7. Not that I do not know the answer, but I must nonetheless emphasize that we are wrong, gravity does not cause the tides, nor does it raise the waters to a rise thereof.  For if anything, gravity constrains the tides, pulling them down rather than up.  

  8. Anything going upwards cannot be said to do so because something is pulling it down.  

  9. If then we take away gravity as a cause to the tides, we are left with nothing to explain it, or are we?  Are we again going to ignore that one and only force that has the power to move stars and planets, and by which tons of water may be drawn up in no more than a reshaping of itself?  We are idealistic, but not idealistic enough.


  1. When we say; “the acceleration of gravity is 32-ft/sec”, it has its reference to the potential of gravity to impose upon, and accelerate any mass that is exclusively subject to it (wherefore it is called “in free fall”) by a velocity of thirty two feet per second.  Meaning, that at each successive second the velocity of the mass will increase by 32 feet - to a terminal velocity that apparently has not yet been established, and at which I can only make an assumption.

  2. The acceleration of gravity thus is a limited acceleration, to merely advance by 32 ft a second, and not like other acceleration multiply like from 32 to 64 and 64 top 128 etc., or chain reaction as it might be called.  

  3. If assuming the angular precession of the inclination of gravity has an amplitude of 1-cm, the terminal velocity of the acceleration of gravity could come to 95,541-km/sec, which is based on the speed of light at which all e/m waves are known to travel.

  4. The acceleration of gravity is then not to be confused with the force of any other acceleration.  For when an object is accelerated at a velocity other than 32-ft/sec, it can not then be said to be of gravity.  A so-called centrifugal and/or centripetal inducement may be found or accomplished at any rate of change in velocity.  Another example is gunpowder, or rocket fuel to accelerate things beyond 32-ft/sec. 

  5. This sum then of the acceleration of gravity, in effect shows gravity for its force as well.  For it means that gravity in the full of its force is such to forge an acceleration of a specific measure, which by consequence is likewise - to the measure of the inertia of the mass.  

  6. This may be demonstrated by the balls in illustration figure 27-5.   Gravity pushing down on the 10-lb ball by a force of 10-lb and pushing down on the 2-lb ball by a force of 2-lb.  Or in saying, the inertia of the 10-lb ball resisting by a measure of its mass in the 10 -lb thereof, and the same for the 2-lb ball in 2-lb.

  7. And so we must come to a question, that while gravity displays a force of a certain fixed measure - if this measure may for its quality and/or its quantity rest in the inertia of the mass, specifically the in-based angular inertia.  Or that it is an exclusive agent of gravity.  

  8. If it were of inertia, then the measure of 32-ft/sec may be used in all formulas to establish quantities of force and/or resistance.  

  9. If however this measure is exclusive to gravity, it should not be used except for gravity.  If on the other hand it proves to be a combination of the both, then it can be used as we do now, for all formulas.  

  10. One way to find out is to compare the acceleration of gravity on the moon if it be the same as on the earth.  If it is the same then the measure may very well be a product of inertia, since inertia, the in-based angular inertia, is the same for any mass be it on the moon or on the earth.  While the gravitational draw of and on the moon is less than that of the earth.

  11. Enhancing the former, illustration figure 27-4, shows a vehicle making a turn within a 100-ft radius at a velocity of 15-mi/hr, to what force of resistance this would come to.  

  12. The vehicle then is not under the duress of gravitational force, but simply making a turn in a horizontal plane, which is at right angle to the force and direction of gravity.  Yet the formula of gravity (32-ft/sec) is used to find the sum.  

  13. Or by Figure 27-4 again, example 4a, to accelerate a mass from zero to 60-mph in ten seconds, which is nothing more than a straight forward movement acting against the inertia of the mass, in which also the formula of gravity is used - while gravity plays no role therein.

  14. It is obvious from example 4b in figure 27-4,  that if we wish to come up with an amount of force equal to a mass (weigh) to accelerate it to a certain velocity, any desired velocity, as per example 50-ft/sec.  We simply divide the weight by that figure, and multiply it by the same.  

  15. The question of course remains, if that amount of force can in fact accelerate the like amount of weight to that 50-ft/sec in the time specified.  If indeed it can, then gravity does not enter in to these calculations.  If on the other hand the 32-ft/sec is the correct formula, then it seems obvious that the acceleration of gravity is factually a factor of inertia, or engraved thereon.

  16. And to return once to figure 27-4, to the textbook example of the car making the turn, the textbook states quote:  “The “centripetal” acceleration on the vehicle in angular change at 15-mi/hr in a 100-ft radius" unquote.

  17. What is wrong here, is the vocabulary to say “acceleration”, rather than force, or impact, or resistance.  For the change being a “constant” change, is not therefore an acceleration.  Nor is it accurate to mention “force” with centripetal or centrifugal, for the better term of inertia.  

  18. Figures of speech then are one thing, but “acceleration for what is not at all acceleration is in error, a fundamental error in our way of thinking.


  1. Backtracking to figure 27-5, the two balls in each their specific inertia, and gravity in the specific or constant of its force pushing or drawing upon them, can not accelerate them faster nor slower than 32-ft/sec.  For this reason they will both come to the ground at the same instant, both accelerating at the same rate.  

  2. If on the other hand a force of 20-lb were exerted on the 10-lb ball, its acceleration would evidently come to twice 32-ft/sec to multiply by 64-ft/sec, but then the force would not be of gravity.

  3. In the case with the rocket in order to lift 5000-lb, a force of 5000-lb is needed, for which we do not need a computation after gravitational acceleration.  But applying 5000 lb to a 5000-lb the rocket would not move it, nor lift it.  An additional force is therefore required in order to overcome that 5000-lb "plus" gravitational force holding the 5000 lb rocket down whereby to accelerate it upwards.  

  4. In the latter then the computation in the acceleration of gravity can be used in the resistance that the rocket proves to be by its weight as a measure of gravity, and by gravity.

  5. In the case of the car that is on a horizontal plane, there is no gravitational force acting against the forward motion of it, but only its inertia, the potential of which for the example is 1500-lb.  We need not therefore (as in the case with the rocket) apply a force in excess of 1500-lb to get the car moving any quantity in excess of its friction will make it move. 

  6. Here is a teaser: the law states; action equals reaction."  Very well then take a 3000 lb weight somewhere in the USA, and by a force of 3000 lb under it, it should equal, and given a single lb more of upwards push it should move -- the action exceeding the force of the reaction since that is man's law.

  7. Wrong, it will not move, because that 3000 lb is its fundamental inertia, being drawn by an equal 3000 lb of g force. And it being let us say 3000 miles from the earth's axis, there is an additional amount of g force on it of, let us say 330 lb. The total g force therefore is 3330 lb, wherefore to move it upwards, the force must exceed 3330 lb.


  1. Now let us look into the mathematics by which force and gravitational attraction are measured.  By illustration figure 27-6  a vehicle with weight 3000-lb is at radius 3964 miles from the center of the earth, its velocity at the equator comes to 1521-ft/sec, how then by that data do we calculate the gravitational force on that vehicle?

  2. My first clue would be to throw the calculation out of the window, since if we know the weight, we know the g force on it (weight being a measure of the force of g).  So why bother?  Then you might say, Yes, but we might want to know what the vehicle in terms of g force is at a different velocity, or at some different radius.  So, all right let us humor ourselves.  

  3. And given the experts opinion on the matter, they concluded to take the mass times the velocity squared, and divided by the radius.  This all sounds fine, we understand mass, and that velocity squares the inertial force, and we understand the need for radius, but where in all of this does the computation of the acceleration of gravity enter into?

  4. If by example # 2 in figure 27-6 we take the mass of 3000-lb, and multiply it by the velocity twice over to square the force, and divide it by the radius.  We will end up with a very wrong answer; namely 331.6-lb/g for a mass that we know must come to a g force of at least 3000-lb.  

  5. We can’t very well say; O the thing has 331 pounds of gravitational force, that is its weight, which is equal to, and also a measure of g force.  For that sounds like we had too much to drink.  And if indeed we do divide the weight by the acceleration of gravity (example # 1) the outcome is even worse.

  6. So either our mathematics are wrong, or our scales are in grave error, or we have not been measuring gravity.  My question in the first place is; how can we apply the acceleration of gravity - which is firm at 32-ft/sec, to a mass traveling 47 times as fast?  (1521: 32=47.5).  For as I surmise; its uniform angular velocity in time and distance by the circumference in which it travels, should be accounted as a movement into the circumference (direction Y).  

  7. And that only part of that distance in the time given can be accounted for as a movement by the pull of gravity towards C, the gravitational center.  But before we dig deeper into that, let us debate more examples to discover what in fact we are measuring, and where gravity in its acceleration may be found useful and appropriate.

  8. Examples # 3 and # 4 of figure 27-6  show a decreasing quantity of force in pounds of weight as radius decreases from 3964 miles to 2000, and 100 miles.  These sums appear to me that we have been measuring centrifugal inertia rather than gravity.

  9. Entering more debate, we know that the car weighs 3000-lb, and that gravity in free fall accelerates by 32-ft/sec, and the car for all practical purposes is in free fall (no other force pushing down on it).  For if we opened the ground under it, we can safely predict that it will fall down by a velocity to square 32-ft every second.  

  10. But we don’t give a rats-tail about how fast this car may fall down to the center of the earth, but rather what the g force may be on this vehicle at a so called free-fall-rest condition (not under acceleration)

  11. If by example we do not know the weight of the vehicle, but wish to compute it from the formula of g force, given the mass, radius, and angular velocity.  And we start out by saying;  “the mass is 94 slugs,” then why bother computing gravitational force from all this hoopla of radius and velocity etc. when we already know the mass as 94 slugs, which is 3000-lb?

  12. Adding reality into it; Gravity should pull on an object to fall by 32-ft/sec/sec even when it is not in uniform angular motion traveling with the earth.  Or, should it?  If for a trial run we set aside the velocity of its uniform circular motion, and also the movement of its second uniform angular movement, that of our orbit around the sun (Figure 27-6 P).  Will then a mass be drawn to the earth at the same force and rate of acceleration?  

  13. If so, then gravity acts indeed with full measure upon the (so-called - in-based) angular inertia of the mass.  Yet we also know it to be a fact that any movement contributes towards inertia, and will then that inertia, which may be said to be upon its whole, have no relevance or add to the in-based angular inertia?

  14. To put it in other words, the atomic gyros always spin at a constant rate, in fact far too constant for us to begin to imagine.  When therefore these gyro's are taken in flight - is then this inertia one of the whole of the gyro to affix it in space, or into a trajectory, or not also additive to the first inertia?  For is not the inertia of a gyro the whole of the gyro?  

  15. The fact now that gravity will indeed draw mass at its affixed rate - even when there is no uniform angular motion to be recorded for it, is clearly found with the mass that resides at the two polar caps (the earth’s axis of rotation).

  16. In the last example (Figure 27-6, example #5),  the 3000-lb divided by the acceleration of gravity to slugs and again multiplied by the 32-ft/sec velocity, comes to 3008-lb.  Meaning that the acceleration is not quite 32-ft/sec, but slightly more, or the computation would be equal.  This formula then serves to find the value of the acceleration, but otherwise is of no practical value.

  17. Thus we remain with these important questions; if the formula named after the acceleration of gravity is indeed of it, or of inertia?  And if gravity draws exclusively by the in-based angular momentum of inertia, yea or ne’er?

  18. Conclusively I am implementing learning along with the revelations that I am promoting.


  1. It stands to reason that gravitational force in the weight of any given mass, is not the full inertia, nor the full of the force that gravity places upon it, but that it compensates (augments) for every ounce of centrifugal impact.  There is however some caution in this last sentence, that the augmentation is not invariable - at which we shall come back later.  

  2. Meanwhile, what are we to determine for the force of gravity to show a weight of 3000-lb on the vehicle - when the computation reflects only upon inertial resistance into the centrifugal direction?  Must we for the vehicle at the equator add the 3000-lb to the 331.6-lb to arrive at a total gravitational force of 3,331.6-lb?

  3. It may be obvious from centrifugal implementation that when an object in uniform circular motion imposes any strain of force from the centripetal direction, that in essence the object floats weightless, the object balanced between two forces.  

  4. Hence, it should be obvious that if there is 331.6-lb of inertial force attempting to drive the vehicle into the direction X (Figure 27-6) there must be an equal amount of force at right angle to the center of its radius.  

  5. And lastly, the fact that the vehicle with the sum of that strain still comes out with a force of 3000-lb pressing towards the center of the earth - suggests - that the gravitational force must be the total of both.

  6. If therefore this shall be correct, that the inertia by velocity is a figure apart and additive from what gravity implements upon the fundamental inertia, the in-based angular momentum, then how is that latter part drawn so congenial with the first?  

  7. This is a very good question for which we have to dig deeper into inertia and into the cause towards gravity.  And which by the way also serves to confirm that The Lord did not instruct me in error when so many years ago I wrote down the cause and nature towards gravity.


  1. Searching out more examples, by illustration figure 27-7, let us have a satellite moving around the earth in a stable orbit by a velocity of 18.000-m/hr.  This 18.000-m/hr at the fixed radius has given the satellite an inertial value that is acted upon equally by both the gravitational force and the centrifugal impact, wherefore the orbit is called stable.

  2. How much then may the inertial value of this satellite be?  If we divide the weight by the acceleration named after gravity (in the example F-1), it comes close to the actual weight.  If instead we used a lower rate of acceleration (gravity diminishing as radius increases) a value of 31.43-ft/sec, it would come to 95.45-slugs, (figure 27-7. F-2) which by the rest of the calculation comes to the 3000-lb of its actual weight (upon the earth).

  3. If then we brought the thing back to earth at sea level and used the same configuration to determine its weight in gravitational force, it would come out (as in F-2 -figure 27-7) at only 10.3-lb.  How is that then possible?  

  4. Is there a meaning here that when the satellite reaches its full weight in space it will have attained to a stable orbit?  And if we use the standard for the force of inertia (centrifugal inertia) ( figure 27-7, F-4), the 3000-lb satellite will have an inertial strain of 94,286-lb.

  5. Next, Let us then slow down that satellite in which case the orbit of the satellite will decay to fall into the earth, the gravitational pull becoming greater than the centrifugal inertia of the object.  Or in reverse with an increase in velocity the satellite will depart, the centrifugal inertia becoming greater than the gravitational pull.  How are we to explain this knowing that - “reaction must equal action”?  

  6. For if by the same token we drive a vehicle to a high speed its inertia increases, but the vehicle will not become defiant of gravity. The reason is, since for every ounce of inertia that the vehicle multiplies to itself - gravity on the opposing end presents a force equally as much to keep it down to earth.

  7. Thus we might say with the satellite up 200-miles above the earth and traveling at 18,000-mi/hr, we found the limit of gravity, the top end of its power to overcome it.  As good as this may sound it nonetheless is also embarrassing - in the fact that a stone cast in the air, not traveling anywhere near 18,000-mi/hr, will also come to defy gravity.  And a gyro with its added angular momentum is another example.


  1. One might say (as the textbooks do), “Gravitational and centrifugal forces are balanced wherefore the orbit is stable.  Yes, indeed, a lot of good that does.  This is like telling me that the automobile rolls with its wheel upon the road.  It does not tell me the how and the why.  So let us reason: 

  2. The satellite puts on its afterburners, which is the action to increase velocity.  The reaction is the inertia of the satellite.  The added inertia then (as an action) has gravity for its reaction.  And so it seems there is not sufficient force within the reaction (gravity) to equal the added inertia.  

  3. Yet before this point the power in the reaction (gravity) to the inertia always exceeded the inertia.  Thus it seems there is a variable factor here that is to be inserted and accounted for.

  4. Shall we then blame it on a diminishing value of gravity?  No doubt gravity diminishes as we gain altitude, and we are told that by 200 miles up it is but a 10%.  When however we divide the 200 miles into the 4000-mile radius of the earth, it comes to only 5%.  

  5. And if we look at the mathematics, examples F1, and F2, of figure 27-7, the value for the acceleration comes to 31.43-ft/sec, which is a difference of only 2%.  (32.14 to 31.45 = 2%). 

  6.  It may seem strange that while we need an 18,000-mi/hr velocity to keep a satellite in suspense of gravity, it is all so simple here on earth to throw a ball or a stone into a parabolic curve, causing it - if only temporarily, to defy gravity.  Or to spin a gyro and it will not move from its position as if held by gravity.

  7. Let us then attempt to explain the ball in this way, using illustration figure 27-8.  What is foremost to be understood of the ball is that while it lies motionless on the ground, it is under the duress of acceleration with a fixed value of 32-ft/sec.  

  8. If then a force in excess of the inertia (weight) of the ball is used to cast it away by which it will follow a parabolic curve (points A through D), and we divide it in three sections.  

  9. In the first section (A/B) the ball accelerates, for the example 3-ft/sec added to the existing 32 to 35-ft/sec.  The velocity also, which of course is converted into inertia went from 1000 to 1010-ft/sec In the next section the ball loses acceleration for a constant velocity still moving at 1010-ft/sec.  But with the acceleration gone it is now subject to 32-ft/sec again (32 + 0 = 32-ft/sec).  

  10. Nor does this change in the last section as far as the rate of acceleration is concerned - since it has fallen back in the domain, the grasp of gravity.  If then the velocity changes back into acceleration on its way down - this does not exceed the existing 32-ft/sec of and by g, but rather that acceleration is of g.  

  11. The ball (at point C having lost its momentum) simply proceeds according to a free falling body by the 32-ft/sec/sec acceleration of gravity. The only time that the ball actually overcame the acceleration of gravity was in the first section.  So that we might coin to say “that it is by an acceleration over the acceleration of gravity whereby an object will come to move in defiance of the gravitational pull.  

  12. This word “acceleration” is important here since the instant it leaves off for a constant velocity, or downgrading velocity, it will no longer be in defiance of gravity, having lost the rate of its inertial moment that placed it in excess of the gravitational rate.

  13. This velocity of constant (which for the ball comes but for an instant) even if it should continue for a long distance, would remain subject to the acceleration of gravity, and only wings with a force to maintain its constant would keep it above the ground. 

  14.   For a constant velocity presents no acceleration to come to exceed the ever present 32-ft/sec of gravity, as even the acceleration in the last section is not one of in excess of gravity, but of gravity drawing it down.

  15. We might liken the 32-ft/sec acceleration of gravity to a curve along which the ball travels.  The line of the curve then is the 32-ft/sec, and gravity can neither depart to the right or the left of that line, and the ball travels with that acceleration.  If then for a short duration the ball is given a boost to exceed that acceleration - it will come free of the line, free of gravity so to speak, noted direction T.    

  16. The ball in fact must come free since the curve cannot exceed 32-ft/sec.  Any force to the contrary then will only dig the ball more firmly into the 32-ft/sec curve.  

  17. When we accelerate anything within a gravitational field, the acceleration poses a relevance to the formation and format that is already existing by the draw of gravity.  It as such may be said; is engraved upon the fundamental angular inertia of the mass.

  18. This may explain why it is so simple to have a stone follow a parabolic curve, and why on the other hand it takes a velocity of no less than 18,000-mi/hr to keep a satellite at “constant” velocity in defiance of gravity.  

  19. All this is not much different from sending a rocket into space, where in order to lift a ten-ton vehicle in defiance of gravity, a force in excess of ten tons is required.  And with a ½-lb ball a force in excess of the ½-lb is required to make for a home run.

  20. But how may a gyro defy gravity when it can not take on an acceleration in excess of what is ever present upon it?  If the gyro could, it would fly off of its base and be like the ball.  The movement of the gyro is however more fundamental than that of the ball, it depicts the very atom in its planetary system, the very root of all substance, and the first movement, and momentum of all that follows.  

  21. And it is also the first and primary movement upon which gravity attaches itself to bring it down by way of a unique overall precession, instigated by a torque upon its axis.  As then the atom fixes itself in space and becomes inertia by its angular momentum, so does the gyro in secondary format.

  22. The movement of the gyro thus is nothing more than greater inertia to affix itself in place, with all its weight in “apparent” defiance of gravity.  I said “apparent” since for its angular momentum the gyro makes itself even more subject to gravity.   

  23. This in some way is similar to our satellite affixed in orbit, where instead of angular momentum, a great deal of linear momentum comes to test gravity for its very power and momentum.  The gyro then will affix itself either vertical or horizontal, since it is not in the relative position to the direction of gravity, but in that it affixes itself by its own inertia.


  1. In conclusion it appears plausible that in the computation for centrifugal inertia the formula for the acceleration of gravity (32-ft/sec) does not enter, yet the radius is used to the axis of rotation.

  2. In the calculation of gravity there is no radius to be used.  This is so because neither gravity, nor inertia is a force of center.  There is no axis, nor a central point in their format.  There is however the value of force in the measure of the acceleration of gravity, which is valid for gravity as well as for inertia.  (At least for the inertia residing within a gravitational field).

  3. That which is called “the center of gravity”, in reality is the central intersecting point of the force of magnetic.  And it is by the force of magnetic that all substance - in and by the inclination of gravity - is drawn downwards.  Therefore, logically, it is also that central intersecting point to which gravity (as such) is seen to be moving.

  4. By figure 27-9a, a principle illustration is given in the way substance is drawn downwards by the combination of magnetic force to a torque upon the angular inertia of which no substance is exempt, that then turns into an unique precession - which is the very essence of a nut being rotated (turned) on a long threaded rod.  

  5. The rod is then the magnetic force in its countless lines of force, the same essence as any wavelength be it of light, radio waves, or whatever e/m wave one wishes to name.  The magnetic wavelengths being the longest of them all, so long in fact that they appear more likes lines of, yet are waves, a single wavelength inhibiting the body it serves.

  6. Inertia then may be categorized in two: The angular inertia of the mass by the movement of its sub atomic parts.  And second, what is classified as “linear inertia", when an object is in uniform circular motion, the inertia is that of its velocity, which is “linear inertia”.  

  7. This as such is distinct from the first inertia, even though in fundamental terms that inertia might also be called linear (the atomic parts in uniform circular motion).  It is upon the first inertia that gravity takes hold, but since the second inertia plays for its measure upon the first as well - so gravity comes subject to both. 

  8. It then is in the unique difference in these two inertia’s that we are able to throw a ball up into the air, and have our satellites pursue an stable orbit.  The first having the impression of gravity’s 32-ft/sec, initiated and augmented to by the second.  

  9. We may then illustrate how the acceleration of 32-ft/sec said to be of gravity is engraved in the inertia.  Figure 27-9b, the driving power (motor) is the magnetic force (3M) which by angular, and uniform angular movement comes to the transmission, which as such is said to be gravity, or the inclination to gravity.

  10. The outcome for which (on earth) is 32-ft/sec that is directly attached to all substance within its reach, specifically the inertia thereof.  Wherefore the transmission (gravity) putting out 32-ft/sec - so inertia is driven, and will react by the same 32-ft/sec.

  11. In regards to gravity and how to calculate the same.  This is simply “its measure in weight”, for once knowing the acceleration, all substance - regardless of how fast it may be traveling away from, or at right angles to the direction of gravity, it is at all times subject to the acceleration of gravity.  

  12. It is thus as I explained before of no practical value to calculate g force in mass or earth-weight with the weight already known.


  1. And to further explain this.  By illustration figure 27-10, the block of mass may be likened to an engine that is running and in gear pushing downwards into direction A at a constant momentum of 32-ft/sec.  Its force is not directed to directions C, and D, wherefore they stand at zero ft/sec.  

  2. The force is however away from direction B, wherefore it is minus 32-ft/sec.  Any movement then to the directions C and D is irrelevant to the downward momentum of the mass, for the “ACTIVE” inertial momentum is not in these directions but only into direction A.

  3. There remains however the “POTENTIAL” of inertia for any movement into the directions C and D.  Which will become active (resist) at any movement of the mass into the directions C or D, that for their force in resistance may be calculated at 32-ft/sec (the inertia being at that value of resistance).  

  4. This resistance then, or the force required to overcome it by any velocity, is not of gravity (gravity being the A/B only).  But as it is properly called “a centrifugal inertia”, and the force, one of velocity, that is the same whether it be in a straight line (to accelerate), or at uniform angular velocity (a movement at a rate of change in direction).  Here then the value in feet/sec is to be squared to ft/sec2 (velocity to square the force).

  5.  The force of gravity thus in the inertia of the mass (Figure 27-10) is the inertia poised at a constant downward (A) momentum of acceleration.  Any amount of force then that exceeds the inertia (weight) of the mass in an upwards (B) directions - will raise the mass, as to say; to bring it free, or lose of the plane on which it rests, or, is pressed onto.

  6. Consequently, the horizontal movement of the mass does neither add nor subtract from the velocity of the acceleration of gravity.  It does not change gravitational force to mass.  The inertial momentum of the mass however in the greater inertia by the movement into C, or D, is nonetheless reacted upon by the force of gravity in a measure above and beyond its fundamental, or constant, of 32-ft/sec.  

  7. We might also explain it this way: Gravity in its fundamental constant acts only upon the “Angular inertia” (noted figure 27-10)  of the mass (wherefore its weight remains unchanged at any movement in velocity).  

  8. When therefore the mass acquires “Linear inertia” (movement in velocity noted figure 27-11) gravity acts upon that linear inertia ounce by ounce until that inertia becomes too great for the potential of the force of gravity.

  9. And so it is that gravity cannot be computed by a radius, nor by any velocity excepts by its own constant.  While centrifugal must be by radius, its inertia being of the second classification.  And one adds to the other, for both being inertia - gravity acts upon inertia, and upon inertia only.  

  10. And inertia in its most fundamental concept is “displacement of matter".  And gravity for its initiation comes about by “displacement of matter in angular change".  Which also reveals that it is by the very movements of the earth, its rotation, and its orbital track around the sun that gravity in the embrace of the one and only force in nature (3M - Magnetic force) is initiated and instigated.

  11. Once more, by figure 27-11, assume the mass of 100-lb in motion direction F, at a constant horizontal velocity that is even greater than the acceleration of gravity which at 32-ft/sec is holding the mass downwards direction E.  The mass will not come off the ground since there is no acceleration above the 32-ft/sec in a direction away from the direction of gravity. 

  12. And even if that were accelerating on a horizontal plane in excess of 32-ft/sec, it would still be subject to gravity in its downward movement.  And if we raised the mass above the ground and gave it an acceleration in excess of 32-ft/sec, it would only remain above the ground in a horizontal plane until the acceleration, and consequently, the linear inertia dropped down to a constant at which instant gravity takes over to a downward direction.

  13. In order therefore to raise the mass of 100-lb, we need a movement away from the perpendicular to gravity, away from the horizontal by a force that must exceed 100-lb. 100+lb therefore to add 1-ft/sec would give us an acceleration of 33-ft/sec.  The 100-lb for the 32-ft/sec, the + for the one ft/sec.


  1. Taken the example (Figure 27-6)  of a 3000-lb vehicle at the equator, the weight multiplied by the velocity of 1521-ft/sec, and squared, then divided by the radius leaves us with a resistance of 331.6-lb into the centrifugal direction.  And since 3000-lb divided by 32.14 is 93.34-slugs multiplied by 32.14-ft/sec is 3000-lb.  

  2. It gives us the acceleration of gravity at 32.14-ft/sec, and force at 3000-lb in weight added with the linear inertia of 331.6-lb.  The total and actual force of gravity on that vehicle therefore is 3331.6-lb.

  3. And for the next example, our satellite at 200-miles above the earth (Figure 27-7) having a gravitational weight of 3000-lb (minus 2% in acceleration of gravity) has an inertial resistance into the centrifugal direction of 94,286-lb - within which the -3000-lb is exceeded.  

  4. We cannot therefore say that the g force on the satellite is zero, since the actual gravitational force on the vehicle is 94,286-lb. but we can say that; “theoretically” the “fundamental” inertia of g is at zero.  I then said “theoretically” since it in fact is still there but overcome, exceeded, absorbed, as such.

  5. Now let us see how much gravitational force there is on the average person of 165-lb in earth-weight.  In order to accomplish this we need to know the mass in “earth-weight”, and we must compute centrifugal inertia to add to the earth-weight.  The force of the inertia then is the mass multiplied by the square of the velocity, and divided by radius.

  6.  165-lb X 1521 X 1521-ft/sec : 20,929,920-ft = 18.24-lb/c    (18.24 + 165-lb = 183.24-lb/g)

  7. The force of gravity (g) therefore on a person at the equator is about 10 percent more than its earth-weight.  And let us also see how much gravitational force is exerted on an astronaut 200 miles in radius above the earth.  We take again his 165-lb to add to velocity square by radius.  (Velocity at 18,000-mi/hr)

  8. 165-lb X 26400 X 26400-ft/sec : 22,176,000 = 5,186-lb/g

  9.  Because of his great speed moving around the globe, if this man had to do without the force of gravity to make him feel as if he were weightless, he would be crushed to death by a pressure on him in excess of 5000 pounds, about the weight of a automobile resting on him.

  10. And how do we know the person is weightless?  We simply divide the sum of the gravitational force on him by the factor of gravity.  And that factor (erroneously termed acceleration) at his radius being 31.34-ft/sec, 5,186: 31.34 = 165-lb.  

  11. The resultant at 165-lb means that 165-lb of his weight has been exceeded by, or absorbed into the inertial weight of the linear momentum.

  12. If for a next example, we send a man of 165-lb into space to the same 200-mi/radius, accelerating him into a velocity of only 26000-ft/sec, then165 X 26,000 X 26,000-ft/sec : 22,176,000 = 5030-lb/g

  13. When therefore we divide that sum of gravitational force by the acceleration (factor) of gravity in 31.34-ft/sec the resultant is 160.5-lb. meaning 160.5 of his weight has been exceeded.  Consequently since the person had a weight of 165-lb, he is 4.5-lb short of coming a state of weightlessness which of course will draw him back into the earth - the person still having 4.5-lb of weight.

  14. Next, let us decrease the radius for the astronaut to 22,000,000 ft but keep the velocity at 18,000 mi/hr, the resultant is: 165-lb X 26400 X 26400-ft/sec : 22,000,000 = 5,227.2-lb/g       (5,227.2: 31.34 = 168-lb or : 31.45 = 166.2-lb)

  15. If we divide by the acceleration of gravity at 200 miles above the earth, the outcome is 168-lb less 165-lb, tells us that the astronaut is +3-lb away from attaining to a proper weightless orbit.  (Or 1.2-lb if the acceleration of 31.45 is used)  The +3 not being a measure of his gravity but of linear inertia in excess of the gravitational pull on him.

  16. If then for one more example, we assume the astronaut happily going along in his weightless orbit, and we increased his velocity to 1535-ft/sec, the resultant would be: 165-lb X 26,600 X 26,600 : 22,176,000 = 5264.6-lb/g       (5264.6 : 31.34 = 168 - 165 = +3-lb.)

  17. The +3 pounds here is the same as before, an excess of linear inertia on him over the gravitational pull - due to the increase of his velocity.  

  18. This I hope should give us more insight in how and why a stable orbit might be attained, and that weight upon the earth is not the full measure of gravity, except for where there is no centrifugal inertia to be accounted for.  These places however are very cold to stand upon (earth’s axis).  

  19. And to clarify the following once for us:

The acceleration of gravity is written 32(+/-)lb  ft/sec/sec

The force of gravity is written 32(+/-) lb

The force of inertia is written 32(+/-) lb

The force in velocity to inertia 32(+/-) lb ft/sec2


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