Variable Acceleration

Abstract: Dressed in clown suit and dunce cap (his standard garb), Mathis makes a comical attempt to turn the calculus entirely upside down.

Don’t be fooled by the title, this is not really about variable acceleration at all. It actually involves turning the calculus on its head; a full one-eighty:

A Study Of Variable Acceleration by Miles Mathis

According to Mathis, physicists have got it all backwards; “upside down”, as he puts it:

“Students are currently taught to differentiate a distance to get a velocity and differentiate a velocity to get an acceleration, when that is upside down. In other words, if you are given an acceleration and you want to find the velocity, you differentiate. The entire modern interpretation of the calculus is upside down!” — Miles Mathis

Alright, so Mathis is telling us that everything is backwards. To find the velocity, we must take the derivative of the acceleration. This is not a joke; that is his claim. As Mathis defines it, velocity is the rate of change in acceleration:

“The derivative of an acceleration is a velocity.” — Miles Mathis

However, this would automatically imply that if the acceleration is constant (unchanging), then the velocity would have to be zero. Hard to believe, but that’s his harebrained theory. So let’s try differentiating the acceleration, just to see if Mathis has made a new revolutionary discovery.

Assume an object has been dropped from the top of a building and falls in a straight line with a constant uniform acceleration due to gravity of 9.81 m/s2. In other words, the acceleration does not change over time; it is a constant, and can be described by the following equation: a(t) = 9.81 m/s2 (See graph below)


Now according to Mathis, to find the velocity we must differentiate this equation. So let’s do that, and see what happens:

\bf v(t) = \frac{d}{dt}a(t) = 0

The velocity is zero? That’s right; the derivative of a constant is zero. So Mathis would have us believe that an object dropped from the top of a building will have a zero velocity as it falls. The sheer stupidity of this man is virtually unbounded.

In closing, Mathis has proved again (as if there were any doubt), exactly why he has become the laughing stock of the science world.

Note: This paper was last revised on 06/24/2013


32 thoughts on “Variable Acceleration

  1. I come here searching for Variable Acceleration . Now, Mathematics comes from
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  2. This question is for Steve:

    You are plotting a graph that shows a(t) = 9.81 m/s^2. However, if a(t) = 9.81 m/s^3 the graph would be identical, yet the “instantaneous” velocity would be quite different at t=1s. How would you explain or calculate that?

    • Their graphs will not be identical; the units are different. And the velocity will also be different, depending on whether the acceleration is constant or varying.

      Constant acceleration:
      v(t) = (9.81 m/s2)t
      a(t) = v'(t) = 9.81 m/s2
      a’(t) = 0

      Variable acceleration:
      v(t) = ½(9.81 m/s3)t2
      a(t) = v'(t) = (9.81 m/s3)t
      a’(t) = 9.81 m/s3
      a’’(t) = 0

      Note: Assume that the constant of integration is zero.

    • This question is for Steven Oostdijk:

      According to Mathis, “if you are given an acceleration and you want to find the velocity, you differentiate”. [Here]

      Please show us how that’s done? Assume that the acceleration is a constant 9.81 m/s2. In other words, the acceleration as a function of time would be:

      a(t) = 9.81 m/s2

      Now, please demonstrate how to compute the velocity using differentiation? I want to see this hat trick.

  3. Attention: Steven Oostdijk
    This comment has been brought to the top of the stack and flagged for your attention. It also appears in the comment section below.

    Michael Norris on April 25, 2013 said:

    Generally in science, we make measurements and then build mathematical models that can predict future measurements. So Steve U. has a valid question. You stipulate that G = ~10 m/s^2, but that figure comes to us from the existing model. Suppose that your table were a list of experimental observations. Can you show us how one would arrive at G = ~10 m/s^2 from these results, using Mathis’ model alone and nothing from the “backwards” version of calculus?

    • I think if you would understand that your above assumption and question is upside down this whole blog would be superfluous ;)

      • Steven O., let me try asking another way. Earlier you wrote, “Do you think this gravitational acceleration ‘constant’ G came into being through textbook math or by actually measuring the trajectory of falling objects?” By this rhetorical question, I assume (correct me if I’m wrong) you’re saying G *was* derived from actually measuring the trajectory of falling objects. But even if you aren’t a fan of G or the current model, you have to admit they are useful for predicting the future trajectories of falling objects, and I’m guessing you used them in constructing your table (again, correct me if I’m wrong).

        So my question is, suppose you were Newton and you wanted to get it right. You have some measurements such as those in your table, and you want to go about formulating a *useful* and *predictive* law that employs correct calculus as opposed to “backwards” calculus. Can you please explain how that would be done using the Mathis approach?

        • I guess you repeat questions so you can add your “correct”, “backwards”, “rhetoric” and other adjectives to hide lack of analytical content.
          The table can be used in opposite direction to integrate values identical to what is done in the incorrect calculus. It is very clearly described in the paper that is discussed.

          • Admittedly I may be a bit dense — I don’t see the table in Mathis’ essay, nor do I see any similar list. So I don’t know how one would go about it, using the numbers you provide. I assume you know what you’re talking about, so rather than just sending me to Mathis’ essay to figure it out (anyone could do that), I’ll ask for help again nicely: Can you step me through the process please, starting with these specific observational results and ending up with a useful equation derived from those results? I don’t know any clearer way to phrase my request. You seem to be really on top of Mathis’ work, so if it can be done, I’m sure you can do it! Thank you.

            • I’m asking you to explain how he would do this particular problem, because I honestly don’t understand how he would do it, and suspect that he couldn’t. Prove me wrong. You seem to understand his mathematics, so I was hoping maybe you could show me, and also school the blogger Mr. Urich at the same time. But seeing that you’re unwilling to demonstrate the procedure, and instead you just repeatedly dodge my question, I can only assume that either (1) you don’t understand it either or (2) it simply can’t be done.

  4. Reading the comments… someone noted that it would be far easier to just show where Miles has been right. His mistakes could span volumes… but his triumphs won’t fill a postcard.

    • That’s good advice; and well put too. However, this blog would then shrink to a single blank page. But maybe that’s your whole point; the less Mathis the better.

      • Hahaha, Miles seems pretty popular for somebody you want to marginalize if you dedicate a blog to him…

        Since you are not inclined to answer any of my questions, I will finish with a quote from Miles:

        “As a fair debater myself, I will tell you the number one rule of debating, according to the current handbook: always keep your opponent on the defensive. It is not a device I use, since I prefer the more subtle and less used method of actually knowing what I am talking about. But the mainstream preferentially follows the number one rule, since in most cases it is so effective. They know that with most audiences, facts and truth mean almost nothing. Everything is judged on form, and if your opponent can be made to look awkward, you will have won more points than could ever be won by being right. For this reason, you are taught always to ask questions; never to answer them. Yes, this is the technique of the mainstream in dealing with any resistance. Always attack. Always go for blood. If there is a threat, do not address the substance of it. Attack the person.

        Therefore, if you get into a debate with anyone over physics (or anything else), I suggest you remember what your opponent is up to at all times. Do not allow him or her to put you on the defensive. If your opponent is part of the status quo in physics, he or she should be able to answer questions. He is claiming near-omniscience, not you, so he should be able to answer all questions with ease, like the god he claims to be. The standard model is the one making the money and getting the attention and taking all the jobs, so they are the ones that should be answering questions, not you. They are the ones getting all the magazine and journal articles, all the book publishing, and all the government funding, so they are the ones that should be answering questions, not you. You are on the fringe, an independent researcher, a person just trying to help for free, so it is no surprise if your ideas are incomplete. No one should find that out of the ordinary. But what IS extraordinary is that mainstream physics, which has been gathered and culled by thousands of geniuses over centuries, and is defended by all the top people now, is full of huge awful holes and embarrassing fudges. Even more extraordinary is that these self-styled geniuses and top-of-the-field people do not have the intellectual honesty or the scientific acumen to see these holes and fudges for what they are, and to want to correct them. Remember that always.”

        • Debating techniques? Mathis? When has Mathis ever debated the content of his theories? He disabled the comment feature at his own website, for Pete’s sake. Gee, I wonder why?

          No, Mathis doesn’t have the cojones to step up to the plate himself; he’s still cowering behind his mother’s skirt, terrified to let go of her apron strings.

          His minions and servile flunkies do all the debating (parroting, to be precise), while Mathis sits at home and sings his own praises.

  5. Steven O,

    You have a rather strange way of phrasing things. “Is calculating a differential an integration operation or not?” By that, do you mean, is finding the derivative of a function the same thing as integrating a function? Which, of course, the answer would be no; one is the inverse operation of the other. But I’m not really sure if that’s what you’re asking.

    And regarding your math, is this alone, the full solution?

    Δx = x(t) – x(t-1)

    That’s it; nothing more? This was my solution:

    x(t) = ½ (9.81 m/s2)t2
    v(t) = x’(t) = (9.81 m/s2)t
    a(t) = v’(t) = 9.81 m/s2
    a'(t) = 0

    And this is your solution:

    Δx = x(t) – x(t-1)

    Well, yours is certainly the shorter of the two. But that’s only because it’s incomplete. Would you please fill in the rest?

    Also, how is x(t) defined; is it a function? Also, how is ΔΔx defined? I need more information. Please show all calculations and include the units.

    • Would you agree there is a difference between “measurement”, “calculation” and “function”?

      Anyhow, I’m happy you also start from a displacement “function” so we can now compare methods. Please notice you follow the same method as Miles in your new description, although the addition of the m/s^2 term as a constant could only be done by non-physicists.

      BTW: my formula describes all math that was used in creating the differentials in the displacement table. Δx is in meters, Δt is in unit seconds. ΔΔx = Δ(Δx(t)-Δx(t-1)), etc.

      Now I have a question: suppose your x(t) “function” would be At^3, would x'(t) then still express a velocity? Isn’t a velocity a constantly increasing displacement?

      • Yes, x'(t) would still express a velocity. And velocity is a change in displacement over change in time. No one disputes that.

        x(t) = (A m/s3)t3 + (B m/s)t
        v(t) = x’(t) = (3A m/s3)t2 + (B m/s)

  6. Steven O,

    The quote given above came directly from the Mathis article:

    “If you are given an acceleration and you want to find the velocity, you differentiate.” — Miles Mathis

    In other words, according to Mathis the velocity is the derivative, or rate of change in acceleration. It doesn’t matter whether the acceleration is varying or constant. His definition for velocity is the rate of change in acceleration; that is the whole premise of his article. If the acceleration doesn’t change, then the velocity is zero. Which incidentally, is proof positive that Mathis is a certified moron; a blithering idiot; someone to either laugh at or pity, depending on your point of view; but then, you knew that already. Anyway, here is the solution to the problem “x=At3+Bt” that you requested.

    Assuming that x = distance traveled in meters vs time (which automatically implies that A must have units of m/s3 and B has units of m/s), then the solution for the acceleration is as follows:

    x(t) = (A m/s3)t3 + (B m/s)t
    v(t) = x’(t) = (3A m/s3)t2 + (B m/s)
    a(t) = v’(t) = (6A m/s3)t
    a’(t) = 6A m/s3
    a’’(t) = 0

    • Hi Steve,

      Did’nt see you continued at the top of the comment. I think your problem originates from not appreciating the difference between a mathematical function and a real physical curve. Anything physical is a differential or interval and the concept of “point” as something with no extension has no place in physics.

      This is described in this article:

      Miles is not talking about differentiating an equation but differentiating a curve.

      • To quote from Miles calculus article (

        “A physical curve and a mathematical curve are not equivalent. They are not mathematically equivalent.
        This is of utmost importance for several reasons. The most critical reason is that once you draw a graph, you must assign variables to the axes. Let us say you assign the axes the variables x and y, as is most common. Now, you must define your variables. What do they mean? In physics, such a variable can mean either a distance or a point. What do your variables mean? No doubt you will answer, “my variables are points.” You will say that x stands for a point x-distance from the origin. You will go on to say that distances are specified in mathematics by Δx (or some such notation) and that if x were a Δx you would have labeled it as such.
        I know that this has been the interpretation for all of history. But it turns out that it is wrong. You build a graph so that you can assign numbers to your variables at each point on the graph. But the very act of assigning a number to a variable makes it a distance. You cannot assign a counting number to a point. I know that this will seem metaphysical at first to many people. It will seem like philosophical mush. But if you consider the situation for a moment, I think you will see that it is no more than common sense. There is nothing at all esoteric about it.
        Let us say that at a certain point on the graph, y = 5. What does that mean? You will say it means that y is at the point 5 on the graph. But I will repeat, what does that mean? If y is a point, then 5 can’t belong to it. What is it about y that has the characteristic “5”? Nothing. A point can have no magnitudes. The number belongs to the graph. The “5” is counting the little boxes. Those boxes are not attributes of y, they are attributes of the graph.
        You might answer, “That is just pettifoggery. I maintain that what I meant is clear: y is at the fifth box, that is all. It doesn’t need an explanation.”
        But the number “5” is not an ordinal. By saying “y is at the fifth box” you imply that 5 is an ordinal. We have always assumed that the numbers in these equations are cardinal numbers numbers [I use “cardinal” here in the traditional sense of cardinal versus ordinal. This is not to be confused with Cantor’s use of the term cardinal]. The equations could hardly work if we defined the variables as ordinals. The numbers come from the number line, and the number line is made up of cardinals. The equation y = x² @ x = 4, doesn’t read “the sixteenth thing equals the fourth thing squared.” It reads “sixteen things equals four things squared.” Four points don’t equal anything. You can’t add points, just like you can’t add ordinals. The fifth thing plus the fourth thing is not the ninth thing. It is just two things with no magnitudes.
        The truth is that variables in mathematical equations graphed on axes are cardinal numbers. Furthermore, they are delta variables, by every possible implication. That is, x should be labeled Δx. The equation should read Δy = Δx². All the variables are distances. They are distances from the origin. x = 5 means that the point on the curve is fives little boxes from the origin. That is a distance. It is also a differential: x = (5 – 0).
        Think of it this way. Each axis is a ruler. The numbers on a ruler are distances. They are distances from the end of the ruler. Go to the number “1” on a ruler. Now, what does that tell us? What informational content does that number have? Is it telling us that the line on the ruler is in the first place? No, of course not. It is telling us that that line at the number “1” is one inch from the end of the ruler. We are being told a distance.
        You may say, “Well, but even if it is a distance, your number “5” still applies to the boxes, not to the variable. So your argument fails, too.”
        No, it doesn’t. Let’s look at the two possible variable assignments:
        x = five little boxes or
        Δx = five little boxes
        The first variable assignment is absurd. How can a point equal five little boxes? A point has no magnitude. But the second variable assignment makes perfect sense. It is a logical statement. Change in x equals five little boxes. A distance is five little boxes in length. If we are physicists, we can then make those boxes meters or seconds or whatever we like. If we are mathematicians, those boxes are just integers.

        You can see that this changes everything, regarding a rate of change problem. If each variable is a delta variable, then each point on a curve is defined by two delta variables. The point on the curve does not represent a physical point. Neither variable is a point in space, and the point on the graph is also not a point in space. This is bound to affect applying the calculus to problems in physics. But it also affects the mathematical derivation. Notice that you cannot find the slope or the velocity at some point (x, y) by analyzing the curve equation or the curve on the graph, since neither one has a point x on it or in it. I have shown that the whole idea is foreign to the preparation of a graph. No point on the graph stands for a point in space or an instant in time. No point on any possible graph can stand for a point in space or an instant in time. A point graphed on two axes stands for two distances from the origin. To graph a line in space, you would need one axis. To graph a point in space, you would need zero axes. You cannot graph a point in space. Likewise, you cannot graph an instant in time.
        Therefore, all the machinations of calculus, all the dx’s and dy’s and limits, are not applicable. You cannot let x go to zero on a graph, because that would mean you were really taking Δx to zero, which is either meaningless or pointless. It either means you are taking Δx to the origin, which is pointless; or it means you are taking Δx to the point x, which is meaningless (point x does not exist on the graph–you are postulating making the graph disappear, which would also make the curve disappear).”

      • Please provide a mathematical solution using the Mathis method, so we can all see exactly where “the calculus is upside down”. Use the example from his article: x=At^3+Bt.

        We have all seen quite enough written commentary and handwaving; talk is cheap. So put up a real, actual solution. You have given us enough background theory, now show the calculations.

        • Hahaha, back to the old “shut up and calculate!” intimidation?

          We are discussing the difference between mathematical graphs and physical curves, if you like to believe there is no difference then there is no need to spend any more time discussing. Your choice if you are not interested in learning the difference.

          But to make that clear on your example: the straight line graph you plot does not represent anything physical, it’s a diagram of a mathematical constant.
          To make it a description of physical reality we could make a table of the motion of the dropped object. Let’s assume the building is 100 meter high and that x is the height of the dropping object.

          Then we could get a table that reads like this (assuming G ~10m/s^2):

          Time, x: Δx: ΔΔx: ΔΔΔx:
          0s 100m 0m
          1s 95m 5m 5m
          2s 80m 20m 15m 10m
          3s 55m 45m 25m 10m
          4s 20m 80m 35m 10m
          5s 0m 100m 20m 15m
          6s 0m 100m 0m 20m
          7s 0m 100m 0m 0m

          (Actually, to be precise, x should already read “Δx” since any physical value represents already a differential. E.g. if the sound pressure of your speakers is 0dB the air pressure is still around 1000mBar. The sound pressure level is a Δ wrt. to the equilibrium pressure.)

          Now, Δx represents the displacement curve, ΔΔx the velocity curve and ΔΔΔx the acceleration curve. As you see the velocity curve is a constantly increasing differential (until the object hits the ground) and the acceleration curve is a constant (until the object hits the ground).

          Now would you call taking the second or third order differential a differentiation operation or an integration operation?

          • That data table didn’t just create itself; you had to use equations to come up with the data points for displacement, velocity, and acceleration. Will you please show the equations you are using to derive the table?

            This is the third time I have asked to see the math. Is there a reason why you keep refusing this simple request?

            • Is asking your opponent to use your arguments a distraction to hide your lack of arguments or analytical capabilities? Anyway, if it helps you: the math is Δx = x(t) – x(t-1).

              Now can you please anwer my repeat question in return: is calculating a differential an integration operation or not?

              Do you think this gravitational acceleration “constant” G came into being through textbook math or by actually measuring the trajectory of falling objects?

            • Generally in science, we make measurements and then build mathematical models that can predict future measurements. So Steve U. has a valid question. You stipulate that G = ~10 m/s^2, but that figure comes to us from the existing model. Suppose that your table were a list of experimental observations. Can you show us how one would arrive at G = ~10 m/s^2 from these results, using Mathis’ model alone and nothing from the “backwards” version of calculus?

  7. Maybe you should use a variable acceleration example here yourself instead of a constant one, that would at least prove you can read :)

    • Damn, it’s Oostdijk himself!

      You must have overlooked the quote. So let me repeat it again:

      “Students are currently taught to differentiate a distance to get a velocity and differentiate a velocity to get an acceleration, when that is upside down. In other words, if you are given an acceleration and you want to find the velocity, you differentiate. The entire modern interpretation of the calculus is upside down!” — Miles Mathis

      By the way, welcome aboard.

      • Repeating mistakes does not make them correct.

        Perhaps you have overlooked that you present an acceleration curve here. Miles’ article analyses an distance curve.

        • This is the way I would solve the problem. Where r = distance traveled in meters vs time:

          r(t) = ½ (9.81 m/s2)t2
          v(t) = r’(t) = (9.81 m/s2)t
          a(t) = v’(t) = 9.81 m/s2
          a'(t) = 0

          Will you please provide a similar solution using the Mathis method; showing the derivative of the acceleration?

          • Yes, you have shown that your curve represents a constant acceleration.

            That’s not the topic of the paper you quote.

            How would you calculate the acceleration of the curve x=At^3+Bt as analyzed in the paper?

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