Re: Magnetic propulsion theory revisted.
Reactor,
You are to be commended for deciding to study this, and learn (or re-learn as the case may be) the powerful tool that is calculus. It can help you not only in your magnetic ideas but also in your AI experiments.
Do I understand the intergral and derivative in calculus? I don,t remember about the intergral but I remember the derivative is a rate of change. I also remember being shown the hard way to compute a derivative then the easy way.
1) The integral is merely the mathematical function that "undoes" or reverses the derivative. It is most instructive to view and describe these functions of calculus within the domain of analytic geometry, which includes simple X-Y graphs...which we will do below.
2) I have a feeling (not sure) that when you say the "easy way" to compute a derivative you might be referring to the finite difference method, which is actually only an
approximation to the derivative. And in some cases it can be a very good approximation, especially if the quantity you wish to take the derivative of is linear in nature. But let's not get ahead of ourselves...
Volocity is a derivative with respect to time of position. It seems to me volocity is also a rate of change per time and position.
Let's make sure we are clear here. The first statement is definitely true. I am not sure exactly what you are saying in the second (i.e. an equation might clear it up). But let's write the finite-difference
approximation for velocity in terms of position and time:
Velocity = (x2 - x1)/(t2 - t1) = Delta(X)/Delta(t) = Difference in Positions/Difference in Time
This is an
approximation because depending upon how large the granulaity is for the time difference (t2-t1), the numerical value of the velocity could change within that time interval and this equation would not represent that change. The way we go from the finite difference approximation to the instantaneous derivative is through the application of the calculus principle of "allow the difference in time to approach zero". Symbolically, this is represented as:
(t2-t1)->0
Where we represent the finite-difference approximation of a derivative as Delta(x)/Delta(t), when we wish to represent the precise, calculus version of the derivative we write it as:
velocity = dx/dt
This notation probably refers to what you have termed the "hard way" to compute the derivative. I agree it may be hard, but it is hard because it is computing the precise derivative, rather than just the approximation. The classic example is if we assume position can be represented by the function "x squared" (x^2):
Position = x^2
The "hard" way to take the derivative would be to move the exponent down to be a multiplier of "x" and then reduce the exponent by 1. Hence:
dPosition/dt = 2*x (we brought the exponent "2" down as a multipler on "x" and reduced the power of "x").
I know that this seems complicated. Let's leave that behind for now and address a few other points you made:
At t1 the volocity is # and position is #. At t2 the volocity is <font color="red">#[/COLOR] and position is #.
I added the "#" which I think you omitted. But yes, this is correct. This represents the finite-difference approximation of the derivative that I explained above. Very good!
The volicity itself can speed up or slow day per rate of change which is a derivative. Does it go something like this?
Yes. Again, this is very good. And when the velocity itself goes up or down, this is what we call
acceleration. Acceleration (linear) is nothing more than the derivative of velocity with respect to time. And since velocity is the derivative of position with respect to time, we could also desribe acceleration as "the second derivative of position with respect to time". In other words: If we apply the calculus derivative function with respect to time TWICE on the measurements on position, then we arrive at the measure of the body's acceleration.
I saved what you wrote above to review later. I will need to spend some time with it to understand it. I have been looking into software for reviewing and then studying up on calculas. I have not found much yet though. I figure I will come across something.
Very good. Now let me give you an example that uses the domain of analytic geometry I mentioned above. Taking the derivative of a quantity is nothing more than figuring out the instantaneous slope of a line (or curve). Here is a graphical example I use in my ARO 101 course to teach students one of the most important
stability derivatives used in aerodynamic analysis of an airfoil or wing... We call this stability derivative the
Lift-Curve Slope and this diagram may help you understand it a bit better. (but maybe not!) /ttiforum/images/graemlins/smile.gif
In this graphic we are using the finite-difference approximation to calculate the slope of the lift coefficient curve. As the angle of attack of an airfoil increases, the total lift increases... up to the point where the airfoil stalls. So the lift-curve slope is a derivative of the lift coefficient with respect to angle of attack in much the same way that velocity is a derivative of position with respect to time. You could also draw an "x-y" plot of position (y values) vs. time (x-values). The slope of that line at any point represents the velocity.
We will deal with the integral in a later post, if you wish. But as I said above, the integral is just the "undoing" of the derivative. So if the derivative is equivalent to finding the slope of a curve anywhere along that curve, we will come to find that the integral seeks to find the area underneath a curve.
Hope this helps more than it confuses...
RMT