We return to our fabled circle, which can be drawn as such:
X = 100 + 50 * Cos(0.1 * T)
Y = 100 + 50 * Sin(0.1 * T)
This goes clockwise starting from the rightmost end of the circle (3:00)
Upon experimentation, you may find that we also get a circle even if we do this:
X = X + 50 * Cos(0.1 * T)
Y = Y + 50 * Sin(0.1 * T)
By accidentally placing X and Y instead of our 100, we get a humongous circle. Now, each
point of the circle is defined with respect to the previous circle.
The circle now has its center where X is 0 and Y is at some value. It obviously starts at
(0,0), which is now the topmost end of the circle (12:00), but continues counterclockwise.
Why does this make a circle?
(This is closely based more with calculus than with trigonometry, and I will avoid using
calculus for now.)
This is because, when we add our previous value to the next value, we are using our
functions (Cos() and Sin()) as an antiderivative of another function.
(The antiderivative of a function is another function, FYI.)
So, the antiderivative of +Cos() is +Sin(). The plus sign means that we are adding the
result (50 * Cos()) to the previous coordinate.
This is not all when taking the antiderivative, you also have to divide the 'radius' by the
angular frequency to get the actual radius. This number that appears in the additive version
as the radius, is not really a radius. It's what I like to call a 'radispeed'. I'll bet
there's some word that already exists for this term.
So, our first part of the circle:
X = X + 50 * Cos(0.1 * T)
is equivalent to
X = 500 * Sin(0.1 * T)
50 divided by 0.1 gives us 500, the actual radius.
X = X + radispeedX * Cos(afreq * T)
is equal to
X = radispeedX / afreq * Sin(afreq * T), where radispeedX / afreq = radiusX
The antiderivative of +Sin is -Cos. This means that the plus will change to a minus.
And our second part of the circle:
Y = Y + 50 * Sin(0.1 * T)
is equivalent to
Y = -500 * Cos(0.1 * T)
X = 500 * Sin(0.1 * T)
Y = -500 * Cos(0.1 * T)
That is fine so far, but we did not accomodate for the original placement of the circle.
The old circle's centerpoint is at (0, ???)
This circle has the centerpoint at (0,0).
How do we find the centerpoint?
This is surprisingly algebra.
The problem is that
X = X + 50 * Cos(0.1 * T)
Y = Y + 50 * Sin(0.1 * T)
does not actually define where it started, so you can specify an initial value for this
to work, and it will start the circle at that point (Xi,Yi as our initial point).
In our new modified version, we actually have to solve for it.
Xi = origX + 500 * Sin(0.1 * T)
Notice that our result point is assumed to be Xi, the initial point, and we now solve for the
origX. Our time variable T will be replaced by 0, and Sin(0) = 0, leaving us with
Xi = origX
The initial point Xi is 0, so the origin is 0.
It's correct so far. We did find the old circle's centerpoint to be (0, ???)
The Y point will be found the same way, using our recently developed expression with the minus
sign in it.
Yi = origY - 500 * Cos(0.1 * T)
Cos(0) is 1, so the - 500 survives the first round. 500 * 1 = 500
Yi = origY - 500
origY must be = Yi + 500, and since Yi = 0, origY must be 500.
Our new expressions:
X = 500 * Sin(0.1 * T)
Y = 500 - 500 * Cos(0.1 * T)
(Looks funny, but that's it)
Then why does your white circle (additive version) look bigger than the red circle (our
newly calculated result)?
That is because the original version is not making its calculations often enough to move as
fast as the real circle. You can get the circles closer together by decreasing the angular
frequency to slow both circles down (this will change the radii, so the radispeeds may need
to be changed as well.
The smaller the angular frequency, the closer these circles are. You can't set the frequency
to zero though, or you'll just get a dot.
Also, computer roundoff calculations in our Fmd function can mess things up as well, but the
principle behind it should still hold strong.
What about a phase shift?
The phase shift is passed right through with the function:
X = X + radispeedX * F(afreqX * T + phaseX)
Y = Y + radispeedY * F(afreqY * T + phaseY)
is equal to
X = origx + radispeedX / afreqX * antiF(afreqX * T + phaseX)
Y = origy + radispeedY / afreqY * antiF(afreqY * T + phaseY)
You will have to consider the phase when solving for the origx and origy
origx = radiusX * antiF(phaseX) - Xi
origy = radiusY * antiF(phaseY) - Yi
When will the additive version be useful? It is useful anytime where you are not in direct
control of the shape's position. This is often the case with an .Offset function, which
is really just a function where you supply an X and a Y, and the object that you offset will
move by that amount. So, .Offset(6, 3) moves the object 6 to the right, and 3 down.
Similarly,
.Offset(radispeedX * F(afreqX * T + phaseX), radispeedY * F(afreqY * T + phaseY))
uses our additive version to move our object in a circle.
The reverse of this process would be useful as well, since we may actually want to get
something that we can use the .Offset function for, in case we have to use .Offset.
In calculus, this would be called finding the derivative of a function
(perhaps I should've mentioned this first?).
The derivative of +Sin() is +Cos(), and the derivative of +Cos() is -Sin().
We follow the same protocol as we did above, except that now, we have to multiply by the
angular frequency. So:
X = origx + radiusX * F(afreqX * T + phaseX)
Y = origy + radiusY * F(afreqY * T + phaseY)
would be written in the additive version as
X = X + radiusX * afreqX * dF(afreqX * T + phaseX)
Y = Y + radiusY * afreqY * dF(afreqY * T + phaseY)
dF() is the derivative of the F() function.
Note that the origx and origy disappear. This is why the additive version has to have a fixed
starting point, which is probably applied to our object before we are able to .Offset() it.
We can also take derivatives of our other functions that we've used for making squares,
triangles, etc, and use those in our additive version as well.
The derivative of the SquareWave is mostly zero (I mentioned that it's pretty useless already,
but...). Our other functions have some meaningful derivatives: the TriangleWave has the
SquareWave as its derivative, as:
Trgl(afreq * T + phase) --> -2 * SquareWave(afreq * T + phase)
A use for our SquareWave!
The -2 comes from the TriangleWave beginning with a slope of -2, whereas the starting value
for the SquareWave is 1.
The explanation for all of these other functions comes with command of calculus
(and remember, I'm skipping that for now), so I'll just give out the answers for now.
The derivative of the HexagonWave is something that we'll have to define.
Similarly, with TrapezoidWave and ScaleneWave.
Public Function Hxgndif(N As Single) As Single
Dim Quotient As Single, Res As Single
Quotient = N / 4
Res = Fmd(Quotient, 4)
If Res >= 0 And Res < 1 Then
Hxgndif = 2 'Rising edge of hexagon wave.
ElseIf Res >= 1 And Res < 2 Then
Hxgndif = 0 'Top part.
ElseIf Res >= 2 And Res < 3 Then
Hxgndif = -2 'Falling edge of hexagon wave.
Else
Hxgndif = 0 'Bottom part.
End If
End Function
Public Function Trpzdif(N As Single) As Single
Dim Quotient As Single, Res As Single
Quotient = N / 3
Res = Fmd(Quotient, 3)
If Res >= 0 And Res < 1 Then
Trpzdif = 0 'Flat part.
ElseIf Res >= 1 And Res < 2 Then
Trpzdif = 2 'Rise to the point.
Else
Trpzdif = -2 'Fall from the point.
End If
End Function
Public Function Sclndif(N As Single) As Single
Dim Quotient As Single, Res As Single
Quotient = N / 3
Res = Fmd(Quotient, 3)
If Res >= 0 And Res < 1 Then
Sclndif = 2 'Steep leading edge of wave.
Else
Sclndif = -1 'Long edge of wave
End If
End Function
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Note that Trpzdif and Sclndif are very prone to floating point errors with Fmd due to
division by 3. Computers do not like division by 3. (You might
see 0.875 coming in as 0.873, for instance.)
But using these functions provide an additive version equivalent to the former functions we've
been using. Also, the additive version's functions are much more simpler (but lots of theory
is involved in them... kind of).