Reply to Drex's Math of Poi

Rem's picture

Hi Drex,
I've found your last tech blogs about Math of poi to be interesting, and they even motivated me to play a bit in matlab. I've found that I actually can not totally agree with your ideas, so I decided to explain you my point of view on this issue, which I hope you will find to be interesting.
Cool

To those, who are not familiar with the topic I suggest to watch the tech blogs #246 and #217

Thus resuming these blogs, there are equations proposed to describe trajectories of poi:
X=A*sin(t)+B*sin(m*t)
Y=A*cos(t)+B*cos(m*t)

where A,B and m are the parameters.
Using these formulae, we may obtain such kind of pictures, where A=B=1, m=-3 and m=3:

So obviously m is just a number of rotations of a poi during one circle. That sounds fine for me and I agree with that.
But in the second blogs it leads to wrong conclusion that varying these parameters we can obtain "isolation". In fact, the parameters a and b appears to be just the lengths of arm and the lengths of poi, but to speak about isolation we actually need to modify the phase.
So I propose the following formulae:
X=A*sin(t)+B*sin(m*t+d)
Y=A*cos(t)+B*cos(m*t+d)

where:
A is the lengths of arm
B is the lengths of poi
m is a number of rotations of a poi around the hand
d is a phase shift between the poi and the arm.
In these formulae the first part (with coefficient "A") describes the rotation of an arm, and second part (with "B") describes the rotation of a poi.

So the isolation is simply B/A=1/0.5, m=1, D=180° and cat eye B/A=1/0.5, m=-1, d=0°:

Here the trajectory of the hand is represented by green color, meanwhile trajectory of poi by red color.

Next, you are talking about "zero points" (also known as star figures), and there you think that you can get it simply keeping A=-m. Well, in fact, that is true, but in my opinion it is just a coincidence, and these are not zero points we are used to deal with. I think real zero point are only possible with very specific conditions (for instance A=-m)(but I can be wrong).
In your case, you just modify the length of the poi. For instance to perform 4pt star, according to your assumption, one should have a poi thrice as short as an arm Jokingly
But we are actually interested in the quite a long poi, so from now on, I shall assume that the length of the poi "B" is equal to the length of the arm "A" and they are equal to one.
So to deal with these zero points we should add one more part into our formulae, which will describe the rotation of a hand around an elbow.

X=L*sin(t)+B*sin(m*t+d) + C*sin(n*t)
Y=L*cos(t)+B*cos(m*t+d) + C*cos(n*t)

here L+C=C - the length of an arm (L - distance from shoulder to elbow, C- from elbow to wrist), n is a number of rotations of a hand around the elbow.

So now have three parts: X,Y= elbow part + poi part + hand part

With this new approach we have ( n=1, m=-3 ):

where yellow and blue colors represent trajectories of an elbow and a hand respectively.
Using this additional degree of freedom, we may explain even zero point, assuming that a hand does a kind of spin flower:

However, it is not a real star, because as you may see there are actually 6 (not 3) zero points (it is not very clear from this picture but there are zero points inside the red curve), but I guess it is the best you can have with the same length of poi and arms, using circle kind of hand trajectory (not a square or triangle).In the reality, I think, the trajectory of the hand (for zero points) is not as simple.

So that was my general idea of how we could use these equations and what do these parameters really means, in my opinion. Playing with various parameters I actually had a lot of fun and got many crazy curves and interesting findings) (may will be posted later)
May be I'm wrong at some point, in fact I did not check very carefully. And I'm sorry about possible errors in English, I did not check as well )

And finally ANTI-ANTI-SPIN !!!! Happy

UPDATE: if one wants to change the phase (i.e. to make antispin in box for example), you just need to alter the equation by adding a component to the phase:

X=A*sin(t)+B*sin(m*t +4*phase )
Y=A*cos(t)+B*cos(m*t +4*phase )

i.e. for the antispin example the "phase" would be pi/4, so you just need to add pi to make the box mode instead of horizontal petals.

Best regards,
Rem

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