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ABSOLUTELY!
I do all my programming on QB45 and QB64
Everything writ in qb45 is runnable in qb64 - which is faster and more powerful and plays GREAT music!
So I write and debug everything in qb45 and then run the working prog to qb64.
This forms a GREAT topic (ability) called "Experimental maths", where you simply TRY IT - and see what results you get! There is almost no task that we cannot simulate in qb64.
Mathematics is an artifact of the human mind, and a tool whereby the human mind wraps itself around reality.
Cheers!
digits
Yes, exactly right, digits.
EVERYTHING we think and talk about - even you, digits.
The brain, we think (imagine?), receives things we call "sensory nerve signals".
This is the stuff from which the BRAIN dreams up the "world" and calls it "real" (in order to prolong its sanity!) The ones who thought the sabre tooth tiger "not real" got eaten.
If you doubt this, tell me how you know right now you are not dreaming!
Please explain what is going on here
The big circle is fixed?
The mid-size circle ROLLS on it without slipping?
OR
Is attached to it at a fixed point - while the big circle rotates
The small circle rolls on the mid-size circle? At what rate?
OR
The small circle is attached fixedly to the mid-size circle (which spins?)
OR
The small circle spins - at what rate?
The pen is attached to a fixed point
ON the small circle
OR
Within the small circle
Or rotates around the circumpherence of the small circle - at what rate?
Many thanks
A good 3-d plotter would give you a model of the dots - a model you could hold in your hands and twist about to see really what was where.
So what are the BEST ways of simulating "twist about to see really what was where" on a flat computer screen?.
Hi;
You would need a 3D plotter.
Is there a good way to simulate a 3-d plotter on a computer screen?
If the bar were, by some immense stroke of luck, to remain straight the axial pressure to reduce its length by one inch would be a small fraction of a pound per square inch.
In practice it would buckle under gravity if vertical or sag under its own weight if horizintal, without requiring any axial pressure at all.
This question seems to be about bending of a bar.
But such a bar will bend under its own weight if horizontal or buckle if vertical
There is no limit to the ways it could bend depending upon
Which DIRECTION the one inch is - sideways or lengthwise.
HOW the bar ends are held - clamped angle or free to pivot
How easily the steel of the bar compresses under axial force.
The bending moment at any position x of the bar is the sum of the axial forces times the y deflection at that point. In this we should add any forces due to the weight of the bar.
A useful formula is the deflection of the bar at point x is
Bending moment M at point x is E I d2ydx2 , where E is Young's modulus and I is the moment of area of the beam's cross-section at position x.
Avagadro number (6 times 10^23 maybe)
times
6 /22.4 vol of lung
times
1/N fraction-fill (1/4 maybe)
times
20% oxygen
The simplest idea so far is this:-
For N>2 points, each point has a neighbour that is nearer than of all the others.
Lets make this distance as great as we can without going outside the sphere.
VERY SMALL increments are best!
We now REPEAT this for every single point in turn.
Now each point ends up max-spaced from its nearest neighbour
This is ONE variety of evenly-spaced.
But is it the only one?
To SEE this we need a way to visualise the pattern of points (on a flat screen)
For example for N=5 there is one pattern (of several?)
When we add a point we now have N=6 points.
If we could SEE how the points moved when we added that point it would help us begin to understand "what is happening" and why!
PLease suggest some ideas of how to visualise points scattered on or inside a sphere.
Many thanks
Bobby, thanks for being all three
Help
Sounding Board
Someone to talk to
Thanks for your comments
So you are saying that "Truth" cannot be found by calculations.
You are saying it depends what sort of answer you want!
The sort of answer I'd find helpful is this:-
For the value of k that gives a stable 3-cycle from the iteration nextx=kx(1-x) do YOU say this is "sensitive to intital conditions" or not?
For the STRUCTURE is the same (a stable 3-cycle and an attractor) whatever tiny changes you make or suffer, for 0<x<1. Start in this range and you stay in this pattern for ever - no escape! If you start outside the 3-cycle range (but with 0<x<1) you soon get attracted into it
Does the iteration nextx=4x(1-x) repeat or does it not?
On any MACHINE eventually it gets an x that it cannot tell different to an earlier x.
As nextx depends ONLY on x, from then on the whole list of xs repeats.
So HOW CAN WE TELL if the x-sequence "really" does nover repeat or not?
Not real planets!
The "mathematically pure" set of point-but-massive bodies moving relatively by Newton's Inverse Square Gravity Rule-of-Thumb without collisions.
There are a few position-velocity states into which, if placed, a third body will seem to be stable rotationg "motionlessly" with respect to the other 2.
A very-slightly-disturbed such body will move in an orbit about this so-called Trojan Point.
Are the calculated "strange attractors" imaginary artefacts due to computer error
How would we know?
The general question "How do I tell when my calculations are rubbish (even when totally reproducible) is most interesting and the opposite of chaos!
With chaos as nextx=4x(1-x) there is chaos spread throughought 0<x<1.
But for less than 4 there are interesting BANDS of permitted x and some periodic results.
The results are NOT evenly-distributed: their structure SEEMS stable against very small errors, but of course if we look at detail sufficient to "show errors" we will surely find them.
It is this ability to "accept and forgive" tiny errors while preserving the structural integrity for me to SEE - that is what fascinates me.
Ok, thank you
So how do I know if the present calculated position of a satellite in a Trojan orbit between earth and moon is valid
What does "stable orbit" mean
Can it be closed BECAUSE OF computer error whereas really it is unstable around the Trojan point no matter how close we are to that point.
I think that some of the trojan points ARE unstable in that sense whilst others are stable within certain bounds.
Guess the volts at the point where the series part joins the parallel part
Then find the currents in the 3 branches
When they add to zero you've got the volts right.
Successive guesswork
When I calculate a path, or movie, often each state is calculated from the previous state.
Thus the ERRORS may tend to build up - even to the "nonsense" level.
For example if I roll a marble down a child's playground slide it goes down wandering from side to side but usually NOT escaping over the side.
So how big, and what KIND, of errors give
(a) an unstable result - escape beyond the slide edges
(b) PERIODIC meanderings about the valley bottom as we roll down the slide
(c) NON-periodic but bounded wanderings down the valley
(d) Settling to a steady descent down the valley bottom
These questions are vital to the whole concept of computer calculation.
And even IF the slide has no bends, changes of gradient or changes in cross-sectional shape.
The question I am asking is WHEN do calculations of a "strange atteactor" give results "of the same structure" - and thus USEFULLY INFORMATIVE.
Beyond what point are they nonsense?
Can they ever recover and become "accurate" again?
How would we know?
I suspect the idea "of the same structure" is meaningless as "structure" is a variable in infinite-dimensional space.
So HOW CAN I TELL before I waste weeks calculating meaningless trajectories!
Only SOME results are stable enough to "forgive" small-enough errors.
SOME resuts - not only those that repeat steadily and periodically!
For HOW LONG do the calculations of planetery orbits stay accurate enough not to be totally invalid.
How do we know when?
All ideas welcome
Please help.
So for N=2,3,4,6,8.12,20,30 there is ONE best pattern of dots for each N
But for all other N there are as many "best srrangements" as the number of ways we care to define "best even placement" - how many neighbours are at WHICH distance.
Yes indeed - we must say what we decide we mean by "evenly spaced"
For the cases N=2,3,4,6,8,12,20,30 it is plain enough
We mean maximally spaced but in groiups as large as possible the distances to be equal
For example for N=8 we want every dot to be equal distance from every other of 3 other dots and equal spaced but more distant from three other dots.
For N=4,5,7,,9,10,11 etc we can make up similar rules
But for all N>4 we need to say what weight we give to the dots closest compared to those farther away. In that way we decide which dot-arrangement is best for each N in turn.
This is where I am stuck and need your ideas and suggestions
I was trying "Quick Reply" and after it had already acknowledged I was signed on
Quick Reply is denioed with idiot message endinmg "Say when"
I was trying to send you a reference illustrating the vast difference between random and uniform
Random means irregular!
The ref is now lost and I have spent an hour trying to recover it - sorry abiut that.
To DENY crystalline's claim to be more uniform we need a way to MEASURE uniformity of spacing
testing why it refuses to allow me to reply
Good idea!
By the way I like your "In mathematics, you don't understand things. You just get used to them." but for me it is "Just try them out!"
OK, so for a certain sphere, 1256200 points, the surface points nearest to one point vary depending on what we mean by nearest
Of ALL points only 2 (one pair) are nearest of all.
But for every point there are:-
A nearest point
A next nearest point
and so on
For example for point A its nearest dot is distance 3.12 away
The next nearest are 3.2,3.33,3.7 and 4
Is that "good"
What would be better?
For point B the nearest neighbours were
3.13, 3.19,3.3,3.73, 3.9
Is B better that A?
Might it pay to skip point B (or A) and replace it with a "better" point
How do we judge?
HOW do we begin to puzzle this out?
Yes, Bobbym, very helpful
Many ingenious ideas, but no solution
Spheres COMTAINED WITHIN a sphere
Points on the SURFACE of a sphere
In BOTH cases there are two oft-suggested ideas:-
A kind of "crystal structure" as when we "grow" close-spaced spirals of increasing latitude and longitude.
A kind of "random" structure in which dots are placed more likely the bigger the area that dot gets placed in.
The "random" has very much more variation in distances (and risks overlaps)
I suspect there IS a computer approx method
Start with the 8 dots of a cube vertices
or the 12 midpoints of its edges.
We now want to add ONE more dot.
The obvious place for it is equidistant from as many other dots as possible.
For example for the N=12 case we get n=13 by placing that one extra dot midway in between 4 other closest dots
We mow have UNevenly-spaced dots.
So we move the nearest dots to the new dot a small distance d away from the new dot.
And for each of those dots we move their nearest-meighbours away by d/2
And so on.
The computer will happily work away on this all night
Reminds me of R V Southwell's brilliant "Relaxation of Constraints"
The general plan is "Always reduce the biggest errors" and keep going!
The site that says "May be impossible" is a mere "pure" maths site!
I simply want the solution that gives for closest neighbours the minimum distance and how much these minimum distances vary
For N as small as 13 this can be listed exactly and fully
For N millions we will have to think carefully how we WEIGHT the importance of nearest-dots agains those progressively farther away!
ALL ideas welcome. that is what I care about - new IDEAS!
Great Stuff!
Can you add a way to halt the drawing so we can see how the wiggles are formed (the "stars" for example.
Did you find the wheel ratios by trial and error?
> I am trying to put N dots equispaced on a spherical surface
>
> For N=2 the answer is clearly `Poles apart` - distance pi times R
> For `The Platonic Solids` there are exact answers
>
> But HOW do we handle N=17 etc and how do we get there?
> The `most evenly-spaced configuration` and the VARIATION in the nearest-neighbour gt circle distances.
>
> For N=millions once again the answer should become computable and be yery near that for a hexagonal array in 2-d
> But how near?
>
> Many thanks
> John
>
> Might this work:-
> Place n dots at random lat/long
> For each dot in turn, Find the ONE nearest-neighbour that is FARTHEST
> Find a new position for it that reduces that distance
> Then find the ONE nearest-neighbour that is NEAREST and move it farther away