our example file you use key name (such as my : MyEquation )
then you use Equation > (x^2 + (9*y^2)/4 + z^2 - 1)^3 - x^2*z^3 -((9*(y^2)*(z^3))/80)
where do you find all those goodies ?? all those expressions and equations
thnx…
Added by AJ to Volvox at 4:28am on August 22, 2017
oning behind using the equality component to test for even numbers is flawed because of the data matching used by gh. It is testing like this:
0==0 True
2==1 False
4==2 False
6==3 False
etc
.............
Where as a Modulo 2 would work like this
0%2 = 0
1%2 = 1
2%2 = 0
3%2 = 1
4%2 = 0
5%2 = 1
6%2 = 0
7%2 = 1
8%2 = 0
9%2 = 1
......
Also I notice you have some errors in your expressions producing Nulls.
If you want it to be twice the value then you should have 2*D in the Expression and 10*D in the other
....
I attach a working version.…
13 5 15 6 17 7 ... …
But it seems I have not been clear:
the input values (3,5,7,9,11,13) should be presented in the way of "number slider" parameter. So by increasing the value of the slider from minimum 3, to 5, 7, 9, 13, 15, 17... I should get an output numbers of 0,1,3,4,5,6,7...
So I if choose value 3 on the input "number slider", on the other end, I want 0 as an output.
Or if I choose value 5 on the input "number slider", I want 1 as an output, and so on.
Is this possible?…
the u value should correspond to the number of points per row (3), not the total number of points (9). in this case, with your total of 9 points arranged 3x3 set the U input to "3" and it will work.
Integer = 0 To 9
val *= 2
lst.Add(val)
Next
Since val is a ValueType, when we assign it to the list we actually put a copy of val into the list. Thus, the list contains the following memory layout:
[0] = 2
[1] = 4
[2] = 8
[3] = 16
[4] = 32
[5] = 64
[6] = 128
[7] = 256
[8] = 512
[9] = 1024
Now let's assume we do the same, but with OnLines:
Dim ln As New OnLine(A, B)
Dim lst As New List(Of OnLine)
For i As Integer = 0 To 9
ln.Transform(xform)
lst.Add(ln)
Next
When we declare ln on line 1, it is assigned an address in memory, say "24 Bell Ave." Then we modify that one line over and over, and keep on adding the same address to lst. Thus, the memory layout of lst is now:
[0] = "24 Bell Ave."
[1] = "24 Bell Ave."
[2] = "24 Bell Ave."
[3] = "24 Bell Ave."
[4] = "24 Bell Ave."
[5] = "24 Bell Ave."
[6] = "24 Bell Ave."
[7] = "24 Bell Ave."
[8] = "24 Bell Ave."
[9] = "24 Bell Ave."
To do this properly, we need to create a unique line for every element in lst:
Dim lst As New List(Of OnLine)
For i As Integer = 0 To 9
Dim ln As New OnLine(A, B)
ln.Transform(xform)
lst.Add(ln)
Next
Now, ln is constructed not just once, but whenever the loop runs. And every time it is constructed, a new piece of memory is reserved for it and a new address is created. So now the list memory layout is:
[0] = "24 Bell Ave."
[1] = "12 Pike St."
[2] = "377 The Pines"
[3] = "3670 Woodland Park Ave."
[4] = "99 Zoo Ln."
[5] = "13a District Rd."
[6] = "2 Penny Lane"
[7] = "10 Broadway"
[8] = "225 Franklin Ave."
[9] = "420 Paper St."
--
David Rutten
david@mcneel.com
Poprad, Slovakia…
Added by David Rutten at 6:26am on September 9, 2010
0;0;1}
{5;0;1}
{9;0;1}
Applying a simplify operation would result in:
{0}
{5}
{9}
If we only had the first branch in each case they would be the same (to wit, {0;0;1}), what should it reduce to? You can only say which parts of a branch are redundant when you have something to compare it to.
--
David Rutten
david@mcneel.com
Tirol, Austria…
ap value = True
Shift List = 1 --> (B,C,D,A)
Shift List = 2 --> (C,D,A,B)
You can also use negative values.
Shift List = -1 --> (A,B,C)
Shift List = -2 --> (A,B)
and with Wrap = True
Shift List = -1 --> (D,A,B,C)
Shift List = -2 --> (C,D,A,B)
The most useful Shift List action I use is to either get rid of the first or last item in a list and sometimes both.
Shift list = -1 --> (A,B,C) Shift list = 1 --> (B,C)
In the example posted above you are creating a shift list value equal to its location along the curve. The first section = 0 doesn't get shifted, the second section gets a shift = 1, third = 2, forth = 3 and because the wrap value is set to true the fifth section gets back to 0, sixth = 1 etc etc. creating the twisting effect.
The "one more stupid question" answer is Mass Addition. You will find the component on the Math tab or you can type it into the Keyword search feature (by double clicking the canvas). This component has two outputs a total amount for each list and a partial set of results giving:
List (3,6,9,12)
{0} = 3
{1} = 3+6 = 9
{2} = 3+6+9 = 18
{3} = 3+6+9+12 = 30…