Francois Normandin

The best strategy will indeed be of calculating your way through this. Since you have three circles that represent your mass, you have their centers and radii, I assume. 1- Calculate the center of mass (xc, yc) of the whole system. There must be a way to do in Vision. Else, use a ponderated way using the radius of each circle. (A smaller circle would move the center of mass less than a larger circle...) 2- Find the extrema for each directions: i.e. for each circles, calculate the point furthest in each direction. (You need only the four cardinal directions if you're dealing with circles.) For example: center of circle1 = (x1, y1) and radius = r1 then left bound = (x1-r1, y1), low bound = (x1, y1-r1), etc. Calculate for each circles and keep the maximum in each directions. 3-Your minimum radius circle that encompasses all circles will be of the same radius as the maximum distance between the center of mass and any of those bounds. 4- Draw your circle from (xc, yc) with a radius rc = distance between (xc, yc) & largest (xn±rn, yn±rn). I hope this doesn't sound too complicated...

You are using three different data types... Your slider is SGL-precision, your indicator DBL-precision and your conversion to string uses Integer... Try to match your datatypes (right-click: representation) to get consistent results or to use "Round to -Infinity" to get 3,51 to be rounded to 3. It is definitely not a precision error to round 3,51 to 4.

QUOTE(shoneill @ Feb 15 2008, 07:00 AM) I agree that the traveling salesman problem is anything but trivial. However, I don't think this problem is an equivalent. The traveling salesman's solution is to find the shortest route possible to visit all points on a map. It's been extensively studied in many engineering problems such as reducing time to solder thousands of components on a circuit board. But if I understand Daku's problem, I tend to confirm Yuri33's assessment that we can discard any points in the middle as they will not give a result anywhere close to the maximum distance between any two points on the periphery.

QUOTE(tcplomp @ Feb 12 2008, 12:34 AM) It is very similar to tree control, but only because most of the methods and properties look alike. I only used MCL control type, no type casting needed. The difference between these two typecasts is that the tree control gives you much more ease of use in icon management. You can hide/show each icon individually identified to a particular tag name. The MCL allows for a whole list of icons to be shown quickly rather than individually controlled. http://lavag.org/old_files/monthly_02_2008/post-10515-1202824855.jpg' target="_blank">

Hi PJM, The key is in the item symbol number. They start at 1000 and there are 141 of them. Select the active cell to change which column you're working on.

QUOTE(Aristos Queue @ Feb 6 2008, 11:15 PM) Well, that's very instructive. When you put it this way, it makes sense! Flexibility at the cost of performance...

1- I noticed the same thing with Dynamic Dispatch... 2- I added a "byReference" comparison to jfazekas's VIs and I end up with a ten-fold decrease in speed... :headbang: Is it to be expected or I completely missed the point of byReference architecture? I would have thought it to be quite comparable. Anyone has a benchmark on that? Download File:post-10515-1202338932.zip

That's a very basic question. Are you sure you tried to look for it before posting? A simple Google search would yield the answer right away... http://digital.ni.com/public.nsf/allkb/E3E...EA?OpenDocument

QUOTE(Dirk J. @ Jan 30 2008, 04:42 AM) Good idea Dirk... it could be as simple as a RLC type circuit equivalent. More info on the system that produces this step response would be required.

QUOTE(jbrohan @ Jan 29 2008, 03:06 PM) If you have an 8th-degree polynomial fit, why not do a 2nd derivative on the algebraic solution? It all depends on how good is the fit compared to original signal, but algebraic solution wouldn't have bumps... y = ax^8 + bx^7 + cx^6 + dx^5 + ex^4 + fx^3 + gx^2 + hx + i dy/dx = 8ax^7 + 7bx^6 + 6cx^5 + 5dx^4 + 4ex^3 + 3fx^2 + 2gx + h d(dy)/dx^2 = 56ax^6 + 42bx^5 + 30cx^4 + 20dx^3 + 12ex^2 + 6fx + 2g Use an algorithm to find the zeroes of your second derivative (inflection points), then keep only those values whose first derivatives are positive. (i.e. where the slope stops increasing and starts decreading). After finding "x", the slope can be found by solving dy/dx for this value... Hope this helps!

QUOTE(Gabi1 @ Jan 25 2008, 04:24 PM) To create a strict type reference of a waveform chart, right-click on waveform chart and create a reference. Then, in the block diagram, right-click on the reference you just created and create a constant. This newly created reference will be "strict" type.

Is that what you're trying to achieve?