how to rotate polygon pentagon dial

I never remember whether x should be sin or cos, but if you’re consistent nobody will know if you have it backwards...

However: one of them is r * cos the other is r* sin and a loop through all five points gets you a polygon. 

Something like

var r = 90;

var cx = dc.getWidth()/2;

var cy = dc.getHeight()/2;

var alpha = 0; // set to start angle

var points = new [5];

var delta = 2 * Math.PI / 5;

var i = 0;

do {

var x = cx + Math.sin(alpha) * r;

var y = cy + Math.cos(alpha) * r;

points[i] = [x,y];

i++;

alpha += delta;

} while(i<5);

dc.drawPolygon(points);

Then you adjust the start value of alpha to rotate. 

Not sure how you plan to use this, but an option is to build a table for the polygons you need.  That eliminates the need to do math all the time, and it's a simple table reference.  It will take more memory but would help battery and the power budget if you are doing a 1hz WF.

There is sample code in samples/Analog/source/AnalogView.mc that does exactly what you are looking for. Here is a snip:

var result = new [4];

var cos = Math.cos(angle);
var sin = Math.sin(angle);

// Transform the coordinates
for (var i = 0; i < 4; i += 1) {
    var x = (coords[i][0] * cos) - (coords[i][1] * sin) + 0.5;
    var y = (coords[i][0] * sin) + (coords[i][1] * cos) + 0.5;

    result[i] = [centerPoint[0] + x, centerPoint[1] + y];
}

It is hard-coded to transform 4 coordinates, but you could easily just take in an input array of coordinates and generate an output array of transformed coordinates.

I'm trying to make better hand dials and yes its for a 1 hz WF. and the power budget is important. when i use polygon with math it exceeds the power budget.

i'll check this out. thanks

Could you show me an example of the table ref?

If you're using this for 1hz, I'd use  whats in the analog sample as  Travis suggested.  I use that in a few things for an analog style second hand.

How close to the power budget are you?  Take a look at how you are setting your clip regions as that alone could be the issue.  The number of rows is really important (the display  is updated by row)  In a typical analog face with a second hand, there will be peaks at 12 and 6 o'clock, and low usage at 3 and 9 o'clock.

I'm using the analog clock example

Heres my code. I have other dials that i'm using but i want better ones :-) 

I'm very close to the budget unfortunatly.

function drawSecondHand(dc, xPos, yPos, clockTime) {
var secondHand = (clockTime.sec / 60.0) * Math.PI * 2 - RAD_90_DEG;
drawHand(dc, xPos, yPos, secondHand, 108, 2, cSecondHandColor);
// clean up the center
dc.setColor(cSecondHandColor, Gfx.COLOR_TRANSPARENT);
}

function onPartialUpdate(dc) {
// no need to do anything if partial updates aren't allowed or if we
// don't have an off-screen buffer allocated
if (!$.gPartialUpdatesAllowed || dcOffscreenBuffer == null) {
return;
}
drawBackground(dc);
middleY = 120;
middleX = 120;
clockTime = System.getClockTime();
drawHands(dc, middleX, middleY, clockTime);
dcCurClip = calculateSecondHandClip(clockTime.sec, 108);
dc.setClip(dcCurClip[0], dcCurClip[1], dcCurClip[2], dcCurClip[3]);
drawSecondHand(dc, 120, 120, clockTime);
drawHandCenter(dc, middleX, middleY, true);
}

function calculateSecondHandClip(seconds, SecondHandLength) {
var angle = (seconds / 60.0) * Math.PI * 2 - RAD_90_DEG;
var points = [
middleX+(Math.cos(angle) * SecondHandLength),
middleY+(Math.sin(angle) * SecondHandLength),
middleX+(Math.cos(angle+Math.PI) * SecondHandLength * 0.17), // add PI to get 180 deg (opposite dir)
middleY+(Math.sin(angle+Math.PI) * SecondHandLength * 0.17) // add PI to get 180 deg
];
var bbox = [
(points[0] < points[2] ? points[0] : points[2])-2,
(points[1] < points[3] ? points[1] : points[3])-2,
(points[0]-points[2]).abs()+4,
(points[1]-points[3]).abs()+4
];
points = null;
// need to adjust the bounding box to handle the tail and the hand centers
if (bbox[0] > middleX-13) {
bbox[0] = middleX-13;
bbox[2] += 13;
}
if (bbox[1] > middleY-13) {
bbox[1] = middleY-13;
bbox[3] += 13;
}
if (bbox[0]+bbox[2] < middleX+13) {
bbox[2] = middleX+13-bbox[0];
}
if (bbox[1]+bbox[3] < middleY+13) {
bbox[3] = middleY+13-bbox[1];
}
return (bbox);
}

I suspect it may be the size of your clip region.  Try making the second had something like 1/2 the length, with the clip region needed for that and see how you're doing with the power budget (make the 108 maybe 70)