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Take the thermostats cover off and expose the low voltage wiring, Y or Yellow is the AC circuit, Y1 will be 1st stage cooling, Y2 will be 2nd stage cooling, You may also look at the furnaces low voltage terminals for the various circuits. You will see screws/terminals marked, R /Red, W /White, Y /Yellow, G /Green and C /Common. The R terminal is where the 24 volt "hot leg" of power enters the thermostat , for the thermostat to send the 24 volts to your desired mode/ circuit. W = Heat, Y = Cool, G = Fan and C is the "common" or neutral leg of 24 volt power. Common is akin to the negative on a battery, it is called Common as every 24 volt circuit terminates on Common to complete the circuit. If you had a 2 stage heat system there would be a W1 and a W2. Same with the AC mode. Y1= 1st stage cooling and Y2 = 2nd stage cooling.
#include<stdio.h> #include<graphics.h> #include<math.h> #include<conio.h> #include<dos.h> #include<alloc.h> #include<stdlib.h> #define RAD 3.141592/180 class fig { private: int x1,x2,y1,y2,xinc,yinc; public: void car() { xinc=10;yinc=10; x1=y1=10; x2=x1+90;y2=y1+35; int poly[]={x1+5,y1+10,x1+15,y1+10,x1+20,y1,x1+50,y1,x1+60,y1+10,x1+90,y1+17,x1+90,y1+20,x1+5,y1+20,x1+5,y1+10}; setfillstyle(SOLID_FILL,LIGHTGRAY); setlinestyle(SOLID_LINE,1,2); setcolor(4); drawpoly(9,poly); line(x1+15,y1+10,x1+60,y1+10); line(x1+20,y1+10,x1+20,y1); line(x1+35,y1+10,x1+35,y1); line(x1+50,y1+10,x1+50,y1); floodfill(x1+18,y1+8,4); floodfill(x1+28,y1+8,4); floodfill(x1+36,y1+8,4); floodfill(x1+52,y1+8,4); setfillstyle(SOLID_FILL,4); floodfill(x1+18,y1+12,4); setfillstyle(SOLID_FILL,BLUE); bar(x1+5,y1+20,x1+90,y1+25); setcolor(DARKGRAY); circle(x1+20,y1+25,8); circle(x1+20,y1+25,6); setfillstyle(1,8); floodfill(x1+21,y1+25,8); circle(x1+70,y1+25,8); circle(x1+70,y1+25,6); floodfill(x1+71,y1+25,8); int size=imagesize(x1,y1,x2,y2); void far *buf=farmalloc(size); getimage(x1,y1,x2,y2,buf); while(!kbhit()) { putimage(x1,y1,buf,XOR_PUT); x1+=xinc;x2+=xinc; if(x2<(getmaxx()-10)) putimage(x1,y1,buf,COPY_PUT); else { cleardevice(); x1=10;x2=x1+90; y1+=yinc;y2+=yinc; if(y2<(getmaxy()-10)) { putimage(x1,y1,buf,COPY_PUT); } else {y1=10;y2=y1+35;} } delay(200); } farfree(buf); getch(); } } } } void main() { int gd=DETECT,gm; initgraph(&gd,&gm,"d:\\cplus"); fig f; f.car(); cleardevice(); closegraph(); }
Line (x1, y1, x2, y1); Line (x2, y1, x2, y2); Line (x2, y2, x1, y2); Line (x1, y2, x1, y1);
Drawing a line from (x1, y1) to (x2, y2) using a default Graphics object: graphics.drawLine(x1, y1, x2, y2);
formula for determining potential difference
On the board there is a R terminal which is usually the red wire and a common. Make sure you have 24v AC power to them even with the thermostat inside is turned off. This powers the board and defrost timer in the heat pump mode. After you've checked that turn the thermostat on in the cooling mode. Check voltage between Y1 and Common. You should have 24VAC. This means the thermostat inside is telling the outside unit to run. Keep in mind some thermostats and all defrost control boards have a 5 minute time delay. Once you have verified the 24VAC coming from the inside unit on Y1 and Common and the five minute delay has passed Check the Y1 out terminal on the board to common you should have 24 volts there as well if you do not then the board is bad as long as you have the 24 VAC between R and Common I mentioned earlier.
There are many calculations that could be done: =SUM(Y1:Y10) =AVERAGE(Y1:Y10) =MAX(Y1:Y10) =MIN(Y1:Y10) =COUNT(Y1:Y10)
5x-y1 = 4
Take the thermostats cover off and expose the low voltage wiring, Y or Yellow is the AC circuit, Y1 will be 1st stage cooling, Y2 will be 2nd stage cooling, You may also look at the furnaces low voltage terminals for the various circuits. You will see screws/terminals marked, R /Red, W /White, Y /Yellow, G /Green and C /Common. The R terminal is where the 24 volt "hot leg" of power enters the thermostat , for the thermostat to send the 24 volts to your desired mode/ circuit. W = Heat, Y = Cool, G = Fan and C is the "common" or neutral leg of 24 volt power. Common is akin to the negative on a battery, it is called Common as every 24 volt circuit terminates on Common to complete the circuit. If you had a 2 stage heat system there would be a W1 and a W2. Same with the AC mode. Y1= 1st stage cooling and Y2 = 2nd stage cooling.
#include<stdio.h> #include<graphics.h> #include<math.h> #include<conio.h> #include<dos.h> #include<alloc.h> #include<stdlib.h> #define RAD 3.141592/180 class fig { private: int x1,x2,y1,y2,xinc,yinc; public: void car() { xinc=10;yinc=10; x1=y1=10; x2=x1+90;y2=y1+35; int poly[]={x1+5,y1+10,x1+15,y1+10,x1+20,y1,x1+50,y1,x1+60,y1+10,x1+90,y1+17,x1+90,y1+20,x1+5,y1+20,x1+5,y1+10}; setfillstyle(SOLID_FILL,LIGHTGRAY); setlinestyle(SOLID_LINE,1,2); setcolor(4); drawpoly(9,poly); line(x1+15,y1+10,x1+60,y1+10); line(x1+20,y1+10,x1+20,y1); line(x1+35,y1+10,x1+35,y1); line(x1+50,y1+10,x1+50,y1); floodfill(x1+18,y1+8,4); floodfill(x1+28,y1+8,4); floodfill(x1+36,y1+8,4); floodfill(x1+52,y1+8,4); setfillstyle(SOLID_FILL,4); floodfill(x1+18,y1+12,4); setfillstyle(SOLID_FILL,BLUE); bar(x1+5,y1+20,x1+90,y1+25); setcolor(DARKGRAY); circle(x1+20,y1+25,8); circle(x1+20,y1+25,6); setfillstyle(1,8); floodfill(x1+21,y1+25,8); circle(x1+70,y1+25,8); circle(x1+70,y1+25,6); floodfill(x1+71,y1+25,8); int size=imagesize(x1,y1,x2,y2); void far *buf=farmalloc(size); getimage(x1,y1,x2,y2,buf); while(!kbhit()) { putimage(x1,y1,buf,XOR_PUT); x1+=xinc;x2+=xinc; if(x2<(getmaxx()-10)) putimage(x1,y1,buf,COPY_PUT); else { cleardevice(); x1=10;x2=x1+90; y1+=yinc;y2+=yinc; if(y2<(getmaxy()-10)) { putimage(x1,y1,buf,COPY_PUT); } else {y1=10;y2=y1+35;} } delay(200); } farfree(buf); getch(); } } } } void main() { int gd=DETECT,gm; initgraph(&gd,&gm,"d:\\cplus"); fig f; f.car(); cleardevice(); closegraph(); }
Line (x1, y1, x2, y1); Line (x2, y1, x2, y2); Line (x2, y2, x1, y2); Line (x1, y2, x1, y1);
If you mean: (y2-y1)/(x2-x1) and (y1-y2)/(x1-x2) then either works out the same.
{3x +y =1 {x+y= -3
if we take the (x1,y1),(x2,y2) as coordinates the formula was (x-x1)/(x2-x1)=(y-y1)/(y2-y1)
You can use Pythagoras to solve this to get the co-ordinates. from Pythagoras we have a^2 + b^2 = c^2 For this a and b will be the x and y distance and c the distance between the 2 points so we get (7-2) ^2 + (8-y1)^2 = 13^2 So we need so solve this to get a correct solution which is (8-y1)^2 = 144 so 8 - y1 = + or - 12 so y1 = -4 or y1 = 20
First substitute the coordinates of (x1, y1) into the equation, then simplify the equation so it has y in terms of x. y - y1 = m(x - x1) y - y1 = mx - mx1 y = mx - mx1 + y1 y = mx + (y1 - mx1) y = mx + (C)
(y -y1)=(x -x1)(y2 -y1)/(x2 -x1) defines the line containing coordinates (x1,y1) and (x2.y2).