Regards, I have problems with ticking shoes on the grid. For example using the following code.
public void setup(){
size(1000, 600, JAVA2D);
background(17,34,51); // Color de fondo principal
rect(-1, 95, 1001, 355,1); // Tamaño del rectángulo blanco donde se grafican los datos
stroke(196,196,196); // Color de la línea del cuadriculado
fill(258,258,258);//Color del Texto
textSize(13);
for (int i = 0; i < 1000; i=i+7) { // Dibuja el enrejado vertical de la zona de graficación
line(7+i, 95,7+i, 449);
}
for (int i = 0; i < 350; i=i+7) { // Dibuja el enrejado horizontal de la zona de graficación
line(0, 98+i, 1000, 98+i);
}
stroke(98); // Color de la línea del cuadriculado
for (int i = 15; i < 999; i=i+34) { // Dibuja el enrejado vertical de la zona de graficación
line(20+i, 95,20+i, 450);
}
for (int i = 0; i < 350; i=i+34) { // Dibuja el enrejado horizontal de la zona de graficación
line(0, 95+i, 1000, 95+i);
}
}
But as it is possible to appreciate, in the end the grid does not fit perfectly, it is important to emphasize that I need that there are 30 large black squares in the dimensions specified in the code.
Hello, your answer can definitely help me achieve what I need, but how could I get 30 frames to the end of the screen? In his example there are 28 frames and if I modify the number at the end, the frames vary a lot “for (int i = 0; i <1000; i = i + 6)”, for example if I put a six, it returns 33 frames and no 30.
I do not understand how he managed to reach 30, however in his example despite the fact that there are 30, they shrunk and did not reach the end of the screen, how could he do to achieve 30 and reach the end of the screen?
It was not clear to me where I should multiply “i”, I try it in different places but I don’t have good results.