my sharp do not perform their function together

an apology for the writing, use the google translator. What I want to do is use 3 sensors Sharp GP2Y0A21YK0F that activate a led, if the first sensor sees, that the led turns on, if the second sensor sees, it turns on the same led, and the same with the third one.

When only one sensor is detected and it detects, the LED is activated, all right up here

#define LD 13
int SENSOR1 = A1; // pin A1
int lectura1;
float conversion1;

void general()
{
digitalWrite (LD, LOW);
}
void setup ()
{
pinMode (LD, OUTPUT);
}
void loop ()
{
lectura1 = analogRead (SENSOR1);
conversion1 = pow (3027.4 / lectura1, 1.2134);

if (conversion1> = 0 && conversion1 <15) {
digitalWrite (LD, HIGH);
}
else
{
general();
}
}

However, when it declares the 3 Sharp when the SENSOR3 detects, the LED lights up completely, but when the SENSOR2 or SENSOR1 detects, the LED turns on very little

#define LD 13
int SENSOR1 = A1; // pin A1
int lectura1;
float conversion1;
int SENSOR2 = A0; // pin A0
int lectura2;
float conversion2;
int SENSOR3 = A2; // pin A2
int lectura3;
float conversion3;

void general ()
{
digitalWrite (LD, LOW);
}

void setup ()
{
pinMode (LD, OUTPUT);
}

void loop ()
{
lectura1 = analogRead (SENSOR1);
conversion1 = pow (3027.4 / lectura1, 1.2134);
lectura2 = analogRead (SENSOR2);
conversion2 = pow (3027.4 / lectura2, 1.2134); // conversions to cm
lectura3 = analogRead (SENSOR3);
conversion3 = pow (3027.4 / lectura3, 1.2134);

if (conversion1> = 0 && conversion1 <15)
{
digitalWrite(LD, HIGH);
}
else
{
general();
}
if (conversion2> = 0 && conversion2 <15)
{
digitalWrite (LD, HIGH);
}
else
{
general();
}
if (conversion3> = 0 && conversion3 <15)
{
digitalWrite (LD, HIGH);
}
else
{
general();
}
}

The sensor that fully activates the LED is the last “if” in the program
I would like you to help me solve this problem, since each sensor activates the LED with different brightness. I await an answer, thank you.

the diagram
https://plus.google.com/photos/photo/114612218997856866388/6611636528582990546?authkey=CLnZnczVyaLsTg

Sometimes I get tired of asking: please show a drawing, schematic, or a picture of how you have this all wired, the pins that are being used and the power supply.

Paul

Why did you create a new account to cross post something you already have posted twice?

DON'T CROSS POST!!!!!!!!!!!!!!!!!!!!

I HAVE REPORTED THIS THREAD TO THE MODERATORS

Duplicate posts waste the time of the people helping you. I might spend 15 minutes writing a detailed answer on this thread, without knowing that someone already did the same in the other thread. This behavior is very disrespectful to the people you're asking for assistance. Just because we give our time freely doesn't mean it has no value.

In the future, take some time to pick the forum section that best suits the topic of your question and then only post once to that forum section. This is basic forum etiquette, which you would already know if you had bothered reading the sticky "How to use this forum - please read." post you will find at the top of every forum section. It contains a lot of other useful information. Please read it.

I offer an apology for what happened, advertising cartography does not appear, and regarding the diagram
https://plus.google.com/u/0/photos/photo/114612218997856866388/6611636528582990546?authkey=CLnZnczVyaLsTg

Other threads removed.

Dulux’s picture:
Untitled Sketch_esquemático.jpg

Consider your LED problem may be because you turn them off so quickly after you turn them on.

Paul

Also consider adding a current limiting resistor in the LED circuit.......220ohms. :o

but pin 13 has an internal resistance

Dulux:
but pin 13 has an internal resistance

Pin 13 has never had an "internal resistance" that can be relied upon to limit current. None of the pins have.

At one time pin 13 had an external series resistor. That has not been true for more than a year.

Unless you can find an actual external series resistor connected to pin 13 you will need to provide a resistor. In any case, adding such a resistor will cause no harm. But, failing to include one when it is necessary will damage the processor.