IR not working

Development Boards Arduino
1

So I made an equation to get a very close match to the actual GP2Y0A02YK0F sharp ir sensor voltage curve. It is:

y=(0.011x-sqrt(2.25)-0.16)^2+0.45

where y is volts and x is distance. Trying to make it into x= form, I got this:

x=(1000*sqrt(y+2.3056))/11

So then i make an Arduino script that looks like this:

#include <SoftwareSerial.h>

#include <math.h>

void setup(){

int IRpin=0;

Serial.begin(9600);

Serial.println(“Hello world!”);

delay(1500);

Serial.println(“This is a test.”);

delay(3000);

}

void loop(){

float volts = analogRead(0)/694;

float distance_a = (1000*(sqrt(volts+2.3056)))/11;

float distance_b = (1000*(sqrt(volts+2.3056)))/11;

float distance_c = (1000*(sqrt(volts+2.3056)))/11;

float distance_d = (1000*(sqrt(volts+2.3056)))/11;

float distance_e = (1000*(sqrt(volts+2.3056)))/11;

float average_distance = (distance_a+distance_b+distance_c+distance_d+distance_e)/5;

Serial.println(average_distance);

Serial.println(volts);

delay(1000);

}

And it gives me junk data. The 694 in

“analogRead(0)/694” in line 14 is based on peak voltage outputs. Any ideas?

2

Does analogRead(0) return a float or an unsigned int?

694 is a constant.

These should be cast to a float before assigning to volt.

Try:

float volts = (float)analogRead(0)/694.0;

and see if volt is a correct value now.

3

I tried reading the volts before and after converted into distance. The before is right, the after is wrong. There is something wrong with the conversion equation i came up with.

4

One way to debug math is do one step at a time and output the intermittent calculations.

So instead of:

float distance_b = (1000*(sqrt(volts+2.3056)))/11;

Do:

float temp1 = volts+2.3056;

float temp2 = sqrt(temp1);

float temp3 = 1000*temp2;

float temp4 = temp3/11;

Serial.println(volts);

Serial.println(temp1);

Serial.println(temp2);

Serial.println(temp3);

Serial.println(temp4);

This way you can tell which of the math operations didn’t work.

5

ah… Okay, I’ll try that. But could i possibly do this?

(0.011distance-sqrt(2.25)-0.16)^2+0.45=volts

And get distance?

You can try getting it so that it’s distance = (equation here), but i tried that.

6

Great, now i’m not even getting any analog input. AT ALL. >:(

7

Selenaut:
So I made an equation to get a very close match to the actual GP2Y0A02YK0F sharp ir sensor voltage curve. It is:

y=(0.011x-sqrt(2.25)-0.16)^2+0.45

where y is volts and x is distance. Trying to make it into x= form, I got this:

x=(1000*sqrt(y+2.3056))/11

For reasons known to only the malevolent gods, the detailed response I just wrote went to the Great Bit Bucket in the Sky.

In brief, you appear to have made one or more algebraic errors in inverting the equation. I tried substituting your second equation into your first, but didn’t get an identity.

When I try to invert the first equation, I get:

y=(0.011x-sqrt(2.25)-0.16)^2+0.45

y=(0.011x-1.66)^2+0.45

y-0.45 = (0.011x-sqrt(2.25)-0.16)^2

sqrt(y-0.45)= (0.011x-1.66)

sqrt(y-0.45) + 1.66= 0.011x

(sqrt(y-0.45) + 1.66)/0.011 = x

which appears to work if you substitute it into the original equation.

Eric

8

esklar81:
(sqrt(y-0.45) + 1.66)/0.011 = x

I checked it by hand and then ran it through a CAS (since my algebra-fu is underutilized) and got the same result.

-Bill

9

Okay, I’ll try that later when i have time, now that I got the analog working again. Thanks for the correction.

10

Is the conversion equation based on the Sharp’s curve? If so may also want to try this one:

x = 194.0/((7.14 + (y - 4.69)y)(0.18 + (y - 0.28)*y))

x: distance

y: voltage [0.45…2.75] ->x~ [150…15]

Which comes out from a least square fit of an expression of the curve in:

http://sharp-world.com/products/device/ … 02yk_e.pdf

This fit has a deviation (absolute error) of less than 2% for most regions, except for x=150cm (y=0.45) where the error is at about 5%. The mean absolute error is about 1.3%

Furthermore if you want accuracy it is a good idea to create your own curves as they may be different from Sharp’s.

One last thing: what do you mean “peak voltage” outputs? Isn’t your output similar to that appearing on Sharp’s datasheet? If so the voltage should be in the region: 0.45…2.75.