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VL53L5CX/8CX LookUp-Table

Hartm_Sve
Associate II

Hi,

i know that there are already a few posts on this topic, and i know the VL53L8CX has the same table as the VL53L5CX. But i just wanted to know, if anyboy could post the calculations for the LookUp-Table?

Thanks!

1 ACCEPTED SOLUTION

Accepted Solutions
John E KVAM
ST Employee

I can't find the spreadsheet where we did the calculations, but it's basically just trigonometry.

There is a discussion of the math in Calculate x,y,z coordinates of each points of the ... - STMicroelectronics Community, but it's really hard to read. 

The light goes out in a pyramid of light 43 degrees on a side. If you are using an 8x8 resolution, then the width of each zone is 5.4 degrees. But we are considering that the distance you get is at the center of each zone. So 2.7 degrees to the right and 2.7 degrees up, would be the center of one of the zones nearest to the mid point. You also get 2.7 degrees left and 2.7 degrees down. 

So if you take a string pointing from the sensor 2.7 degrees right and 2.7 degrees up and at a distance given by the sensor you get a point in space. But that's not a very good coordinate system. 

Some people like polar coordinates - meaning an angle (think 1:30 on your clock) and a radius of some distance and then a height. 

But you can translate that point in space to an X, Y, Z coordinate system as well. But it takes a bit of thinking and book on trigonometry. 


If this or any post solves your issue, please mark them as 'Accept as Solution' It really helps. And if you notice anything wrong do not hesitate to 'Report Inappropriate Content'. Someone will review it.

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1 REPLY 1
John E KVAM
ST Employee

I can't find the spreadsheet where we did the calculations, but it's basically just trigonometry.

There is a discussion of the math in Calculate x,y,z coordinates of each points of the ... - STMicroelectronics Community, but it's really hard to read. 

The light goes out in a pyramid of light 43 degrees on a side. If you are using an 8x8 resolution, then the width of each zone is 5.4 degrees. But we are considering that the distance you get is at the center of each zone. So 2.7 degrees to the right and 2.7 degrees up, would be the center of one of the zones nearest to the mid point. You also get 2.7 degrees left and 2.7 degrees down. 

So if you take a string pointing from the sensor 2.7 degrees right and 2.7 degrees up and at a distance given by the sensor you get a point in space. But that's not a very good coordinate system. 

Some people like polar coordinates - meaning an angle (think 1:30 on your clock) and a radius of some distance and then a height. 

But you can translate that point in space to an X, Y, Z coordinate system as well. But it takes a bit of thinking and book on trigonometry. 


If this or any post solves your issue, please mark them as 'Accept as Solution' It really helps. And if you notice anything wrong do not hesitate to 'Report Inappropriate Content'. Someone will review it.