LabVIEW
"relative" real time
@kramerfrog wrote:
If I am sampling at 10000 HZ and taking 3000 samples then in 0.3 seconds of data I could have 2 and sometimes 3, 5-volt spikes. I am able to pull the RPM out of this already from my program.
I would like to calculate future position, i.e. the next 2 spikes based on the last two spikes RPM and position. I would like I think some sort of counter that either counts how many milliseconds it takes to get from one peak to the next. If I have that then I believe I can predict the future position of the shaft pretty accurately.
Just count the amount of samples and multiply by 100µs (1/10kHz).
But that is complete nonsens!!
Why try to calculate the future position??
To do something at two different angular positions like you suggested in the other post??
You said it yourself, in one acquisition you get 2 or 3 spikes, read 2 or 3 rotations of the shaft, and now you ask for a way to do something in between two spikes, read during 1 rotation?
Those positions you look for are long gone by the end of the acquisition!!
You need a triggerd acquisition system so that your software can act the moment one pulse comes in.
Even then it will be hard to calculate/estimate the position of the shaft since both systems, hardware = the shaft vs. software = your code, has complete different, non synchronized timings.
In this or the other thread (can't remember) you talked about a windtunnel (hugh, expensive stuff bla bla bla), is there really no way two add one more switch so that your software can be triggered by those switches.
You also never answered my question about digital acquisition instead of analog 
@kramerfrog wrote:
i tried counting number of samples and then multiplying by 1/10khz but it wouldn't count.
Consider the array of 3000 samples.
First rising edge is at location 683.
Second rising edge is at location 2475
So time between the two edges is (2475 - 683) * 100µs = 0.1792 sec = 5.5807 rpm
That wasn't so hard, right?
@kramerfrog wrote:
i'm new to labview and data acquisition (as you can tell) so i don't really understand what i benefit by taking an analog signal and converting it through software to a digital signal.
You don't have to convert anything!
The signal from the switch is or 0V or 5V right?
Then that can be seen by a digital I/O pin as "0" and "1" respectively.
So you can use that signal as a digital input signal.
If by "swith" you really mean a switch, then you should use a pull up or pull down resistor to ensure both levels.
I mean, you can't leave the digital I/O pin "free". It has to be or 0V or 5V never floating.
Check with the MAX if the hardware detect the levels correctly.
Once you got that right start with the triggered digital acquisition.
Just take 1 sample with a sample rate of e.g. 10kHz
The hardware will check the rising edge of the switch for you at high speed.
The software will do nothing while waiting for a trigger from the hardware.
Don't forget that at 1000rpm you have only 60msec in between 2 pulses.
So if your code takes more than 60msec to execute things will go wrong 
@kramerfrog wrote:
At a certain shaft degree say 20 degrees, a trigger needs to go off which will start a VI, and at say 30 degrees a trigger will go off to end a VI. This is why I wanted to find the position of the shaft.
Those 20° and 30° are, I suppose, after the trigger from the switch, right?
Suppose speed is 1000rpm, then 360° = 60msec, 1° = 0.16666msec
Trigger switch is time 0, 20° is at t= 3.3333msec and 30° is at t= 5 msec
You want to run a VI for only 1.666 msec 
What is that VI doing for 1.666msec?
This is definetly not a task for a PC!
This is a task for a stand alone µC like this one for example.
