12/28/2017 8:08 AM  
Joined: 6/29/2012 Last visit: 1/20/2022 Posts: 877 Rating: (7) 
In cams where slave position is function of master position, I should want to know how to calculate the acceleration of the slave. For velocity I know that 1 means velocity of the slave is the same as the velocity of the master. But I don't know how to interprete the acceleration. If for instance the master is moving at constant speed of 200 mm/sec, I should want to calculate from the cam curve what at a certain point the slave acceleration will be in mm/sec². I should need a formula for this, so I am able to calculate this in runtime in the controller. 
12/28/2017 9:34 AM  
Joined: 9/23/2005 Last visit: 1/21/2022 Posts: 3692 Rating: (578) 
x_{m}(t)  position of the master as a function of time (i.e. the master motion profile) x_{s}(x_{m})  position of the slave as a function of position of the master (i.e. a cam) The slave acceleration  second derivative of its position over time x_{s}''(t)=[x_{s}(x_{m}(t))]''=[x_{s}'(x_{m}(t))*x_{m}'(t)]'=x_{s}''(x_{m}(t))*(x_{m}'(t))^{2}+x_{s}'(x_{m}(t)*x_{m}''(t) However, if one wanted to calculate an instantaneous acceleration of the moving slave then it'd take few past positions to be collected and the derivative calculated numerically directly from the position values. 
Regards, 

1/9/2018 10:45 PM  
Joined: 6/29/2012 Last visit: 1/20/2022 Posts: 877 Rating: (7) 
I observed that at least for constant master speed, the actual acceleration of the slave was : 
1/10/2018 10:38 AM  
Joined: 9/23/2005 Last visit: 1/21/2022 Posts: 3692 Rating: (578) 
Yes. When (x_{m}(t))'=const then (x_{m}(t))''=0 and the second term disappears. 
Regards, 

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