drbitboy
Lifetime Supporting Member
PID Trainer, issue on setup
- Thread starter geniusintraining
- Start date
MaxK
Member
agree
can't you explain?
Hey, let's make it interesting: fill the ping pong ball with water; now Navier and Stokes are involved!
You are fond of perverse, I see. )))
I'm in
This should solve the problem of the rod playing, but will make the PID trainer less scientific-like and, as a result, less attractive to potential buyers
drbitboy
Lifetime Supporting Member
drbitboy
Lifetime Supporting Member
drbitboy
Lifetime Supporting Member
Maybe we could think about this another way: sell it with the slop (backlash) built-in to the linkage, but also with the rubber band.
Teaching either how to tune a PID* or how to think about the mechanics is valuable; teaching how to integrate, and see the interaction between, both of those is far more valuable.
We can't count the number of times @Peter Nachtwey has been shaking his head in this forum, where designers blithely wrote (PID) on the P&ID without thinking about the process, leaving some poor schlub to commission it and they post in this forum for help.
* yes, I know we are tuning the system, but PID makes sense there for the point I am making.
Teaching either how to tune a PID* or how to think about the mechanics is valuable; teaching how to integrate, and see the interaction between, both of those is far more valuable.
We can't count the number of times @Peter Nachtwey has been shaking his head in this forum, where designers blithely wrote (PID) on the P&ID without thinking about the process, leaving some poor schlub to commission it and they post in this forum for help.
* yes, I know we are tuning the system, but PID makes sense there for the point I am making.
Peter Nachtwey
Member
I think what you guys are trying to get at is the inertia for a hollow sphere like a table tennis ball is greater than that of a solid sphere with the same mass.
hollow sphere inertia = (2/3)*m*r^2
solid sphere inertia = (2/5)*m*r^2
This is why a table tennis ball works better than a solid steel ball.
The mass cancels out but the ratio fractions don't.
drbitboy's toy has a lot of inertia because of the mass is distributed away from the center of rotation.
However, I don't know where those acceleration numbers come from. I would need to revisit my Mathcad worksheet.
hollow sphere inertia = (2/3)*m*r^2
solid sphere inertia = (2/5)*m*r^2
This is why a table tennis ball works better than a solid steel ball.
The mass cancels out but the ratio fractions don't.
drbitboy's toy has a lot of inertia because of the mass is distributed away from the center of rotation.
However, I don't know where those acceleration numbers come from. I would need to revisit my Mathcad worksheet.
drbitboy
Lifetime Supporting Member
The external force vector on the ball is its weight, mg (product of mass of the ball, m, and the gravitational acceleration constant, g), downward.
Any force normal to the beam is held by the beam and does not contribute to acceleration along (parallel to) the beam.
If the beam is level, then all of the force is normal to the beam and the ball does not accelerate along the beam.
If the beam is off level by some angle θ, then the force along the beam is the Weight scaled by sin(θ), so m g sin(θ).
That force is applied as two forces: i) a force that causes translational acceleration FLinear = m aLinear; ii) a rotational force i.e. torque that causes angular acceleration because we assume the ball's surface does not slip on the beam FRotational r = Torque = α Inertia = aLinear Inertia / r, where α is angular acceleration and r is the radius of the ball.
Since the forces must balance, m g sin(θ) = FLinear + FRotational = m aLinear + aLinear Inertia / r2
Inertia, as @Peter Nachtwey notes, is k m r2[/sup], where k is 2/3 or 2/5 for a spherical shell or solid sphere, respectively, so
g sin(θ) = aLinear + aLinear k = aLinear (1 + k)
so
aLinear = g sin(θ) / (1 + k)
getting close now:
aLinear = g sin(θ) / (1 + 2/3) = 3 g sin(θ) / 5 for spherical shells (ping pong balls)
and
aLinear = g sin(θ) / (1 + 2/5) = 5 g sin(θ) / 7 for solid balls
Finally, θ is small, so sin(θ) ≈ θ (θ in radians), and g ≈ 9.8 (SI), so
aLinear ≈ 5.9 θ, shell
and
aLinear ≈ 7 θ, solid
P.S I am ignoring the radial acceleration along the beam due to the angular velocity of the beam, ω, at the distance of the ball from the fulcrum, d:
d ω2
I am also ignoring the buoyancy force on the ping pong ball, which is about 1.5%, so somewhere between negligible and meh.
Last edited:
MaxK
Member
Yep. Or I it is my poor English I misunderstand “acceleration coefficient” or something wrong with my math – I don’t get your numbers
The picture seems to be something useless but funny (although fun is one of the few things that are really valuable in life)
If you're talking about geniusintraining's PID trainers, I guess they can't teach someone how to tune a PID controller, they can only "teach" a person to be sure that he can tune a PID controller (please understand correctly, I don't invest negative connotations in relation to geniusintraining - he is not the first and not the last who makes money from human stupidity, to a certain extent, and I even envy his quick wits)
MaxK
Member
I get it now.
For geniusintraining's plastic channel (90deg)
aLinear = 3 g sin(θ) / (3+2*2^0,5) for spherical shells (ping pong balls)
aLinear = 5 g sin(θ) / (5+2*2^0,5) for solid balls
Thanks
I already know what this idiom means :-///
Last edited:
Peter Nachtwey
Member
I get the same number as drbitboy, 5.887 for a hollow sphere if I combine all the numbers so 5.887 is just multiplied by the tilt in radians.
Also, since this is two pole system I calculate the Kp and Kd values for the tilt as Kp=6.709 rad/m and Kd = 2.136 rad/(m/s). This places two poles at -2*PI. I could calculate the an integrator gain too but that adds another pole to the closed loop transfer function. I didn't assume any friction. That would decrease the Kd value a bit. Since my simulator doesn't have any noise, I can increase the response but I think in reality that noise will limit to how high the gains can go.
Are we done yet?
Also, since this is two pole system I calculate the Kp and Kd values for the tilt as Kp=6.709 rad/m and Kd = 2.136 rad/(m/s). This places two poles at -2*PI. I could calculate the an integrator gain too but that adds another pole to the closed loop transfer function. I didn't assume any friction. That would decrease the Kd value a bit. Since my simulator doesn't have any noise, I can increase the response but I think in reality that noise will limit to how high the gains can go.
Are we done yet?
drbitboy
Lifetime Supporting Member
yes, (3/5 * 9.8) and (5/7 * 9.8) ... um, oh well ...
5.9 Theta, shell
7 Theta, solid.
fixed in the original, thanks for the quick review!7 Theta, solid.
MaxK
Member
We haven't even started yet.
The amount of information we know about the PID trainer = 0
If geniusintraining provides data, it becomes possible to do something.
5.887 is a factor for a flat surface and not for a 90 degree gutter, but it's not really significant.
But for the system to remain stable at Kp = 6.709 and Kd = 2.136, the rate of change of the ramp angle should be at least 0.5 rad / sec - I have vague doubts that the geniusintraining's PID-trainer provides such speed
drbitboy
Lifetime Supporting Member
MaxK
Member
Similar Topics
I have Real PLC but I dont have a real system to have PID Control- any recommendation how can I build such a system- there are some PID trainer...
I've built a little trainer with a Automation Direct DL05, some pilot lights, switches, a ethernet card and some of the new PX remote I\O. The PX...
Hi,
I would like to assemble a simulator/practice booster pump system that uses PID to maintain steady water pressure under various outlet demands...
Hello,
I have a motor that we are sending a RPM Speed Output from 0-100% to the VFD. However, the Motor HP needs to be limited to 6000 HP and the...
I have S7 1512C controler for controlling 48 PID temperature loop, the output is PWM.
Please I need the best, most efficient way to write the...