I'm looking to ease out pneumatics for electronics on part of a machine I'm working on. I'm trying to get in the ballpark with my numbers so I can talk to vendors about a replacement.
It's a transfer head that shuttles back and forth a lot - so far in three months 1.7 million times. The head moves on a two rod slide I intensely dislike. The side is subject to random binding that completely circumvents any attempts to bring the head to smooth stops at the end of travel. This results in nasty slamming.
Here's what it looks like:
Here are two flavors of the same movie in case you want to get a feel for the motion. If either work for you don't bother looking at the other - they're the same.
Transfer.mov 1.7M Quicktime movie
or
Transfer.wmv 2.4M Windows Media Player movie
With that in mind I did some numbers but I'm not comfortable with them since the final number I got seems about 2 orders of magnitude less than my gut feeling predicts. Anybody see an error in here somewhere?
DATA
------------------
3 sec per cycle (back and forth)
1.5 sec per direction
22" stroke
22"/1.5sec = 14.7in/sec AVERAGE SPEED
Preferred would be 11 inches of acceleration followed by 11 inches of deceleration.
Dynamic hardware weight is approximately 32 lbs.
Mass = 32 lbs/32ft/s^2
Mass = 1 slug
1 slug = 15kg
I need to meet the cycle time of 3 seconds. That's out and back.
But we can focus on just one direction. One direction is going to take 1/2 the time or 1.5 seconds.
This boils down to accelerating half way across then decelerating the second half to a stop.
So half of a stroke is 1.5 sec / 2 = 0.75sec.
We need to make the half stroke 22inches/2 = 11 inches
To make the schedule I need to get 11" in 0.75 secs.
Lets go metric:
11in x 0.0254 = 0.28m
Bring in the kinematics:
d = distance
Vi = initial velocity
a = acceleration
t = time
d = Vi + 1/2at^2
But if the initial velocity is 0
Vi = 0 we have:
d = 1/2at^2
a = 2 x d / t^2
a = (2 x 0.28m) / (0.75s)^2
a = 0.996m/s^2
Call it 1m/s^2
F = ma
m = 15kg
a = 1m/s^2
F = 15Newtons
Work = F x d
F = 15N
d = 0.28m
Power = Work/t
t = 0.75s
P = (15N x 0.28m) / 0.75s
P = 5.6W
This number is giving me pause. A fraction of the energy used in a night-light can't be enough to accelerate 32 lbs up to a meter per second in 3/4 of a second can it?!? This thing is run by a 5hp(3700W) air compressor that runs at a high duty cycle. The lion's share of air is used by this cylinder.
Maximum speed at center of stroke (end of acceleration):
Vf = final velocity.
Vf = Vi + at
Vf = 0 + 1m/s^2 x t
Vf = 1m/s^2 x 0.75
Vf = 0.75m/s 3ft/sec Seems reasonable.
Keith Cress
kcress -
It's a transfer head that shuttles back and forth a lot - so far in three months 1.7 million times. The head moves on a two rod slide I intensely dislike. The side is subject to random binding that completely circumvents any attempts to bring the head to smooth stops at the end of travel. This results in nasty slamming.
Here's what it looks like:
Here are two flavors of the same movie in case you want to get a feel for the motion. If either work for you don't bother looking at the other - they're the same.
Transfer.mov 1.7M Quicktime movie
or
Transfer.wmv 2.4M Windows Media Player movie
With that in mind I did some numbers but I'm not comfortable with them since the final number I got seems about 2 orders of magnitude less than my gut feeling predicts. Anybody see an error in here somewhere?
DATA
------------------
3 sec per cycle (back and forth)
1.5 sec per direction
22" stroke
22"/1.5sec = 14.7in/sec AVERAGE SPEED
Preferred would be 11 inches of acceleration followed by 11 inches of deceleration.
Dynamic hardware weight is approximately 32 lbs.
Mass = 32 lbs/32ft/s^2
Mass = 1 slug
1 slug = 15kg
I need to meet the cycle time of 3 seconds. That's out and back.
But we can focus on just one direction. One direction is going to take 1/2 the time or 1.5 seconds.
This boils down to accelerating half way across then decelerating the second half to a stop.
So half of a stroke is 1.5 sec / 2 = 0.75sec.
We need to make the half stroke 22inches/2 = 11 inches
To make the schedule I need to get 11" in 0.75 secs.
Lets go metric:
11in x 0.0254 = 0.28m
Bring in the kinematics:
d = distance
Vi = initial velocity
a = acceleration
t = time
d = Vi + 1/2at^2
But if the initial velocity is 0
Vi = 0 we have:
d = 1/2at^2
a = 2 x d / t^2
a = (2 x 0.28m) / (0.75s)^2
a = 0.996m/s^2
Call it 1m/s^2
F = ma
m = 15kg
a = 1m/s^2
F = 15Newtons
Work = F x d
F = 15N
d = 0.28m
Power = Work/t
t = 0.75s
P = (15N x 0.28m) / 0.75s
P = 5.6W
This number is giving me pause. A fraction of the energy used in a night-light can't be enough to accelerate 32 lbs up to a meter per second in 3/4 of a second can it?!? This thing is run by a 5hp(3700W) air compressor that runs at a high duty cycle. The lion's share of air is used by this cylinder.
Maximum speed at center of stroke (end of acceleration):
Vf = final velocity.
Vf = Vi + at
Vf = 0 + 1m/s^2 x t
Vf = 1m/s^2 x 0.75
Vf = 0.75m/s 3ft/sec Seems reasonable.
Keith Cress
kcress -