Sunday, November 29, 2009

Timing Diagram (Part 4 - Timing Diagrams Comparison using Motion Simulation in Microsoft Excel)

In post [Timing Diagram (Part 1 - No Overlap Movement)], we saw the design requirement that we have to design the die to work together with the indexing mill with the construction as shown below.
Without detailed calculation, we could end up with a very simple timing diagram as shown below.
But it's not good enough. The die has to wait for the indexing to finish its movement before moving. This reduces the indexing time of the die, and get high acceleration on the die.

In post [Timing Diagram (Part 2 - Maximum acceleration calculation)], we calculated the maximum acceleration of Cycloidal motion cam profile and saw the opportunity to reduce the acceleration by extending the indexing time.

In post [Timing Diagram (Part 3 - Cycloid Cam Profile Analysis)], we analyzed the cycloid cam profile and see opportunity of overlap motion. Some calculations have been made and we end up with new timing diagram which is smarter than the original one as shown below.

We calculated the maximum acceleration of the die for this new timing diagram, and we found it was 5 times lower than the acceleration in the first timing diagram.

After all calculations, it's time to see the calculation results. The easiest way is to compare the movement at the same time. The idea is to make simple motion simulation without any timing diagram software or complicated simulation software. I'm using Microsoft Excel again, because it's the main theme of this blog.

We start with plotting the shape of indexing mill and die and add cells called "driver" to change the positions of the indexing mill and the die. Then write VBA code to change the values in "driver" cells to move the parts. Then we can see the movement.

From this example, I don't mean that this is the best timing diagram that give lowest acceleration. We could make it smoother with lower acceleration using "Fifth degree polynomial" together with "Linear" cam profile. But this is enough to show how we can play with timing diagram in the design phase to reduce unnecessary waiting time.

See the result below.



FREE DOWNLOAD EXCEL FILE FROM THIS EXAMPLE

Extract the zip file with password: mechanical-design-handbook.blogspot.com

Tuesday, November 24, 2009

Timing Diagram (Part 3 - Cycloid Cam Profile Analysis)

In previous post [Timing Diagram (Part 2 - Maximum acceleration calculation)], we calculated the maximum acceleration of the die using cycloid cam profile. We found out that this maximum acceleration of the die can be reduced if we can extend the indexing angle (Bm) or indexing time (tm) through overlap motion.

Then, let's see how we can calculate for the suitable indexing angle to reduce the acceleration of the die.

Cycloidal motion cam profile has movement equation as follows.
h = hm x [t/tm - 1/(2 x pi) x sin(2 x pi x t/tm)]

Rearrange the equation to get
h/hm = t/tm - 1/(2 x pi) x sin(2 x pi x t/tm)

The displacement profile can be plotted as shown below (dimensionless).
We can see that at the first 10% of indexing time, the movement is just only 0.65% of the total movement (stroke) and at 90% of time, the remaining movement for the die is only 0.65% of the total movement (stroke). Or we can say that, there is not much movement at the first and last 10% of indexing time.

For this example, the indexing mill has an indexing angle of 150 deg and stroke of 100 mm.
That means at 10% of indexing angle (time) = 10% x 150 = 15 deg, the indexing mill moves only 0.65% x 100 = 0.65 mm. (Let's say we use the margin of 0.65 mm)

The effective angle (time) of the indexing mill = 150 - 2x15 = 120 deg.
The remain cam angle = 360 - 120 = 240 deg.
Since the die has to stay at the bottom for 100 deg to do the job, the remaining cam angle becomes 240 - 100 = 140 deg.

If we divide the movement of the die equally, the angle for moving up and down becomes 140/2 = 70 deg. This is when the die starts moving down from 205o - 70o = 135 deg to 205 deg with the movement of 50 mm. But this is not good enough. We can do more overlap!!

If we reconsider the timing diagram again, at first the die waits for the indexing mill to complete its movement. But we can see that the die does NOT have to wait for the indexing mill to complete the movement. If we take the margin of 1 mm between the die and the indexing mill, the die can move already = 20 - 1 = 19 mm when the indexing mill has the remaining movement of 0.65 mm (at 135 deg).

Thursday, November 19, 2009

Timing Diagram (Part 2 - Maximum acceleration calculation)

Let's calculate the acceleration of the die from previous post [Timing Diagram (Part 1 - No Overlap Movement)]

The die moves using Cycloid cam profile. So first we have to get the formula to calculate the maximum acceleration of cycloid cam profile.

If the machine speed is N (pcs/h) and the indexing angle (degree) is Bm, the indexing time (second) tm can be calculated as follows.

Cycle time (sec) = 3600/N
Indexing time tm (sec) = (Bm/360) x Cycle time = (Bm/360) x (3600/N)

Hence,
Indexing time tm (sec) = 10Bm/N

Cycloid cam profile has the equation of displacement as follows.

h = hm x [t/tm - 1/(2 x pi) x sin(2 x pi x t/tm)]

where:
hm: Maximum displacement (m)
tm: Indexing time (s)
pi: 3.141592654

We can get velocity equation by differentiation.

v = dh/dt = hm x [1/tm - (2 x pi)/(2 x pi x tm) x cos(2 x pi x t/tm)]
v = hm/tm x [1 - cos(2 x pi x t/tm)]

Then, the acceleration is as follows.
a = d2h/dt2 = dv/dt = hm/tm x [0 - (-2 x pi/tm) x sin(2 x pi x t/tm)]
a = 2 x pi x hm/tm2 x sin(2 x pi x t/tm)

The maximum acceleration (amplitude) occurs when sin(2 x pi x t/tm) = 1 or -1
Therefore the amplitude of of maximum acceleration is as follows.

amax = 2 x pi x hm/tm2

We can clearly see from the above derivations that the acceleration is inversely proportional to the square of indexing time. Since the indexing time (tm) is proportional to the indexing angle (Bm), then the maximum acceleration is also inversely proportional to the square of the indexing angle.

That means if we can increase the indexing angle by a factor of two, the maximum acceleration will reduce by a factor of four!!

And we can do this by putting more overlapped motion in the timing diagram design.

From previous post, the die moves 50 mm in 55 degrees using cycloid cam profile.
Let's assume we calculate the maximum acceleration of the die at the machine speed (N) of 2000 pcs/h
So, we have

N = 2000 pcs/h
Bm = 55 deg
hm = 50 mm = 0.05 m

tm = 10Bm/N = 10(55)/2000 = 0.275 s
amax = 2 x pi x hm/tm2 = 2 x 3.141592654 x 0.05 / 0.2752
amax = 4.154 m/s2

Let see in the next post when we put more overlap motion, the indexing angle will be increased as well as the indexing time, thus reduce the maximum acceleration.

Timing Diagram (Part 1 - No Overlap Movement)

When I search in Google for "timing diagram", I found a lot of results about electrical timing diagram software but they're not about what I'm going to tell. Timing Diagram in my meaning is a tool that represents the sequences of movement of mechanisms. It is a very useful diagram for mechanical design engineers to understand how each part of the machine works together.
" By properly design the timing diagram, we can make machine moves smoother even at higher speed. "
We often draw the timing diagram using cam angle (in degree) in horizontal axis and use the movement of mechanism (in mm) in vertical axis.

From the timing diagram, we can find the opportunity to reduce the acceleration (force) of the moving parts so as to reduce the wear in machine.

Experience shows that a lot of mechanisms have been designed without using "overlap" movement. This makes the machine parts move from one point to another point in short period. But if we provide the overlap motion between relevant mechanisms, the machine parts can travel between the same distance, but in longer period. This reduces the acceleration of the parts, that means lower forces exerted on the parts and result in less wear on the parts.

Let's have a look at the example of simple machine that press the die into the hole of the indexing mill.


Operation overview:
The indexing mill has 24 stops and it moves 100 mm during each index. It uses the indexing cam with Cycloid Cam Profile having indexing angle of 150 degrees. After the mill stops, the die then moves down using Cycloid cam profile until it reaches the bottom of the holes with 1 mm gap (traveling distance = 20+31-1 = 50 mm). And the die has to stay at the bottom for 100 degrees, then it moves up before the indexing mill rotate again in the next cycle.

If we make a timing diagram without any overlap, we will come up as follows.

With no overlap movement, the die has to wait until the indexing mill (turret) has finished indexing then it will move down to the bottom of the mill and stay there for 100 degrees. After that, it moves up. Then the indexing mill start indexing again in the next cycle.

The remaining angle for the die movement can be calculated using the following equation:

Cam Angle for die movement = 360 - mill indexing angle - angle for die at the bottom

So, cam angle for die movement = 360 - 150 - 100 = 110 degrees
We have to split the die movement for up and down equally. Hence, the cam angle for die movement up or down = 110/2 = 55 degrees.

The timing diagram is then constructed by using 150 degrees in the timing diagram as a starting point of die moving down. So the die fully moves down at 150+55 = 205 degrees. Then it has to stay at the bottom for 100 degrees, that means the die starts moving up again at 205+100 = 305 degrees. The die finishes move up at 305+55 = 360 degrees. Then the next cycle starts again.

This seems very easy to calculate. But is it good enough?

Let's calculate the maximum acceleration of die when moving up and down in the next post [Timing Diagram (Part 2 - Maximum acceleration calculation)]. Later we can compare this result with the acceleration after making some overlap movement between indexing mill and die. At the end we will write a simple Microsoft Excel program to simulate the motion of indexing mill and die for both cases at the same time.

Tuesday, November 17, 2009

Standards of limits and fits for mating parts (Part 2)

In the previous post (Standards of limits and fits for mating parts), we talked about the definitions of each term related to limits and fits as well as the formulas to determine the values of tolerances. In this post, we're going to convert those information into the real calculation using Microsoft Excel (as usual). As stated earlier, the calculation results may be different from the real values used in general mechanical design handbook. So please use this just for educational purpose only, but use the real table from general limits and fits table if you want to get higher accuracy values.

This is the screen shot of excel file to calculate upper deviation and lower deviation according to the selected shaft diameter and tolerance grade.


Let's see how to manually calculate the deviation values before using the program.

Example: To calculate the upper deviation and lower deviation of a shaft with diameter of 40 mm and tolerance g6.

Please refer to previous post (Standards of limits and fits for mating parts) for more details.

Shaft diameter = 40 mm has Dmin = 30 mm and Dmax = 50 mm as shown in the following table.



The geometric mean of the size range (D) = SQRT(30 x 50) = 38.73 mm
Then "i" can be calculated using the following formula.


i = 0.001 x [ 0.45 x (38.73)^(1/3) + 0.001 x 38.73 ] = 0.00156
Tolerance "g6" has grade = 6
From the table (in previous post), the formula of IT grade 6 is 10i = 10 x 0.00156 = 0.0156
For shafts designated a through h, the upper deviation is equal to the fundamental deviation. Subtract the IT grade from the fundamental deviation to get the lower deviation.

Tolerance "g6" has a = 0; b = -2.5 and g = 0.34
From Fundamental deviation = a + (bDg)/1000
We have Fundamental deviation = 0 + (-2.5 x 38.730.34)/1000 = -0.009


Upper deviation = fundamental deviation = -0.009
Lower deviation = fundamental deviation - IT grade = -0.009 - 0.0156 = -0.025

To use the program calculate the upper deviation and lower deviation of a shaft with diameter of 40g6, do the followings.

1) Enter 40 in cell D5 and the program automatically highlights the row that has Dmin <= d < Dmax as shown below.



2) Select "g" from the drop-down list in cell D6 and "6" from the drop-down list in cell E6. Then the result is displayed as shown below.

password: mechanical-design-handbook.blogspot.com
FREE DOWNLOAD EXCEL FILE TO CALCULATE TOLERANCE VALUES