Tip of the Day 175: Full Integrals vs. Fill and Pack Integral

In Tip #174 you saw how Fill and Pack Integral could be useful for monitoring dimensions in very thin-wall, fast freezing parts. In that discussion we used End of Cavity as the indicator, assuming that the dimension near the end of the cavity was most important.

A couple of readers caught the idea that perhaps End of Cavity is not necessarily the whole story. This is quite true. Some thin-wall parts freeze almost all at once; for example cell phone covers. Others may have some flowable material near the gate well after the time that the material freezes at the end of the cavity. In such cases Cycle Integral or Injection Integral^† near the gate would be more appropriate to find dimensional variation in that area of the part.

^† In the graph above, we chose “Injection Integral / Post Gate #2” instead of “Cycle Integral” to detect a dimensional quality problem in that area. If the whole part is frozen at the end of hold, then the pressure rise due to mold deflection after hold adds non-useful variation to the data we want for dimensions. Using “Injection Integral” ignores mold deflection variation after the end of hold.

Nevertheless, in all cases you should check the correlation to dimensions of both “Injection Integral” and “Cycle Integral”. Sometimes one correlates to dimensions better and sometimes the other.

Tip of the Day 150: The Mystery of the “Integral” (part 3)

In tip #148 we explained that the integral is an area or “area under the curve” as we say in class. At a recent training event Rajen was trying to answer a couple of tough remaining questions.

1. “Why are the units shown as psi-seconds?” (or bar-seconds)

Just as we multiply length by width we also multiply the units: yards x yards. Though this is normally abbreviated “square yards” but can also be written “yards-yards”.


Now a cavity pressure curve is not bounded by lengths but by pressure and time thus:

So when we multiply our little rectangles to compute the area the rectangles have a height in pressure in “psi” (or bar) and width in time in “seconds”. Thus, just as you multiply yards x yards and state it “yards–yards” you do the same here, multiplying the pressure units times time units, psi x seconds and stating it “psi-seconds”.


Note that this is not a division as would be the case in a slope of so many psi per second or “psi / second”. The dash indicates multiplication.
2. Why are the numbers so large?

^† The “2 seconds” is absurdly long for accuracy. We use it here just for easy illustration of the arithmetic. The eDART actually uses slices in the range of 1/500th of a second (0.002).

Tip of the Day 149: The Mystery of the “Integral” (part 2)

With an understanding of what an integral IS, the popular next question is “How do you compute it? I know how to figure the area of a rectangle or triangle but what is formula for the area of this:  ?”

Answer: Let the eDART™ calculate it.

Those with an itch to know more, read on.

To find the area of a shape that has no formula we simply break it into small areas that do have formulas: in this case a series of thin rectangles. Back to the property metaphor: Walk the road and mark every 100 yards with your GPS. Then walk the river and read out distances to the road at each GPS longitude line, breaking the property into 11 parts like this:

Each slice is averaged into a rectangle^† shown here in blue (just one height given as an example):

Then we can calculate the area of each rectangle as width (100 yards) x height. We add those 11 areas and come up with something close to the area of under the curve.



You will, of course, say “That’s pretty sloppy!” Quite so. So why not use 10 yard steps giving us 110 rectangles? Still not accurate enough? One yard, then? ^‡

The eDART only makes rectangles as thin as its sample rate in time steps. “Time” is the road in the property example. This is quite adequate for creating stable values that correlate to dimensions and weights.

Next Tip (# 150 ): Why do integrals have weird units (psi-sec.) and huge numbers (50,000 +)?

^†Making Rectangles from Trapezoids

Standard numerical integration methods get the best accuracy for the selected step size by averaging the two sides of each rectangle to come up with one height like this (using the same slice as in blue above):

The average of 292 and 220 is 256 which comes from (292 + 220) ÷ 2

This is the same thing as calculating the area of each rectangle under the curve and adding the area of each little triangle on top thus (the triangles are hatched for emphasis):

The first method gives good results using the lowest computing time. With a little algebra the computing time can be reduced even further.

^‡ Isaac Newton got the idea to make the rectangles infinitely thin, thus inventing The Calculus (“The” was required by my physics prof.).

Tip of the Day 148: The Mystery of the “Integral” (part 1)

We regularly struggle to explain what an “integral” is and where it comes from. Integrals are numbers calculated by the eDART that are particularly useful for detecting changes in part weight or dimensions.

Generally we say an integral is the “area under the curve,” thinking that should settle it. But it still leaves some people confused. Many people can visualize area for a rectangle as length x width. But what if you are thrown a curve?

Here is a little exercise that may help with this area concept. Let us start with a section of land bounded by a road and some survey lines that looks like the outline here.

It is easy to calculate the area of this land: 440 yards x 1,100 yards = 484,000 square yards (yards x yards = “square yards”). There are 4,840 square yards in an acre so this is exactly 100 acres of land (484,000 ÷ 4,840).

But suppose you only own from the road to the river. This bend in the river looks suspiciously like a cavity pressure curve.

You really don’t want to pay property taxes on the hundred acre rectangle when you only own to the river. The actual area of this property is the “area under the curve”; that is, inside that boundary of the road and the river. The eDART’s™ lower boundary (the road) is zero pressure (bottom of the cycle graph) and the upper is the pressure curve.

So what is the area under that curve or inside that property boundary? It turns out to be about 45 acres.

Next tip ( # 149 ): How do we compute the area inside that curve?

Tip of the Day 99: A Graphical Lesson in Integrals and Process Times

A customer pointed out yesterday that there is confusion about where integrals come from. He suggested having the Cycle Graph on the eDART™ automatically draw shading to show the source of the various integrals and times. Since this would be a rather involved software project with a lot of coding and testing I thought to build something that runs off-line in a simple PowerPoint presentation.

I created a PowerPoint file

    A Graphical Lesson in Integrals and Process Times

that you can save by right-clicking and selecting “Save link” (or “target”).

This presentation contains an example of some data with integrals and Process Time type values. Click on each of the summary graph buttons (when the mouse shows a hand) to see a graphical explanation of how integrals are areas under various parts of each curve. The colors are coordinated. Note that the buttons on the Summary Graph show the summary Type / Location pairs as discussed in tip # 83. The abbreviation is in parenthesis.

Note that the cycle integral for cavity pressure is shown to the end of Mold Clamped. This is the eDART default. It can be changed (as noted) to end the integral at the end of Screw Run if a flipper chute needs time to move or the robot needs an early signal. See tip # 61 for details on integration limits. The Sequence Settings “Other” tab has changed slightly since that tip was written.

Use this tool as a quick interactive training lesson when you need to explain what an integral is and how it is derived.

Comments and feedback are welcome.

art

See our Art’s Tips too find earlier Tips of the Day.

Tip of the Day 67: Injection Integral Error due to Incorrect Machine Sequence Signal

Tip of the Day 66: Injection Integral Depends on Injection Time

Tip of the Day 61: About Integrals and Integration Limits