Interpret a graphic

[DaveO]

I've been reading the Miller TIG manual, and have a question about a graphic. This clip from the manual discusses the AC cycle, and describes the change from electrode positive to electrode negative that occurs while welding.

All well and good, but the graphic labeled Figure 2.14 shows a horizontal arrow pointing right in the positive half of the cycle, and a horizontal arrow pointing left in the negative half. That pointing-left arrow is bugging me: it seems to contradict the implied time continuum of the X axis.

What do those arrows indicate?

[everlastsupport]

I've been reading the Miller TIG manual, and have a question about a graphic. This clip from the manual discusses the AC cycle, and describes the change from electrode positive to electrode negative that occurs while welding.

All well and good, but the graphic labeled Figure 2.14 shows a horizontal arrow pointing right in the positive half of the cycle, and a horizontal arrow pointing left in the negative half. That pointing-left arrow is bugging me: it seems to contradict the implied time continuum of the X axis.

What do those arrows indicate?

I think with the arrows they are trying to show the current flow direction during that phase of the sine wave.

[DaveO]

I thought the sine wave was showing the current flow. "The current is now alternating or changing its direction of flow at a predetermined set frequency and with no control over time or independent amplitude." To me the graphic makes sense without those arrows.

Wait- you're saying the arrows are there just to make the point "current flowing one way, then flowing the opposite way"?

[Rambozo]

The arrows are rather strange and really do not belong on a graphic like that. The sinewave depicts voltage first rising from zero to positive over a period of time, then falling to zero and going below zero to negative and back to zero. It is referenced to this zero point. The current is a constant value if the machine is set that way (CC) but the power follows this curve because power in watts is the product of amperage and voltage. (P=I*E) The actual flow of electrons is from the negative to the positive. They are correct in saying that in a sinewave the arc goes out during the zero crossings, which is why continuous high frequency is required for low amperage sinewave AC welding.
I have been curious as to how these latest inverters that offer sinewave, handle that. I suspect that the sinewave is probably adjusted a little so that there is not enough time during the zero crossings for the arc to extinguish. Being able to weld AC without continuous HF was one of the great advantages of squarewave machines.

[performance]

What you guys have to remember, is that welders aren't electrical engineers. Teaching laymen about things like this, you must simplify, and dumb down theory. Using extra graphics to show the relationship helps customers to understand the relationship. When people talk about AC polarity, being reversed every half cycle, and flowing in the opposite direction, you have to find someway to represent it in a less technical manner. What people with full understanding don't understand, is this information wasn't written for them to criticize. It was written to give the customer a better idea of what is happening, be it completely right or not. Perfectionist need not try to overthink this. A visual understanding that represents a change in flow is 100 times more effectivewith arrows than some sine wave plotted upon multiple points on a grid that represents the increasing and decreasing voltage at the same time representing the same change in polarity. Since a reversal is typically thought of in a linear manner the linear drawn arrows within the half wave represent to the average person the actual change in the directional flow of electriciy.

[performance]

Inverters pass through the "0" point much more rapidly than transformers. Square waves are either on or off with little drop basically between + and - voltage. This transition happens so quickly that the arc really isn't perceived as going out. Sine waves have "flat" spot on the ends of the wave that transitions directly to the opposite equivalent voltage.

[DaveO]

Thanks for responses. I sure wasn't expecting a "less technical manner" of explanation mixed in with the technical explanation: I can't tell you how long I've stared at that thing, trying to figure out how the lower half of that sine wave was moving from 360* to 180*.

[Rambozo]

Inverters pass through the "0" point much more rapidly than transformers. Square waves are either on or off with little drop basically between + and - voltage. This transition happens so quickly that the arc really isn't perceived as going out. Sine waves have "flat" spot on the ends of the wave that transitions directly to the opposite equivalent voltage.

Thanks, I figured the inverters didn't put out a pure sinewave, and just squared up the transitions a bit, to avoid that issue, but I haven't put one on a scope to check.

[Rivets]

Thanks for responses. I sure wasn't expecting a "less technical manner" of explanation mixed in with the technical explanation: I can't tell you how long I've stared at that thing, trying to figure out how the lower half of that sine wave was moving from 360* to 180*.

Hi DaveO,

Just pretend that the x-axis is one of the wires to the torch.

Current goes right during the + half of the cycle then left during the - half. Just a visual thing.

rivets

[DaveO]

Thanks- I finally get that those arrows are an add-on to assist understanding. The graphic could have been expanded just a little, maybe to show a torch with a + sign and a torch with a - sign.

Miller pulled this TIG manual from their website. I read somewhere that it had been replaced with a paid download version, but I don't see it on their site.