Understanding Plunge Rate

[Kabbie]

I'm trying to understand Plunge Rate entered in the tool database. The manual says Plunge Rate is the "rate at which the cutter is moved vertically into the material or during ramping moves"... simple enough. Let's say I used 20 inches/minute. But when adding a smooth ramp to a toolpath, I have to specify the ramp distance, say 1". If I've specified a feed rate of 100 inches/minute, then wouldn't the 1" ramping occur in 0.01 minute? The plunge rate would then be determined by the depth of the plunge rather than the plunge rate entered for the tool. If I was plunging 1/2" during the 0.01 minute ramp, wouldn't it be plunging 50 inch/minute and a 1/4" plunge would be 25 inch/minute regardless of the 20 inches/minute plunge rate assigned to the tool?

[martin54]

No the feedrate won't exceed the plunge rate during the ramp so if you plunge at 20ipm for 1" & the DOC is set to 1/8" then the machine will travel the 1" at 20 ipm & then increase in speed to 100 ipm providing there is enough time to do that before the machine slows for a change of direction or another ramp

Just to add that not being an expert that might not be true for all control software but those that I have used the machine speed, feedrate is set by the slowest axis which in the case of a plunge move would be the z axis

[IslaWW]

Martin is correct... not for all controls.

Some controllers will operate as per your question. Lets say the material is 1" thick and you are cutting in one pass. At 100ipm, the 1" ramp, as you say will cut in .01 minute or .6 seconds. At 20 ipm there would be up to .2" of depth change in .6 seconds. In all cases the slowest axis will determine the time. This means it will take 5*.6 or 3 seconds to cut that ramp. Then 100 ipm will be accelerated to. Actual feedrate is 28.28 ipm for my ramp If Martins example of a 1" ramp to 1/8 depth was cut on this machine, it would not slow down and cut the ramp at 100 ipm unless the plunge rate was reduced below 12ipm.


Actual feedrate for Martins ramp is 20.14 ipm.

Numerous options, if you know how to use them they can be very advantageous.

[Kabbie]

Thanks for the explanations! It makes sense that the slowest axis controls the rate. ISlaWW, it took a bit but I followed all of your math except for the actual feedrate of Martin's ramp of 20.14 ipm. Can you explain how you got that?

[IslaWW]

This may be an oversimplification, but in reality, most of the examples given in this thread will be cut in slightly different times by different controllers. So in Martins example where the horizontal axis was simply moving at 20 ipm, a 1.008" sloped vector was cut, but at a faster than programmed feedrate. That said, it was far below the acceptable 100 ipm, so all is good.