# Why laser cutting speed suddenly decrease when cutting sheets from 1 to 3mm thickness then stabilize?

Please to understand what I mean, see the picture below :

I don't understand why it is not a linear curve (or almost), because at first sight it should be no ?

If you need 10 sec to cut a length of 1mm thickness, why not 20 sec for a 2mm and 30sec for 3mm... I can't figure this out !

If you have an answer, please don't hesitate to share your experience with me !

• I've never cut steel with a laser but I can see that with thinner material the conductive losses are lower and increase as the sheet gets thicker but the proportional increase is less. Eventually you reach a point where there is so much conductance that making the plate thicker makes little difference. Commented Mar 12, 2021 at 16:55
• do you know the type of laser that this diagram comes from? more specifically the duration of the laser pulse? Commented Mar 12, 2021 at 18:59
• Cutting is usually done with cw, not pulsed Commented Mar 12, 2021 at 19:04

## 1 Answer

To see why this graph can only have a hyperbolic shape, it can help to take a look at the energy balance of the laser cutting process with absorbed laser power $$P_A$$.

$$P_A = \underbrace{vtw}_{V_M}\cdot \rho \cdot \underbrace{(c_p \Delta T_P + h_M + \xi h_E)}_{\text{process energy}}+\underbrace{P_L}_{\text{losses}}$$

The term labelled "process energy" doesn't bother us further here, it describes the energy we need to put in to bring the material from environment temperature to process temperature and is constant for a given material. Same with the power losses $$P_L$$, maybe a little unintuitive, but experiments showed that they depend mainly on the thermal conductivity of the material, so we approximate them as constant as well. The material's density $$\rho$$ is of course also constant.

What remains? $$V_M$$, the molten volume per time, which consists of cutting velocity $$v$$, the sheet thickness $$t$$ and the width of our cut $$w$$. The width $$w$$ depends on the laser beam diameter, so we also consider $$w$$ constant.

In the end, only $$v$$ and $$t$$ are left. We consider everything in the energy balance except those two constant, so let's claim:

$$v \cdot t = const.$$

A product being constant? That is just the equation of a hyperbola with the axes as asymptotes.