我真的认為这种理论性的分析对於解决翘曲的问题而言是很重要的。我写了一篇简短的分析（应该也要多阐述一些细节才对）。写这篇回应的同时，我思考更深入关於造成翘曲的因素，显然我的分析与你的重点不同；这让我得到不同的结论，所以我想，在这边留下这讯息是重要的。
I really think this kind of theoretical analysis is important to solve the warping problem. I wrote my own short analysis in this thread: http://forums.reprap.org/read.php?1,55300 but I might do a more detailed write up as well. Writing that post made me consider in more detail the causes of warping. My analysis differs in important ways from yours, though, which leads me to different conclusions. So I feel it's important to comment here.
我想我们两者都清楚，翘曲发生的关键在於切层温度的收缩（thermal contraction）。用计量的方式分析，我们可以想像不管再小的任何线段，都具有两种不同的长度；高温时的长度（Lh）以及低温时的长度（Lc），而且每种塑料都是Lh>Lc。
It's clear to both of us that the key factor when it comes to warping is the thermal contraction of layers. For a qualitative analysis, we can imagine that any short segment of filament has two lengths: The length when hot (Lh), and the length when cold (Lc), where Lh > Lc for any particular filament.
任何时刻只要你在一个冷的塑料上面列印上一条热的塑料，你必须思考四个长度的变化。第一个长度是刚列印出来的塑料（L1h），第二个是这个塑料冷却后的长度（L1c），第叁个是旧的塑料在刚被列印出来时的长度（L2h），以及旧的塑料冷却后的长度（L2c）。
Any time you print a hot filament on a cold one, you have four lengths to consider: The length of the hot new filament when freshly printed (L1h), the length of the new filament after it has cooled (L1c), the length of the old filament when it was freshly printed (L2h), and the length of the old filament when it has cooled (L2c).
根据温度收缩，我们知道 L1h > L1c 且 L2h > L2c。
Due to thermal contraction, we know that: L1h > L1cL2h > L2c
我们也知道，热的塑料是列印在冷的塑料上，所以这意味着后面两者的长度相等：L1h = L2c
But we also know that the new, hot filament is printed on the old, cold filament. This forces their lengths to be equal: L1h = L2c
这是造成问题產生的癥结。当物件冷却之后，因為 L1c < L2c 所以物件捲曲。不均等的冷却并不至於导致翘曲，列印的哪一部分先冷却并不重要。重要的是列印出来的塑料是堆叠在冷却后的塑料上面。一个缓慢的冷却行為不该与快速冷却有任何的差异才对。
That's where the trouble happens. When the object cools, L1c < L2c, so the object warps. Uneven cooling does not lead to warping; it doesn't matter which part of the printed part cools first. What matters is that hot filament is deposited onto cold plastic. A slow cool down shouldn't be any different than a rapid quench.
假如你在均温下拿一个没有翘曲的物件，然后对其增加一个特定变幅的温度（在任何的增温速度下，比方一个渐变的高温差），这物件将会弯曲。但是当你让物件回归均匀的温度时（不论降温的速度如何），它将会回到原本的形体。现在面临的问题是，刚被列印出来的物件是一个尚未捲曲的状态（因為它定形在一种不均温的状态）；当它被带到一个均温的环境下，它就会翘曲了；这件事情变换得快或慢并不重要。例外的状况是，假如温度的变化大到足以造成相位的改变，这样的前提下，内部的结构将会在这个过程之后改变。
If you take an un-warped object at a uniform temperature, and impose a temperature gradient on it (at any speed), it will warp. But when you return it back to uniform temperature (at any speed), it will return to its original shape. The trouble is that a RepRapped part is an un-warped object at a nonuniform temperature. When it is brought to a uniform temperature, it warps. It shouldn't matter how quickly the transition happens. The exception is if the temperature change is high enough to cause a phase change, in which case the internal structure will be different before and after the process.
现在来思考列印速度与物件尺寸对翘曲產生的影响。Lh 与 Lc 的差异是与温度变化呈一个比例关系的（Lh = Lc a dT）。假设一个极度快速或是非常小型的列印，所有的塑料是以急速堆叠的方式（没有时间去冷却），因此此时的 dT = 0，而且没有翘曲的现象发生。反之，在一个极度大型或是非常缓慢的列印状态，先前列印的塑料总是在新塑料列印堆叠之前，冷却到室温下；此时的 dT = max（材料热缩的最大值）。
Now to consider the effect of print speed and object size on warping. The difference between Lh and Lc is proportional to the temperature change (Lh = Lc a dT). For an infinitely fast or small print, all the filament will be deposited instantaneously with no time to cool between layers, so dT = 0, and no warping occurs. For an infinitely large or slow print, the old filament has always cooled to the ambient temperature by the time the new filament is laid down, so dT = max.
一个列印的常态既不是非常快也不是非常的慢，因此有两个关键的因素要思考；物件是从表面冷却（热传递与热辐射），以及热量的扩散（热传导）。热量是由新堆叠上来的塑料所產生的接触面，经由新塑料的本体传递到旧塑料的冷却表面。
For a print that's neither infinitely fast nor infinitely slow, there's two important factors to consider (that I can think of). There's cooling from the surface of the part (convection and radiation) and thermal diffusion inside the part (conduction). Heat is added from the surface, in the form of freshly deposited filament, and then conducts through the body of the part toward the cold surface.
这并不是温度遽变导致翘曲的发生。这裡提到的非均质温度指的是物件的边缘区域，热的塑料是被列印在冷的表面上；然而在加热板中心区块，热的塑料是列印在热的塑料上。这样的现象意味着边界的部分将会有最大的应力；物件内部则是最不售应力影响的，因為 dT 在中心是最小的。然而，就算物件没有温度上的渐变温差（举例，假如它是一致的冷，比方慢速列印的状态），翘曲的状况还是会发生（因為新列印的塑料是热的）。这仅是单纯的考量，然而翘曲的应力甚至会贯穿整个部分。
It is not the steepness of the temperature gradient that leads to warping. What this non-uniform temperature means is that at the edges, hot filament is printed on cold plastic, whereas at the center, hot filament is printed on warm plastic. That means that the warping stresses will be highest at the edges, and less at the center, because delta T is less at the center. However, even if there were no temperature gradient in the object (ie, if it were uniformly cold, as in the case of the slow print), warping would happen because the fresh filament is hot. It's just that in this case, the warping stresses will be even throughout the part.
性质上地说，这两者之间并没有过大的差异；一者是均衡的捲曲（所以它可能是导致底部產生一个拋物线形变的原因），另一个则是不均衡的捲曲（可能是一个更接近於四度曲线，在边缘处有一个落差的翘曲）；不论何者，它们都导致物件的翘曲。
Qualitatively, there's not a big difference between these two cases; one warps evenly (so it would probably form a parabolic shape at the bottom) and one warps unevenly (so it might be more of a quartic curve, with the steepest bend at the edges). But they'd both warp.
就我的想像与理解，一个物件是可以被列印到一个非常高的程度，只要抢在加热板的作用衰退之前。加热板通常都是比物件本身来的宽，热传导的增加和温度的提升是从物件的边缘开始，然而减少对流的冷却效应也是。假如这样的现象变成一种困扰，那麼当列印推升到某一个定点高度之后（这个定点是承受整个结构改变的基础点），只要将加热板的温度视物件的高度作提升，如此就可以让上部的表面维持一个恆定的温度。然而，这样可能需要一些非常尖端复杂的软体。
An object would have to be very tall before the effect of the heated bed could really wear off, I imagine. Because the heated bed is generally wider than the object, there should be warm convection currents rising and warming the sides of the object, reducing convective cooling. If it became a problem, then up to a certain point (the point where the base of the part would undergo a phase change), the temperature of the heated bed can be increased as the object gets taller, so as to maintain a constant temperature at the upper surface. That would probably take some sophisticated software, though.
Jacob
|
楼主
京公网安备11010802043351