There is this one thing called hero forge, and it looks really cool,  but... it really expencive for what im looking for, I also tryed Warhammer but that was expencive and the minis where not what I was looking for, I also tryed just the plane D&amp;D characters they where cheap but not what i was looking for, so I was wondering if anyone new a good cheap place I can find my minis... (I am suplying for mor ethen myself) (about 5 people)
I'm looking to build a collection of fantasy miniatures, preferably pewter - but I don't want to buy them one at a time from my local hobby store... Much too expensive, and they don't often carry anything I'm interested in buying, which makes it a long term venture as well. So I'm looking for ideas on how to obtain many miniatures at once, preferably for cheap. D&amp;D Miniatures aren't a good source for me as they're collectibles, and the prices on those are also prohibitive.  By the way - they don't have to be pewter, that's just my preference. =)
What is the difference between buy and purchase Suppose you have a physical . It is true that as amplitude increases, the period increases. Can we demonstrate this fact without explicitly finding the period (which is pretty involved and pretty messy) in:    an intuitive fashion,  rigorously? When is the sum of divisors a perfect square? Apparently you can easily obtain a client IP address in WCF 3.5 but not in WCF 3.0.  Anyone know how? The sum of powers of two and two's complement – is there a deeper meaning behind this? Probably everyone has once come across the following "theorem" with corresponding "proof": $$\sum_{n=0}^\infty 2^n = -1$$ Proof: $\sum_{n=0}^\infty q^n = 1/(1-q)$. Insert $q=2$ to get the result.  Of course the "proof" neglects the condition on $q$ for this formula, and the sum really diverges. However I now noticed an interesting fact:  If you use two's complement to represent negative numbers on computers, $-1$ is represented by all bits set. Also, sign extending to a larger number of bits (that is, getting the same number in two's complement representation on more bits) works by copying the left-most bit (also known as sign bit) into the additional bits on the left.  Now imagine that formally you sign-extend the number $-1$ to infinitely many bits. What you get is an infinite-to-the-left string of $1$s. Which, using the normal base-2 formula $n = \sum_k b_k 2^k$ (where $b_k$ is the bit k positions from the right, i.e. $b_0$ is the rightmost bit), that infinite string of $1$s translates into exactly the sum above! So in some sense we have an independent re-derivation of that equation.  Now my question is: Is there something deeper behind this? Somehow I cannot imagine it is just coincidential. Why isn't it $E \approx 27.642 \times mc^2$? Why is it that many of the most important physical equations don't have ugly numbers (i.e., "arbitrary" irrational factors) to line up both sides?   Why can so many equations be expressed so neatly with small natural numbers while recycling a relatively small set of physical and mathematical constants?  For example, why is mass–energy equivalence describable by the equation $E = mc^2$ and not something like $E \approx 27.642 \times mc^2$?  Why is time dilation describable by something as neat as $t' = \sqrt{\frac{t}{1 - \frac{v^2}{c^2}}}$ and not something ugly like $t' \approx 672.097 \times 10^{-4} \times \sqrt{\frac{t}{1 - \frac{v^2}{c^2}}}$.  ... and so forth.    I'm not well educated on matters of physics and so I feel a bit sheepish asking this.   Likewise I'm not sure if this is a more philosophical question or one that permits a concrete answer ... or perhaps even the premise of the question itself is flawed ... so I would gratefully consider anything that sheds light on the nature of the question itself as an answer.    EDIT: I just wished to give a little more context as to where I was coming from with this question based on some of the responses:  @Jerry Schirmer comments:      You do have an ugly factor of $2.997458 \times 10^8 m/s$ in front of everything. You just hide the ugliness by calling this number c.   These are not the types of "ugly constants" I'm talking about in that this number is the speed of light. It is not just some constant needed to balance two side of an equation.  @Carl Witthoft answers:      It's all in how you define the units ...   Of course this is true, we could in theory hide all sorts of ugly constants by using different units on the right and the left. But as in the case of $E=mc^2$, I am talking about equations where the units on the left are consistent with the units on the right, irrespective of the units used. As I mentioned on a comment there:     $E=mc^2$  could be defined using units like $m$ in imperial stones ($\textsf{S}$), $c$ in cubits/fortnight ($\textsf{CF}^{−1}$) and $E$ in ... umm ... $\textsf{SC}^2\textsf{F}^{−2}$ ... so long as the units are in the same system, we still don't need a fudge factor.   When the units are consistent in this manner, there's no room for hiding fudge constants. Closing Streams in Java $F$ closed and connected then... Let $F\subset X$ be a closed and connected subset of the metric space $(X,d)$ then for every pair of points $a, b\in F$ and each $\epsilon&gt;0$ there are points $a=z_0,z_1,\ldots z_n=b$ in $F$ such that $d(z_{k-1}-z_k)&lt;\epsilon$ for $1\leq k\leq n$.  Is the hypothesis that $F$ is closed needed? Cycles: How do I exclude the Environment Texture from reflections? Two correlated marginally Gaussian RV, but not Jointly Gaussian We know if two RVs are marginally Gaussian and uncorrelated, they can be not jointly Gaussian. How about two correlated RVs which are marginally Gaussian? Are they Jointly Gaussian or no, there are occasions when they are not Jointly Gaussian? How to make Outlook 2010 minimize when I try to close it? When clicking the X button or ALT+F4, how can I get Outlook to minimze to the tray instead of close?  Several programs have this behavior, and it's annoying that Outlook behaves different... We are planning a two-night trip to the Canary Islands. The cheapest options include Lanzarote and Las Palmas for flights. Travel dates are last week of November, so I guess weather should be okay. I have never been to these islands. I'd like to sleep on the beach in a tent or mattresses. Is that possible, safety, legality, and weather? It would also be nice to know how populated with tourists the islands are. Find the equation of the plane knowing that it passes through 3 points How can I best describe "making a new discovery"? Animations with movie15 and swf files I am trying to use movie15 to embed vector animations in my presentation. The trouble is that embedded SWF files refuse to display on Acrobat Reader X sometimes with a "This Media requires an additional player" message. Eg. this  doesn't work. Animated GIFs won't display either, with no error message.   Maybe this is an Acrobat issue but I tried the flashmovie package with the same SWF files and that does work (but only with the embedded player option so no controls are visible). But flashmovie is supposedly unstable with beamer so I was wondering if someone had run into the same problem with movie15.  I am using Windows 7, Acrobat 10.1.2 and MiKTeX. How can I change my VPN connection settings in Windows 7 from the command line?  I know how to do it using the GUI but I really need to do it from the console.  I need to change the user, password and internet address and then restart the connection. It doesn't matter if I have to install some third-party software in order to do this (from the command line). What's the reasoning behind phrase "dissertation submitted in partial fulfillment of the requirements"? Phrase "dissertation submitted in partial fulfillment of the requirements" seems strange. Can one submit dissertation in full fulfillment of the requirements? Creating unnumbered chapters/sections (plus adding them to the ToC and/or header) The score for the 4 questions are 111, 1218, 261 and 5 respectively, but they display 2 digits per line. This is an obvious bug in design. Shape of a string/chain/cable/rope/wire? The height of a string in a gravitational field in 2-dimensions is bounded by $h(x_0)=h(x_l)=0$ (nails in the wall)  and also $\int_0^l ds= l$. ($h(0)=h(l)=0$, if you take $h$ as a function of arc length) .  What shape does it take?  My try so far: minimise potential energy of the whole string,  $$J(x,h, \dot{h})=\int_0^l gh(x) \rho \frac{ds}{l}=\frac{g \rho }{l}\int_0^l h(x) \sqrt{1+\dot{h}^2} dx$$  With the constraint  $$\int_0^l \sqrt{1+\dot{h}^2} dx- l=0$$ If it helps, it's evident that $\dot{h}(\frac{l}{2})=0$.  Generally, this kind of equation is a case of a constrained variational problem, meaning that the integrand in  $$\int_0^l \frac{g \rho }{l}h(x) \sqrt{1+\dot{h}^2} +\lambda(\int_0^l \sqrt{1+\dot{h}^2} dx- l)dx$$  Must satisfy the Euler Lagrange equation. The constraint must also be satisfied.  But, in truth, by this point I am clueless. $\lambda$ is worked through $\nabla J  = \lambda \nabla(\int_0^l \sqrt{1+\dot{h}^2} dx- l)$. I have tried this , but get nonsensical answers.  Is this method the best? If so, in what ways am I going about it wrongly thusfar?