I recently bought an old Shimano EX crankset, which I'd like to mount new clip-on pedals to (Shimano Dura-Ace PD-R9100). Does anyone have any experience with using new clip-on pedals on vintage cranks? Thank you, Fynn
How do you measure pedal axle/spindle size?  Replacement pedals and accessories specify a size like 9/16" or 1/2".  How do I measure my current pedal to see what I need?  Is it just a matter of putting the pedal threads in a caliper and measuring the outer diameter of the threads across the axle/spindle?  Or do you need to measure the solid part, excluding the threads?
How to find an unknown Steam User I was playing with someone on Ark: Survival recently, I CTD'd and couldn't get his information. Are there any means that I could find this user's Steam account even though he hasn't logged in since?  I know of the function View: Players in Steam, but it only shows me the last 24 hours and I haven't played with this player in 3-4 days. Is there a way I could extend the view recent players function on Steam? Is there a record of this somewhere? Or do I have any further options? Per User Named Credential and External Identity Users My question is almost similar to one mentioned in 'Auth Provider with per-user Named Credential' (). I am trying to authenticate and make a API call when a new or existing Salesforce community users try to log in. The Authentication is provided by 3rd party (Cognito). So, the requirement is :  User to be authenticated using the 3rd party system for login. After successful authentication and before community page is displayed, the API call is to be made to 3rd party system to get more details about the user.  In order to achieve this:  Created an Open ID Connect Auth. Provider Use the registration handler on the Auth. Provider to create/link a new/existing user. Use a login flow for API call. Use Per User Identity Type Named Credential to make API calls.  No, where I am having issue is once the user has authenticated successfully, how do I allow the user to make API call without authenticating it again. I am getting the error: An Apex error occurred: System.UnexpectedException: You don't have permission to view this data, enter credentials for the named credential in your personal settings. Or ask your Salesforce admin for help. I believe this error is due to the fact that the user is not listed under 'Authentication setting for External Systems'. My requirement doesn't allow me to authenticate the user again, so how do I achieve this using either APEX or any other way? I am relatively new to Blender so please excuse me if this is rather amateur!   The bottom of my torus has poor surface quality, I have scaled the vertices down for a smaller (variable) tubular radius all the way around the torus and at the bottom the surface quality is awful, with inaccurate shading. Any ideas as to what is wrong and/or how I could fix this?  Thanks, itsanate Why is FIFA against adding instant replay to the game? Is there any evidence or reason why FIFA is so against adding instant replay to soccer? I find myself seeing many unlucky or just plain wrong decisions made by refs that could be corrected with instant replay. Ok. I was playing mcpe and I noticed that mossy cobblestone isn't in the crafting table. And I wanted to make a house that looks old. So if its possible to get mossy cobblestone without mods, how can I get it on survival? If it is possible can somebody please leave a step by step guide. Thank you. Example 4.21 in Baby Rudin: Why is the inverse of this function not continuous at this point? This question is from [Number Theory George E. Andrews 1-1 #3].     Prove that $$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1}).$$   This problem is driving me crazy. $$x^n-y^n = (x-y)(x^{n-1}+x^{n-2}y+\dots +xy^{n-2}+y^{n-1)}$$  $(x^n-y^n)/(x-y) =$ the sum for the first $n$ numbers and then I added $(xy^{(n+1)-2}+y^{(n+1)-1})$ which should equal $(x^{n+1}-y^{n+1})/(x-y)$ but I can't figure it out    This is a similar problem in the book and I tried this method but it wasn't working out      $\quad$Thereom $\bf1$-$\bf2$: $\,\,\,\,$ If $\,x$ is any real number other than $1$, then $$\sum_{j=0}^{n-1}x^j=1+x+x^2+\ldots+x^{n-1}=\dfrac{x^n-1}{x-1}.$$    $\quad$Remark: $\displaystyle\sum_{j=0}^{n-1}A_j$ is shorthand for $A_0+A_1+A_2+\ldots+A_{n-1}.$     $\quad$Proof: Again we proceed by mathematical induction. If $n=1$ then $\displaystyle\sum_{j=0}^{1-1}x^j=x^0=1$ and $(x-1)/(x-1)=1$. Thus the theorem is true for $n=1$.  $\quad$ Assuming that $\displaystyle\sum_{j=0}^{k-1}x^j=(x^k-1)/(x-1)$, we find that $$ \eqalign{ \sum^{(k+1)-1}_ {j=0}x^j &amp; = \sum^{k-1}_ {j=0}x^j+x^k=\dfrac{x^k-1}{x-1}+x^k \\ &amp;= \dfrac{x^k-1+x^{k+1}-x^k}{x-1}\\ &amp;= \dfrac{x^{k+1}-1}{x-1}. }$$   Hence condition $(\rm ii)$ is fulfilled, and we have established the theorem.    $\quad$Corollary $\bf1$-$\bf1$: $\,\,$ If $\,m$ and $n$ are positive integers and if $m&gt;1$, then $n&lt;m^n.$ What is the difference between kernel drivers and kernel modules? Since the new top bar has been rolled out - a number of different people, established and new users alike, have asked questions on where a specific link has disappeared to with the new top bar design. Where has the link to &lt;X&gt; gone? 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. Do virtual particles actually physically exist? I have heard  pop in and out of existence all the time, most notable being the pairs that pop out beside black holes and while one gets pulled away. But wouldn't this actually violate the conservation of energy? How can I explain how to play D&D to new players without them falling asleep from boredom? Some of my friends just started playing Dungeons &amp; Dragons [5e] for the first time and don't know anything about how to play. I have tried explaining to them the things that are completely necessary and teach them the rest while playing but I could not get them to understand. None of them have a players handbook and they can't afford one and they really want to play. I really want a simple way to explain the basics and since I have a habit of explaining things in ways that are much too complicated. I have already told them the basic rules so all I need to know is a simple way to explain how to play. I have a follow up question to this question:   My question is: when should you add a comma when starting a letter/email with     Hello John,      It was fun to see ...   I am pretty sure that a comma goes after the name, but lately I have seen several people (academics) write      Hello, John,       It was fun to see ...   with a comma after the Hello.  I am wondering when a comma there is appropriate.    I see this question  but I am wondering about the situation where you have a comma after the name also. Converting string arrays into Map Is there an option to create and Opt-in a user in the same API call? I want to upgrade my Ubuntu 13.04 to Ubuntu 13.10. I'm unable to do via GUI.   [ The reason why I'm unable to do so:  Via software updater, I proceed ahead and click upgrade (for upgrading my OS). I authenticate by providing my password. Then it proceeds and downloads upgrade tool. That time only (while downloading the tool) error message comes "system program problem detected". It then doesn't proceed ahead and I click to report the problem ]  So, I tried via command line but here also I couldn't succeed upgrading my OS.  For the same I ran   $ sudo apt-get update   Then I ran  $ sudo apt-get dist-upgrade   And then:  ravbholua@ravbholua-Aspire-5315:~$ sudo do-release-upgrade Checking for a new Ubuntu release Get:1 Upgrade tool signature [198 B]                                                                                                  Get:2 Upgrade tool [1,135 kB]                                                                                                         Fetched 1,135 kB in 6s (32.9 kB/s)                                                                                                    authenticate 'saucy.tar.gz' against 'saucy.tar.gz.gpg'  gpg exited 2 Debug information:    gpg: Signature made Friday 01 November 2013 10:34:27 AM IST using DSA key ID 437D05B5 gpg: Can't check signature: public key not found  Authentication failed Authenticating the upgrade failed. There may be a problem with the network or with the server.  ravbholua@ravbholua-Aspire-5315:~$    Please suggest how to go ahead and upgrade my OS. I am travelling from Panama to Los Angeles and around the States for approx 80 days then flying back to Panama and making my way up Central America for approx 3 months before flying back to LA to fly home to Australia.    Will this be OK with an ESTA?    Does it count as spending too long in the States?    I have been told that Panama is not far enough away. Group of $r$ people at least three people have the same birthday? What is the probability that in a randomly chosen group of $r$ people at least three people have the same birthday?   $\displaystyle 1- \frac{365\cdot364 \cdots(365-r+1)}{365^r}$   $\displaystyle \frac{365\cdot364         \cdots(365-r+1)}{365^r} +{r\choose 2}\cdot \frac{365\cdot364\cdot363 \cdots         (364-(r-2) +1)}{364^{r-2}}$  $\displaystyle 1- \frac{365\cdot364             \cdots(365-r+1)}{365^r} +{r\choose 2}\cdot \frac{365\cdot364\cdot363 \cdots (364-(r-2) +1)}{364^{r-2}}$  $\displaystyle\frac{365\cdot364 \cdots(365-r+1)}{365^r}                 $     My attempt :  (May be typo in option $(3)$ of question !)  $$P(\text{at least 3 persons have same birthday})$$  $$= 1 - \{P\text{(no one has same birthday) + P(any 2 have same birthday)\}}$$  So, option $(3)$ is true.     Can you explain it, please?   It  before,  but I'm not satisfied by explanation. How do you use a variable in a regular expression? Linked account +100 rep bonus also applied to account with 200