monkeyseemonkeydo

Oh dear. Edit: Oh my, what have we here?... Think about it .

No worries chief .

:facepalm: Item 1) You haven't used and "wouldn't ever think of bleeding a brake with water", therefore your opinion on water bleeds is completely irrelevant (and, by the way, wrong!). Item 2) If you really think about it, oil being thicker (higher viscosity) will make it slower and harder to pull than a less viscous fluid like, say, water.

Yes that's what I'd do . Bleed How To Not sure if that'll be applicable to our lever but there are others on the Hope site.

In what way? Oil is thicker than water so it'll be more powerful?

If I were you I'd leave the fluid in and just take the lever off to replace the seals. Use a zip tie or something to keep the hose from flapping around. Once sorted, put the lever back on and bleed normally. By leaving the majority of the fluid in there's less chance of you trapping air in unwanted places.

Have you tried bleeding it? Sounds like you have air in the system, whether that's from dodgy piston seals or not is another matter.

Fair enough! Chainline thing depends on what tensioner you use I guess. But yeah, hub wise it looks good.

Looks good if that's what you're after. Are you? Really?

Without making your own you'll need to buy a Profile one.

It's also worth pointing out that so long as you do the same to both sides, the equation remains correct and you can't really go wrong. Taking the example you give again, 7 - 5x = 5x + 10, rather than adding 5x to both sides, subtract 5x from both sides. That gives: 7 - 10x = 10 Now instead of subtracting 10 from both sides, subtract 7 instead: -10x = 3 You should find that doing it that way will give you exactly the same answer for x as the way you say, it might take a different number of steps but because you've done the same operations to both sides of the equation it will still be true and therefore correct.

Can you not just leave them stealthy black?!

It has nothing to do with what you've done before and only to do with what you need to do in order to rearrange the equation to get all the x's on one side and all the numbers on the other. You're correct- you add 5x to both sides which means all the x's are on the right. You now want all the numbers on one side (for ease get them on the left) so as you say you want to subtract 10 from both sides. There are other ways you could do it but the shortest route is what you've said. Don't think about what you've done before though- that's largely irrelevant. Only consider what's left and what you have to do to get all the x's on one side (so you can eventually solve for a single x) and all the numbers n the other side, remembering to do the same to both sides each time. Edit: An example picked out of thin air to try and show it.. 5x - 3x -10 + 4 = 4x - 12 You probably won't see anything like that because you'll be given problems which have already been simplified but starting there, collect like terms first: 2x - 6 = 4x - 12 I'm going to collect the x's on the left so first I'll subtract 4x from both sides: -2x - 6 = -12 Now collect the numbers on the right by adding 6 to both sides (that will remove the -6 on the left): -2x = -6 So now we just divide both sides by -2 (to get x on its own on the left): x = 3 You can double check if that's correct by substituting x = 3 back into any of the equations above. Just to prove it stick x = 3 back into the original equation: 5x - 3x -10 + 4 = 4x - 12 5(3) - 3(3) - 10 + 4 = 4(3) - 12 15 - 9 - 10 + 4 = 12 - 12 0 = 0 Which is lucky. But proves the answer is correct (i.e. the equation is true).

What sort of parts are you talking about? Exposed bearings/details? Being fitted to other tight toleranced components or stand-alone? What about taking the easy/cheap option of just using paint stripper?

I swear that was mentioned on the first page of this thread and many times subsequently! Following from that (and to take what Luke said and also the way you're looking at it), So long as you do the same to both sides it'll aaaallll be goooood. It's not about what's smallest or biggest (that just makes your life easier) but you can do it any way you like: 9x - 4 = 63 For shits and giggles subtract 63 from both sides and subtract 9x from both sides gives: -67 = -9x Divide both sides by -9: -67/-9 = x x = 7.444... So long as you've carried out the same operations on both sides it'll work out, you just may need to deal with the odd negative in there.

No, if you follow it you've added 4 to the left hand side to remove it but you haven't added 4 to the right hand side. What you should have is: 9x - 4 = 63 9x - 4 + 4 = 63 + 4 9x = 67 x = 67/9 x= 7.444... Not sure what you mean by adding the x's. You don't necessarily do anything to every equation, it all depends on what you have to begin with. In the equation you give: 4x - 9 = 7x + 5 Get all the x terms on the left and all the numbers on the right. To do that we subtract 7x from both sides and add 9 to both sides: 4x (- 7x) - 9 (+ 9) = 7x (- 7x) + 5 (+ 9) A bit too much going on there but you should then get: -3x = 14 We can now divide both sides by -3 to get x on its own on the left: -3x/-3 = 14/-3 x = -4.666... The key to it all is that you have to rearrange the equations to be left with x on the left. To do this you carry out operations but if you do anything to one side you MUST do it to the other. If one side needs 3 subtracted to get rid of what's there to start with you MUST subtract three from the other side of the equals sign to make sure everything remains balanced and correct. If you consider it in just numbers, you know that 5 + 3 = 8. If we subtract 3 from the left and nothing else you end up with 5 = 8. Which is clearly wrong. However if we subtract 3 from BOTH sides, in exactly the same way as you do in an equation containing unknowns like x, you would get 5 = 8-3 = 5 which is correct.

It's punctures As has been said, a little extra pressure (try ~25psi to start with judging by your weight) will definitely help but better tyres with thicker, 'armoured' or dual ply sidewalls will also help prevent snakebites.

I was taught both BODMAS and BOMDAS. I'm also a user of Nofclbrih... it's a good word which is worth remembering in Chemistry.

When you write a post there are the A2 and A2 buttons to make subscript and superscripts next to all the bold, italic etc. buttons .

Which can also be written as (xy)2.

Where did you get that equation from? Work it through: 9x + 81 = 10x + 100 x = -19 Which obviously isn't right and you've lost the y's... 9x + 81 = 10y +100 Substituting for x and y gives: 81 + 81 = 100 + 100 162 = 200 Which also obviously isn't right.

No, it isn't. Trust me. Did you actually read my post? If you have xy2, ONLY the y is being squared. As JT says, that means it can be rewritten as x * y * y. It's really not difficult...

Because it's y2 which is y * y. 32 = 3 * 3 = 9.

The type you describe are some of the easiest you can come across as you're basically given the answer. All you need to do is substitute the numbers given (x=9, y=10) into the formula given (xy2). The only thing you need to remember is what to square. In this case if you were to write the formula out slightly differently you'd have x * y2 Noting that only the y term is being squared. Substituting x=9 and y=10 gives: 9 * 102 Which becomes: 9 * 100 Which = 900. Edit: Not entirely sure how your teacher will have turned it into a proper equation without introducing another unknown (lets say z). In this case we can write z = xy2 Because we know x and y we have one equation and one unknown (z) which means we can solve it. So following what we did before: z = 9 * 102 z = 900. You can always check these things by going backwards. If we pretend we don't know what y is we can start with the same equation as before: xy2 = z So if we first rearrange for y: y2 = z/x y = squareroot(z/x) And now substituting z = 900 and x = 9 into that we get: y = squareroot(900/9) y = squareroot(100) y = 10 Is that taking it too far?