Any question answered...

Where can I get a compact ~65W EU charger in the UK? Ideally with folding prongs, like Anker's Prime 67W GaN Wall Charger.

I'm assuming the chances of picking something like this up at midnight in Oslo Gardermoen airport will be pretty slim.

maybe something like this?

https://www.amazon.co.uk/gp/product/B0BVPYKMYD/

https://www.amazon.co.uk/gp/product/B09X1H88VP

what solution should I use to replace lever hoods that have totally rotted and are non-replaceable. i tried heat shrink today but it seemed not to work because it didn’t grip hard enough/ shape well enough

For the sake of about £4 you could try hockey tape. It seems to stick to anything and you could make a proper pig's ear out of it.

Surely the first petrol station you ride by would have one? Gamble though and your route might not pass many I guess.

Hope you can get sorted.

It seems to be growing around the edge, including the outside edge. Idk why it's doing that.

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Could well be a vending machine with one tbf.

Is there such thing as a sleeve washer if I wanted to put a m5 bolt through an m8 hole and take up the slack.

Why? My motorcycle has after market bar ends with m5 and the expander bolts are stuck. I want to fit bar end mirrors and rather than fucking around with removing them I'm just going to reuse them. However, the mirrors have m8* bolts and holes and I want to keep everything as snug as possible without over torquing given I'm using an underspec'd bolt. The hole has a chamfer and the head of an m5 won't fall down...but still.

*I actually need to double check this

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Short answer is yes, if you search for M5 Standoff Spacer on eBay etc you can get various stand-off lengths with an M5 bore. The tricky bit can be finding the correct outer diameter.

Top hat sections can be harder to find. I usually have to get these made and it’s expensive.

You may be able to 3D print something if it’s just locating the screws and not load bearing. Also search for 8mm brass or aluminium tube with 5mm bore and cut to length.

Cheers - These are the search terms I need.

Can't think of any issues with nylon as a material and that could be ghetto lathed with a drill and sandpaper.

I had a load of nylon spacers come with a screw kit for a TV bracket
Just googled ‘TV bracket M5 spacers’ and a place called accu.co.uk does all sorts of spacers in both plastic and metal.

My daughter is travelling to turkey on Sunday to volunteer with a charity for a month and I’d ordered an additional credit card on my account for emergencies. This card hasn’t arrived in time, so I’m thinking I’ll just give her my main card to take. Is this ok to do? Or could she get into trouble if she had to use it?

What's the ratio in terms of pi of the blue and red shaded areas in this quarter circle? Thought it would be easy but...

(am not sending this mid-GCSE exam)

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Not confident in this, but:

B:R = (5.pi - 4): 4.pi

Or something else

It's been over a decade since I did any integration, so frankly I completely forget how to go about it. But, I'm guessing you want to find the area under the quarter circle for the first unit along by getting the area under the curve (the curve being y = (r^2 - x^2)^0.5 ) between x = 0 and x = 1. This will be the white square + blue area above.

The whole quarter is 9pi/4, minus the 1 for the white square, minus double whatever the result of the integration is (remembering to take away 1 for the white square component).

Blue area 2.9285
Red area 3.1429

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Is that assuming the red area is a quarter circle? As in, the 1/4 * pi * 2^2 term?

Yup, but the question is for ratio in pi.

Stucked converting 1 to a fraction of pi.

Got it, blue area is 82/88 of pi.

Red area is pi.

Working on the basis that pi = 22/7.

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I guess my point was that I don't think the red area is a quarter circle, so can't be calculated by pi*r^2 using 2 as the radius.

[Lots of edits:] I think I have it by working out the generalisation for the red area, as the area of overlap between two quarter-circles of radius r.

If they are right on top of each other, overlap area A = r * pi / 4.
If one is moved exactly r in any direction x, then they stop overlapping and A = 0.
Given that for any* 2D shape, area is directly proportional to dimension, A varies linearly but inversely with x.
[Edit 2: * this isn't right here is it, as the shape isn't just changing size]

So, generally A = - ( pi * x / 4 ) + pi / 4 [as long as r >= x >=0]

In the diagram, the distance between the two quarter-circle centres is sqrt(2).
So red area A = (pi / 4) - sqrt(2) * pi / 4
Blue area B = (pi / 4) -1.

[Edit: FFS I have misplaced something somewhere, it doesn't add up.]

The red area is not a quarter circle of radius (3-sqrt 2), so this calculation doesn't work.

Is there some context to the problem btw? Might shed some light on the expected complexity of the solution.

I have that the red area is 9/4 pi + 1 - 2 sqrt(2) - 9 arcsin(1/3), and the sum of the two blue pieces (assuming they don't overlap with the red piece at all) is 9 arcsin(1/3) + 2 sqrt(2) - 2.

arcsin 1/3 is not a "nice" number I think, so I don't see a natural easy ratio

Well doctor, it started when I dreamt I was in a snowstorm in a Minecraft-like world, and wondered what algorithm would be used to control the settling of cubic snowflakes on the landscape...

If a column of snowcubes start at cell X on a grid, and must each move to a randomly chosen neighbour cell including diagonals, how so you weight the choice so that you end up with roundish, not square snowdrifts?

When I said 'ratio in terms of pi' I'd assumed there was a simple-ish ratio.

Thanks for the exact answer @wence !

I now think a practical way of approximating it would be to draw a bigger version of the original diagram on a grid, and just count squares in the blue and red areas.

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