froggy

Thats absolutely fantastic and plenty for me to get my head around. Thank you for the detailed response 🙂

Looking for a trials geometry explained guide. Have tried the search function but couldn't find anything. Does such a thing exist? I'm keen to understand what things like BB rise, chainstay length, wheelbase, stem length, stem rise, bar rise, bar pitch and other factors have on the feel of a bike. I was also wondering if there's any sources of collated information for frames which include detailed geometry measurements. TIA

I had a £50 amazon voucher lying around for 18 months, decided to put it to use the other week to order a second monitor and a dual monitor stand.

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There's a garage up at MT which you can use. There's also a f**king ski slope. Nothing to do.

I've taken the plunge and bought http://www.eclipsecomputers.com/product.aspx?code=KEF-KB460 £37 inc postage. A standard layout and cherry MX red switches.

What do people think about mechanical keyboards? I’ve got some spare cash and have wanted one for a while but haven’t been able to bring myself to pay £70+ for a keyboard. I was looking at https://www.overclockers.co.uk/ducky-dk2108-zero-mech-mechanical-keyboard-black-cherry-switch-kb-029-dk.html with brown switches. I only play CS:GO and use my PC for uni work

Wouldn't sound work through the HDMI? How difficult would it be to set it up?

Is it possible to set up a raspberry Pi as HTPC? I want to buy my dad one for Christmas which some kind of attached storage and stick a bunch of 50 and 60s movies on there. Is this doable?

Thanks for your all your help guys. much appreciated. I don't know why I approached the question the way I did, I think I just over complicated it. Here is how I ended up solving it

Thanks for your steer, I believe I've cracked this now. The vertical displacement should be 18cos30 = 18sin60

The opposite of the angle(30°) over the adjacent (18m)

Thanks for your reply, although I don't understand how the hypotenuse can be equal to cos30, doesn't cos30 need a coefficient to be a valid hypotenuse?

Thanks for your help So, if I wanted to calculate the vertical component of time, I know that.. S = 18tan30 U = ? V = 0 (final velocity) A = -9.81ms-² T = ? Which leads me to the problem that no equation of motion tends for only the 4 variables of displacement, final velocity, acceleration to give time I’ve got as far as splitting the horizontal component into half to give… So from this I should be able to calculate the initial velocity, as I have the same as before S = 18tan30 U = ? V = 0 (final velocity) A = -9.81ms-² T = ? So, using v² = u² + 2as I can rearrange to find U² = 2as – v² ∴ u² = 2(9.81)(18tan30) – 0² u² = square roof of 203.9 u = 14.3ms-² Although I know this is wrong. Because this velocity only gives a range of 18 meters, and if I was to double the initial velocity it gives a range of more than double EDIT: I could use the value derived for initial velocity in the vertical plane to calculate the T in the vertical. However, the initial velocity is derived and I don't want to carry that error forward, because I'm not happy that that is the correct value for initial velocity. (Or is this initial velocity ONLY in the vertical plane?)