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About froggy

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  1. 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.
  2. There's a garage up at MT which you can use. There's also a f**king ski slope. Nothing to do.
  3. I've taken the plunge and bought £37 inc postage. A standard layout and cherry MX red switches.
  4. 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 with brown switches. I only play CS:GO and use my PC for uni work
  5. Wouldn't sound work through the HDMI? How difficult would it be to set it up?
  6. 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?
  7. 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
  8. Thanks for your steer, I believe I've cracked this now. The vertical displacement should be 18cos30 = 18sin60
  9. The opposite of the angle(30°) over the adjacent (18m)
  10. 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?
  11. 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?)
  12. This is a question on a homework assignment and I just can't figure it out An object a ramp at an angle of 30 degrees from the horizon then lands at the same height after travelling 36m horizontally. What is his his take off speed? (Assume negligible air resistance) So, first of all I’ve tried to calculate the vertical and horizontal components of velocity S = Displacement (m) U = Initial velocity (ms-¹) V = Final velocity (ms-¹) A = Acceleration (ms-²) T = Time Horizontal S = 36m U = Vcos30 V = ? A = 0 (because of no air resistance, or is this wrong?) T = ? Vertical S = 0 U = Vsin30 V = ? A = -9.81 (Gravity) T = ? I’ve got as far as subbing the vertical and horizontal components into s = ut + 1/2 at², giving me 36=Vcos30 * t+1/20 * t and 0=Vsin30* ½ -9.81 * t² But I’m unsure how I can derive a value for either V or T, no matter how much I rearrange s = ut + ½ at² Can someone point me in the right direction?
  13. I've just grabbed a second hand Cannondale trail 27 29" at a great price. Some of the stock parts I plan on upgrading over time. Such as the brakes, wheels and complete drive train. It may seem like an obvious question. I may be over thinking it. Are standard 26" mountain bike parts compatible with 29" bikes? I'm thinking of buying a set of second hand hope hubs, hydraulic disc brakes and cranks. Do 29" bikes require larger disc rotors, longer crank arms and different/stronger hubs? Or is it Ok to grab a second hand pair of old hope xcs and minis and keep the cost down?