|
|
Who own a 10R and a tank cam?
|
That looks like it's on A413 starting at Denham heading towards Aylesbury.
|
That's quite interesting. 2 seconds is quite a long time (relatively
speaking), and to struggle travel 64 feet in that period from a
standing start shows how important, and difficult, the launch is.
In hill climbing, does the clock start when the lights change (as per
drag racing) or when the bikes crosses the beam (as per sprinting)?
|
64 feet in 2 seconds = 32 ft/s/s acceleration = 1G. All about finding
|
Not quite - more like 21 ft/s/s about 0.6G.
|
Not my strong suits Maths & Physics (got an 'O' level pass at 'A'
level Physics and only 'O' levl Maths)
Mixing up average velocity and acceleration aren't I (another hill
climb paddock myth exploded).
|
I'd like to know how you both arrived at those figures.
|
Acceleration is measured in units/s-2 (units per second per second).
Accelerate at 21 ft per second per second and at the end of the first
second you have travelled 21 feet. Then in the next second (the second
second as it were) you do 21 feet again and accelerate by 21 feet per
second per second - giving you a total distance travelled of 63 feet
and a final velocity of 42 feet per second.
|
Hmm.
s = ut + (at^2)/2, so from a standing start s = (at^2)/2
Rearranging that gives a = 2(s/t^2)
Since s = 64 feet, and t = 2s, a = 2(64ft/4s^2) = 32ft/s^2
Your version only works if velocity is quantised. Accelerating at 21ft/s
for one second gives you a terminal velocity of 21ft/s, but an average
speed over that second of 10.5ft/s, not the 21 feet you claim.
|
*ding*
And for the second second, start v is 21ft/s, end v is 42ft/s, average v
is 31.5ft/s, giving a total distance travelled in 2s of 42 feet.
|
That said, I've probably gone wrong somewhere above, and Ivan or Simon
will be along in a second to point out where.
|
You are probably right - I did quantise it to make it easier (for me).
|
|
|
|
Very simplistic sum involving 64 feet and 2 seconds calculated on a
totally false premise.
The correct answer might be available using the equations here:
They relate to constant acceleration though, but in this scenario is
that really the case? There will be an average acceleration, and over
just two seconds I suppose that comes to the same thing - unless
traction is lost and then regained?
|
I think we can say it's quick - life is too short to start calculus
before tea-time.
|
|
|
|
Still impressive though.
|
Shows how important power to weight, traction and, as Champ said,
weight distribution is.
Top fuel dragsters can allegedly pull 3G launches though (straight
line innit).
The big hill climb cars often dip below the magic 2 and still manage
to get round bends at amazing rates - I wonder what the lateral G
forces are.
|
|
just the right amount of grip and not skying it. No 'reaction time'
involved, you break a start beam as in sprinting. Once the light goes
green you start when ready. The timekeepers don't like you fannying
around for too long before going for it though.
|
|
|
I'm sure. Of course, in straightline racing you can offset the
problems caused by lots of power by lengthening and lowering the
chassis. This would not work on a hill climb!
|
Some of the MX based machines have 'launch control' gadgets that keep
the forks compressed off the line. The rider lunges down on the forks
and hooks up catches that keep them compressed. Once they hit the
brakes for the first bend the forks compress even further and the
catches release.
It is reckoned that there is half a second to be gained or lost off
the line. At big meetings the top ten qualifiers can all be within
that margin of each other even over a long hill (Loton = 60 seconds
for riding gods).
|
|
There's a 2-3 mile section of dual carriageway that's usually totally empty.
|
|