Who own a 10R and a tank cam?

That dear old chap, peter tugged on his pipefull of old shag and said:

Not quite - more like 21 ft/s/s about 0.6G.

Still impressive though.

 

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).

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.

 

I'd like to know how you both arrived at those figures.

 

That dear old chap, Eiron tugged on his pipefull of old shag and said:

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(64/4) = 32ft/s

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.

That said, I've probably gone wrong somewhere above, and Ivan or Simon
will be along in a second to point out where.

--
ogden
sv650sk5 (chick wheels) cg125 (dead wheels) zx7r (hasn't even got
wheels)

buy tat here: http://search.ebay.co.uk/_W0QQsassZdr.daifQQhtZ-1

 

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.

That said, I've probably gone wrong somewhere above, and Ivan or Simon
will be along in a second to point out where.

--
ogden
sv650sk5 (chick wheels) cg125 (dead wheels) zx7r (hasn't even got
wheels)

buy tat here: http://search.ebay.co.uk/_W0QQsassZdr.daifQQhtZ-1

 

That dear old chap, ogden tugged on his pipefull of old shag and said:

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:
http://www.ajdesigner.com/constantacceleration/cavelocity.php

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?

 

That dear old chap, peter tugged on his pipefull of old shag and said:

I think we can say it's quick - life is too short to start calculus
before tea-time.

 

*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.

 

Well I read it as "barely gets it in before he shoots his load"