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A little more phun with physics

Discussion in 'The Pub' started by Bravus, Mar 10, 2009.

  1. Here's a PowerPoint I prepared for my class today. Thought I might share it here just for fun: http://www.bravus.com/Bandit.ppt

  2. hehe nice
  3. #3 Bravus, Mar 10, 2009
    Last edited by a moderator: Jul 13, 2015
  4. Nicely done.
    Must try and sneak bikes into some of my lectures, rather than just relating them all back to booze, or smutty humour (like using a pic. of Paris Hilton in a police uniform for a lecture on copper ores ;)).
  5. P=38.5kW assuming constant acceleration.

    W=1/2m(v^2) and P=W/s

    where v = 27.8m/s and s = 3.5


    38.5kW = 51.6HP
  6. Yep, that's about what I got.

    Hence the dyno chart to talk about the fact that it doesn't sit on max power the whole time it's accelerating.

    That explains some but not all of the disparity. Wanna have a bash at the rest?
  7. Probably way off but

    Would inertia becoming into play? Since on the dyno the bike doesn't have to move, it just rotates the 'rollers' which I would assume weigh less then bike + rider

    Or are you talking about eviromental factors like wind resistance?
  8. Glad to see Junglist took the same approach I did. I'm not totally crazy, then. ;)

    On a big bike like this, we can't deploy the full power of the motor until after a certain speed due to the bike trying to power-wheelie.

    (which I guess could be approximately found by considering the C/G of the bike + rider, and solving for moments about the contact patch of the rear wheel... When the inertia moment of the C/G exactly opposes the gravity moment of the C/G (and therefore the front contact patch is weightless), we will know what acceleration rate the front wheel lifts off at. Approximately, anyway.

    From there, Power = Force * Velocity... we can determine at what wheelspeed the acceleration force is low enough that the power-for-a-given-rpm shown on the dyno graph can be safely deployed.


    In fact, we could plot Wheelie_Acceleration_Force * Current_Velocity = Wheelie_Power onto the dyno graph's scale to graphically show what fraction of the engine's power we can safely deploy at a given speed.

    Or to phrase it another way: We can't deploy all the full power shown at any given point on the dyno graph until after, say, 80-90kph anyway.
  9. I'd actually been thinking more about the air resistance stuff as one factor - which is why, despite having the same power on tap, the next 100 km/h (100-200) increase in velocity would take a hell of a lot more than 3.5 s.

    But the wheelie one of course is absolutely correct - without a lengthened swingarm and a wheelie bar it's impossible to get anywhere near full power to the ground on take-off. It's the limiting factor, rather than power.
  10. Yeah, certainly after 100kph the V^2 component of aero drag force gets messy... In fact it becomes V^3 when you're considering Power required to overcome aero drag.

    I'm not sure it's thaaaaat big a factor below 100kph, however, because even a 10 horsepower CBR125R can reach 100kph... if only just. Bandit's a much bigger bike, but a 10-20hp power draw at 100kph (and even less at lower speeds) seems a minor factor in the 0-100kph situation. :)
  11. True, but say it's 20 hp at 100, and therefore (being a squared not linear relationship with velocity) averages 7 or 8 across the whole time... that's actually helpful in explaining why we are getting 50 hp instead of 100, if it's combined with the dyno graph. That is, say across the revs it's averaging something more like 70 horse, 10% of that is significant though it doesn't explain all the difference.

    Part of the point I'll be making is that real life situations are messier and more complicated than your average 'physics world' question!
  12. And thanx to all of you, btw - I'm going to class better prepared for a good discussion than I would have been without this discussion.
  13. Yepyep, no worries. :)

    I'm not sure if I'm reading that incorrectly, but to clarify;

    Aerodynamic drag Force is the squared relationship to velocity,

    But through Power = Force * Velocity, the power requirement is actually a cube relationship to velocity. That's why it only takes a paltry ~120bhp for a car to reach 190kph (MR2 Mk1, Chevrolet Lacetti, etc), a modest 492bhp to reach 322kph (Lambo Diablo), and a staggering 1000bhp to reach 406kph (Bugatti Veyron). :eek:

    All that aside... I'm not sure if you're being cruel or kind to your students by presenting a hairy "real world" problem like this. :LOL: I know years ago I was often heartbroken with how poorly the simplified solutions and approximations we had been taught matched all the intricacies of reality.

    Probably best to do as you're doing, and break it to them now. :)
  14. I kinda miss physics. Studied it through VCE.
    At least with it I didn't have to study business structures like I do with my current course.
  15. additional factors probably include:

    -rotational wheel mass/inertia and rolling resistance +brake drag (though probably significant for only the front wheel in a road vs dyno comparison).

    -damping through suspension. basically energy lost as heat as the whole pike pitches under the off-COG acceleration. (minimal but worth noting. it all adds up)
  16. I'm a firm believer that there are very few engineering problems that can't be approximated as a either a simply supported beam or one dimensional flow :grin: .
  17. ok, so assuming a solid cylindrically distributed mass moment of inertia for a front wheel lumped 70% to the outside of the wheel (fairly crude, but whatever), and the similarly crude method of determining power to get the rotation to match the 100km/h bike speed.... anyway, calcs below:

    I = m(r^2)*.7 (r = radius, m = wheel mass)

    W = I(w^2) (Work = Moment of Inertia, w = angular velocity)



    w=110.7 rad/s (going from a 120/60 17 front tyre/wheel r=0.251m)
    I=0.53 kg m^2
    W=3241 J


    so the total front wheel rotational + my bike calc from earlier power = 39.5kW

    same methodology for the rear gives an additional ~2.3kW but that would be taken into account on the dyno.

    39.5kW ~= 53HP
  18. All valid arguments, but take a step back.

    Rev, fast ide, 4000 rpm.

    Slip Clutch - unused potential energy coupled with energy losses through the clutch plates shearing oil. Safely assume you're not using a fully disengaged clutch for between 1/4 of a second to a full second (for a crap launch)

    Throttle control - likely not using full throttle till you get to a "safe"- I.e non-wheelie dump- speed.

    Gear change from first to second, jam clutch, snap throttle, bang up, peel little flakes of metal off synchro teeth - 1/2 sec tops.

    You're flying all over the rev range - Changing up well after you're hit the other side of the peak. fortunately, second gear places you just on the incline of the peak, max power is coming down.

    So, I'd be guessing valid assumptions:

    1)Not fully engaged clutch for up to a second,

    2)Averaging out the powaaahhh curve (taking the average power value once the clutch is engaged until shift up (Assuming slip of 4000 and shift up at 7250, would give you a maximum "pool" of approximately 75kW peak at your disposal)

    3)Clutch and driveline losses.

    Plus what everyone else said.
  19. Bravus, you throw this one out there, after I have returned from 1400kms (700 a day) of driving, and road designing. You know I love my fizix, but unfortunately, I'm mentally shagged.
    A little more courtesy next time.....