1
$\begingroup$

The answer may seem obvious: 'because they're less massive'. But what's the correct physics relationship here?

If two identical engines were fitted in a bike (say, 200kg) and a car (1000kg), will they:

a) put out identical energy over a given time. So, when converted to kinetic energy via $E_k=\frac{1}{2}mv^2$ the bike's velocity after a given period will be faster than the car's by a factor of the square root of five because $v=\sqrt{ 2E_k/m}$.

Or, will the two identical engines b) put out identical instantaneous force? So, when converted to acceleration via $F=ma$, the bike will accelerate five times faster $a=F/m$.

There's also a wind resistance element, so we could assume the bike has one quarter of the font-facing surface area (ignoring any streamlining design). I assume this is part of the equation that would govern top speed. (Would love a pointer to how to calculate this given known torque (and thus force at tyre) and drag.)

Some aspects of this question relate to engineering. Eg, the tendency for bike engines to be over-engineered cf similarly-priced car engines. There may also be an effect of the bike's power being applied to the wheel/the ground more directly than the car's. But I'm really asking about the physics of the situation.

Assume same engine in each vehicle. Assume bike is a fifth the mass of the car. Will acceleration of the bike be five times that of the car, or square root of five?

I'm kinda interested in difference in braking ability too, assuming half the contact on the road and one fifth of the momentum at equivalent velocity.

$\endgroup$
  • $\begingroup$ Note that you must not confuse acceleration with maximum speed. If you look at this video you will realise that there is no easy answer to your question. youtube.com/watch?v=bfLqa4vlmFI The maximum torque relates to the maximum acceleration whilst the maximum power relates to the maximum speed as you can arrange the gear ratios to achieve a very high acceleration whilst the maximum speed is comparatively low. As the video points out power, aerodynamics and gearing for the motorcycle are all significant factors. $\endgroup$ – Farcher Mar 15 '18 at 9:41
  • $\begingroup$ Fascinating race, but not quite apples and apples, as the car engines have five times the power of the bike's engine and nine times the torque. Thanks for your notes about power (determines max speed) vs torque (determines acceleration) - good point! This link explains it quite well: physics.stackexchange.com/questions/163263/… $\endgroup$ – Errol Hunt Mar 15 '18 at 12:38
  • 1
    $\begingroup$ I have just remembered this answer. physics.stackexchange.com/a/352498/104696 $\endgroup$ – Farcher Mar 15 '18 at 12:40
3
$\begingroup$

There are two important points to be made in the performance of a vehicle

  1. Power over Weight is directly proportional to acceleration. In fact acceleration (without air resistance) is exactly power over momentum $$ a= \frac{P}{m v}$$

    So in the comparison between a car and a motorcycle consider that the power to weight ratio (and hence power over mass) is much higher for a motorcycle.

    This plays a role at mid speeds. At lower speeds traction is important and at higher speeds aerodynamic forces become significant.

  2. Traction forces acceleration to be very similar in 1st gear for most vehicles (with the caveat that motorcycles can do wheelies which increases traction). So if initially acceleration is limited to some $a=\mu g$ what differs between vehicles is the speed at which acceleration is due to power only happens. Let us call this $v_1$, and it usually corresponds to the top speed in first gear. $$ v_1 = \frac{P}{\mu\,m g} $$

    Again, the higher the power to weight ratio is the higher speed can be achieved in first gear (in general). A semi-truck for example might shift from 1st to 2nd gear at 10mph, while a passenger car might at 40mph and a motorcycle at 70mph.

So a motorcycle can accelerate at the maximum traction limit to a higher speed and then transition to acceleration under power. Of course, there are many factors that go into designing drivetrains and the above is a very coarse simplification.

$\endgroup$
  • $\begingroup$ This is a wonderful answer, thank you. I have more questions though, to understand these relationships. 1a) Can you tell me why $a=P/mv$? What is that derived from? And 1b), I assume that means $a$ drops off as $v$ increases (even ignoring air resistance)? $\endgroup$ – Errol Hunt Mar 17 '18 at 16:22
  • $\begingroup$ Also, it's a fascinating note that initially $a$ is dependent on traction. 2a) Can you tell me why traction causes $a$ to be very similar in 1st gear? Ie, why does changing from 1st to 2nd gear correspond with when $a$ becomes dependent on power/weight, not traction? 2b) Does a motorbike's reduced number of wheels reduce it's traction and thus $a$ in this first phase? 2c) how does pulling a wheelie help with traction? $\endgroup$ – Errol Hunt Mar 17 '18 at 16:22
  • $\begingroup$ The definition of power is $P = F\,v$ and force is $F=m a$. Combined they result in $a = \frac{P}{m v}$. When considering air resistance the general form is $$ a = \frac{P}{m v} - \beta v^2 $$ where the coefficient $\beta$ depends on many factors. $\endgroup$ – ja72 Mar 18 '18 at 12:53
  • $\begingroup$ Traction depends on the % of weight on the driving wheels. Compare a moto with a RWD car and you have more % weight on the back on a moto. If a moto pops a wheelie, all the weight goes to the rear wheel increasing traction. Car designers generally don't want their cars to overpower the wheels in 1st gear because it leads to less acceleration, tire wear and possible loss of control. Why have a 300hp engine, if the traction control limits it to 200hp in 1st gear (I am looking at you dodge caliber SRT4). $\endgroup$ – ja72 Mar 19 '18 at 0:53
2
$\begingroup$

the maximum acceleration rate of any vehicle is determined by its power-to-weight ratio. It's common to have sportbikes these days that weigh 500-600 pounds (with rider) that contain 100HP engines, and so they can out-accelerate almost any street-legal car. In fact, because of this even a Honda CM450 can out-accelerate a Porsche 912.

The top speed is set by the power-to-drag ratio. The dominant term in the drag sum is projected frontal area which is small for a bike and large for cars, which is why the superbike class of sportbikes (150HP+) can do almost 150MPH on 150HP.

$\endgroup$
1
$\begingroup$

The situation may have changed with some hybrid cars like the Tesla Model S P100D, with a launch controlled acceleration of 0 to 60 mph in 2.5 seconds, provided there's enough traction (probably a drag racing track with VHT is needed). The Ferrari LaFerari takes 2.4 seconds and the Porsche 918 Spyder 2.2 seconds. This matches or exceeds the fastest bikes with the best riders hunched down over the tank to reduce the wheelie factor. Some bike testing is done with the front end strapped down and the rear shocks set to max stiffness and height or locked into position.

In a street situation, it seems unlikely that even with smooth torque applied by the electric motors to all 4 wheels, there simply won't be enough traction for the cars to achieve sub 3 second 0 to 60 mph times.

However, once past 60 mph, or more like 80 mph, since that's 1st gear on the high end bikes, the bike accelerate harder than almost all cars, due to power to weight ratio.

As for the bikes, the high end bikes are around 200 hp now, but due to wheelie factor, 0 to 60 mph times haven't improved much, stuck at around 2.5 to 2.6 seconds. For example the 175 hp first generation Suzuki Hayabusa matches the 0 to 60 mph time of the 200 hp current ZX-14R, which is 40 lbs heavier, and possibly with a center of mass that is further back and/or higher. The ZX-14R is about 0.25 second quicker than the Busa in a 1/4 mile run (9.5 versus 9.75 with a pro rider on a good track).

$\endgroup$
0
$\begingroup$

First things first. If you have two identical vehicles, except mass, then yes you can calculate F=ma and KE = 1/2 mv^2 and you will see that these two equations are linked. And if you ignore friction they would tell you the same thing. (but you may need a little calculus ) so (a) and (b) in your question are the same, just measured differently.

But bikes don't have identical engines as you say, and the resistance and traction is all very very different. But generally speaking it isn't that motorbikes have magic engines. Even a pretty boring motorbike has the main advantage of lack of mass (as you say) and that is the core of their advantage when accelerating from stationary - even though they are limited by only one wheel.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.