# Why is a $5-60 mph$ time slower than a $0-60 mph$ time for some automobiles?

This doesn't make a lot of sense to me, from a physics 101 point of view. I've read a few blog entries on why this is, but none of them explain it well or are convincing. "something-something launch control. something-something computers." Nothing in physics terms or equations.

For instance, Car and Driver magazine tested the Porsche Macan GTS. The $x-60$ times are:

• Rolling start, $5-60\; \mathrm{mph}: 5.4\;\mathrm{ s}$
• $0-60\;\mathrm{mph}: 4.4\;\mathrm{s}$

That's a whole second - about $20$% faster from a dead stop than with some momentum - which seems rather huge.

edit:
here is the article for this particular example. But I've noticed this with many cars that are tested for $0-60$ and $5-60$ times.

Here is another example - an SUV.

Another example.

And finally, interesting, even for the Tesla Model S (EV) where power doesn't depend on engine RPM, $0-60$ is still slightly faster than $5-60.$

• Can you provide an online link to the article, or upload a copy? I like to be able to verify claims which I am being asked to explain. Commented Aug 29, 2016 at 0:23
• Perhaps the car is driven by a mathematician, who steps on the brakes at 5 mph, reducing the problem to the previously solved 0-60. More seriously, see the Mpemba effect that in some situations, hot water freezes faster than cold water does when exposed to the same environment. en.wikipedia.org/wiki/Mpemba_effect Commented Aug 29, 2016 at 9:06
• As a side-note, the torque and power curves for DC motors are related to speed. See this really terrible looking site with lot of technicals. The upshot is that an electric motor has the most torque at the lowest speeds. I happened upon a video of a guy drag-racing gas-burners in a Tesla and it's a useful demonstrating as to how he leaves them in the dust off of the line but they gain on him as the race goes on. Commented Aug 29, 2016 at 13:40
• The OP would learn by doing some research on anti-lag systems. At 5mph, even in first gear, my car takes a long time to spool the turbo and give me any significant power. But on a standing launch I can have the turbo spinning at full pressure before the wheels start moving. The moment I lift the clutch I will overtake an equal car that has a 5mph rolling start. Commented Aug 29, 2016 at 20:15
• As others have stated, starting at 0 mph usually involves dropping the clutch at optimal rpm, whereas starting at 5 mph usually implies the clutch is engaged and the engine is at low rpm. If you were to be rolling at 5 mph then drop the clutch at higher rpm, the 5-60 time would be slightly lower then the 0-60 time. Commented Aug 29, 2016 at 23:09

Ok, from the link given by @count_to_10, I think the answer is clear from this response:

You can launch from a dead stop at any RPM you want, whereas from 5 MPH it's assumed the car is already in gear at low RPMs.

When you start from a standstill, you can rev the engine to any RPM you like before throwing the clutch to engage the axle. Maybe you could match the static friction of the surface to achieve the maximum possible acceleration. When they start at 5 mph, another answer on that site makes it clear that they assume your RPMs are matched to your motion:

"What about rolling at 5mph and dropping the clutch like a regular launch? Wouldn't that help?"
Yeah, but that's not how they test 5-60 or any other rolling acceleration tests. That's the point of them: to test how much passing power you have while already rolling, in gear without a clutch-drop.

So the engine has to move through the entire range of RPMs, which takes more power.

• I know some guys who run a steam powered truck that was originally built in the 1920s. It has two gears, but you never change gear while moving. It's been modified a bit to take advantage of modern roads and tires - the speed range is about 0-40 mph in low gear and 0-80 in high. The torque of a steam engine is maximum at zero RPM. That beast can leave any car driven by a "boy racer" standing when traffic lights change - and even its even more alarming when you are overtaken by a ancient looking truck which (compared to a gas powered car) is absolutely silent. Commented Aug 29, 2016 at 1:58
• If it were merely bringing up the engine to peak HP rpm, I would think an electric car (where torque and power come on fully at 1RPM and there's no 5000RPM clutch dumps) would have a faster 5-60 time. Except that's not the case. Tesla model S 0-60: 2.8s, 5-60: 3.0s. caranddriver.com/reviews/2015-tesla-model-s-p90d-test-review Commented Aug 29, 2016 at 4:23
• Is the Tesla allowed to burn rubber starting from 0? That would allow its motor to run at some revs. Electric motors DO have peak torque at 0 rpm, but also 0% efficiency, which means there's a LOT of heat to get rid of, so the motor controller has to baby it at low speeds to control the temperature rise. Commented Aug 29, 2016 at 13:59
• Seen elsewhere today ( mechanics.stackexchange.com/questions/5574/… ) Tesla's torque curve is flat to about 40mph, which must be limited electronically. Commented Aug 29, 2016 at 15:56
• Sounds like the problem is really in the name. Starting at 5 isn't the requirement. Starting at idle is the requirement. Calling that a 5-60 test invites abuse. Commented Aug 31, 2016 at 16:23

In the rolling start there is no tire slip or revving on the engine and so the run starts at low rpm where the engine is making less power.

A rolling start might have the engine at 2000rpm making for example 200 lb-ft (or 76hp) resulting in 0.45g of acceleration at 5mph (this example yields the acceleration to be 0.002253 times the torque produced).

With a launch from zero the engine is revved first then its kinetic energy transferred to the car yielding the first 5 mph almost instantaneously. At this point the clutch is either still slipping, or the tires spinning allowing the engine to be at about 4500rpm. The higher engine speed and the slightly higher torque (like 220 lb-ft) results in significantly higher engine power at about 188 hp (Power = Torque × RPM/5250). Some of this power is lost due to the clutch/tire slipping so the wheels see like 50%-60% of it, or 113 hp. At the same 5 mph this power at the wheels means about 0.67g of acceleration (or 0.0030 times torque produced) or 35% more.

In summary,

• Rolling start: Engine bogs down and it takes time to get up to the "power band". Peak acceleration decided by engine torque only.

• Launch: Keep engine spinning in the "mid range" and slip clutch or spin tires enough to match available traction.

• 0.45g / 200 lb-ft is not a dimensionless number. You forgot the units on 0.002253 g / lb-ft. Commented Sep 1, 2016 at 9:17
• No it is not. It is an example to show how clutch slipping changes a value that should be constant. The formula $$\frac{a}{T} = \frac{\omega}{m v}$$ shows this because engine speed $\omega$ and vehicle speed $v$ are proportional to each other normally. Commented Sep 1, 2016 at 16:35

This is not so much a question of physics as it is a question for mechanics.

The 0–60 mph benchmark is commonly quoted in publications for car enthusiasts. As with any benchmark, manufacturers will try to game the system. Fancy sports cars have launch control systems: if the car starts from standstill and the accelerator is floored, then special programming kicks in, with extremely aggressive shifting and engine tuning, without regard for usual considerations such as longevity and emissions.

Basically, it's a bit like Volkswagening a test, but less evil since the test case rarely happens in real life. Arguably, if the tuning technique achieves the desired result of maximizing acceleration at any cost, then it's not cheating.

The answers are correct, but just to look at this from a mathematic perspective (which I think is where the confusion starts):

It indeed seems strange that starting from velocity $v_{5}>0$, the minimum time is higher, since the trajectory starting from initial condition $v_0 = 0$ will forcibly pass through $v_{5}$;
this would indicate that $v_{5}$ is a point lying on the same orbit as $v_0$, and thus should follow the same path and simply be a sub-trajectory (sub-segment) ending at the same point $v_{60}$. This follows from the principle of optimality (see below), or from a dynamics point of view by looking at the state space as a vector field: orbits (and therefore trajectories) cannot cross each other.

The explanation is that $x,v$ (where $x$ is the position, and $v$ is the velocity) does not constitute the entire state-space: to be accurate we need to augment the state-space with things like the angular velocity of the wheels (RPM), as well as discrete switches in dynamics due to gear-changes etc. This would allow the vector-field to change, and therefore avoid the issue of what seemed like the same point in the vector-field flowing to the same terminating condition following two different orbits.

An intuitive explanation of the optimality criterion (straight from the Bertsekas book "Dynamic Programming and Optimal Control I"):

suppose that the fastest route from Los Angeles to Boston passes through Chicago. The principle of optimality translates to the obvious fact that the Chicago to Boston portion of the route is also the fastest route for a trip that starts from Chicago and ends in Boston.

In your example, the points in state-space are equivalent to the cities in the intuitive example.