# "Peak" acceleration

A couple weeks ago on YouTube John Cadogan made the statement “peak acceleration occurs at peak horsepower”, referring to typical IC engines.
I considered that ambiguous and not true in every situation, so I made a slightly less ambiguous statement in the comments.
Every Friday he does a video responding to some of the comments.

What he didn’t mention is that I further explained my viewpoint the previous week.
As follows (also in this weeks comments):

There are two possible scenarios:
Acceleration is proportional to the torque at the wheel. A = F / M (force applied divided by mass of the vehicle). F = T / R (torque applied to the wheel divided by radius of the wheel). Therefore A = T / (R * M). Since wheel torque is proportional to engine torque maximum acceleration occurs when the engine delivers maximum torque, when IN A FIXED GEAR.

After watching Nut-Fest 20 I see you are talking about a situation where VEHICLE SPEED IS FIXED (“any given speed”, as John says) and transmission ratio is variable. In this case picking a gear ratio that delivers maximum power from the engine also delivers maximum torque at the wheel.

I’ll walk through with some actual numbers, first ignoring rolling and wind resistance, on flat ground. Suppose the engine torque peaks at 4000 rpm, power peaks at 6000 rpm. The trans stays in 2nd gear. Engine 1000 rpm per 10 kph. Lets start at 20 kph & 2000 rpm and floor it. Car accelerates stronger and stronger (can be measured by an accelerometer) until it reaches 40kph. Acceleration in 2nd gear is strongest at 40 kph. Then acceleration starts to taper off. At 60kph still accelerating, but not as strong as at 40kph. No other gear ratio can give stronger acceleration at 60 kph because engine is at peak power.

Thoughts?

Yes, peak acceleration IS at peak torque, not at peak horsepower.

F = mass * acceleration As you posted, F = Torque/Radius accel = Torque/Radius/mass

Since neither mass nor radius change, max acceleration is a direct function of max torque.

Maximum power does not occur at maximum torque since Horsepower = torque * RPM / 5252

Higher RPM means higher horsepower but at less than maximum torque

This statement, however is not true because max power and max torque don’t occur at the same RPM…

Purely maximum acceleration occurs, assuming good traction and no clutch slippage, at the max engine torque RPM at the lowest gear (highest multiplication) and the highest acceleration seen through each distinct gear is also at that same RPM regardless of speed.

That’s the math and I have taken data that confirms it on my own cars. John is wrong.

Interesting that when I go to my YouTube account, I cannot find Cadogan’s Nuts 021 listed when I search for it. Maybe he has seen the error of his ways?

It does work from the link in my opening post.

Me too, but I wanted to read the comments under the video so I went to my YouTube account and it is not there.

Let’s see if I can give the link…

IN A FIXED GEAR, max acceleration occurs during max torque but maximum kinetic energy gain happens during the horsepower peak.

If you have an infinitely variable ratio transmission, choosing a gear ratio that puts the engine at maximum horsepower will result in maximum acceleration.

There is a difference between acceleration and kinetic energy gain. If I drop a one pound weight in earth’s gravity, it will accelerate at a rate of 32.2 ft per second per second no matter what the initial velocity was. If the initial velocity is zero, then after one second of free fall, it will be traveling 32.2 ft per second and will have fallen a distance of 16.1 ft and will have gained 16.1 ft-lb of kinetic energy.

During the second second of free fall, it would have accelerated from 32.2 ft per second to 64.4 ft per second and would have fallen a total of 64.4 ft, or an additional 48.3 ft and gained an additional 48.3 ft lb of kinetic energy.

Even though the acceleration is constant, the kinetic energy gain increases exponentially as the object falls faster and faster.

In the real world it’s not that simple. An engine has to accelerate its own flywheel mass as well as the car’s mass and if you use too low a gear ratio, you actually start to get less peak acceleration simply because an unloaded engine will only spin up so fast, the flywheel inertia of the engine consumes as much of the torque as the car’s acceleration does.
That’s why the real world optimum shift points for the lowest gears are somewhat lower than the theoretical shift points based on the horsepower and torque curves of a dyno graph.
My grandpa’s old pickup truck with a four speed transmission that had a “granny” first gear is a perfect example. While that gear was great for pulling stumps or slow speed maneuvering without having to slip the clutch, it was mostly a waste of energy if you were trying to win a drag race. You were better off starting in second gear.

Of course it’s not that simple. But if someone doesn’t get the simplified concept right it’s pointless to add the real world details, like rolling & wind resistance in addition to drivetrain inertia.

Not maximum in an absolute sense. Maximum that can be attained at any particular speed.
Of course when holding max engine power the vehicle acceleration will decrease as vehicle speed rises.

The “granny gear” - what we always called it - has too wide a gap to 2nd to be useful. The best gear spacing keeps RPM’s between max torque and max HP, especially if HP falls off rapidly past the peak.

I disagree. Inserted below is a plot the longitudinal acceleration (negative is accel, positive is deccel, the X axis is time) of a drag strip run I took a while back. Maximum torque for this engine is at 4250 rpm and max HP is at 5200 with little fall-off in HP up to 6500 RPM. Shift points were at about 6200 rpm. Note the highest acceleration (greatest negative number) is at the beginning of each gear, not at the end.

I think you’re arguing at cross purposes. Your graph shows (as said previously) max acceleration at torque peak in any given gear. However, if you had a CVT, you could have a lower “first” gear, launch at the power peak, and thus have more torque multiplication from the git-go (and at every point of the run). The CVT would win the race.

The original assertion was that max acceleration occurs at max HP. My argument is that the statement is wrong because the max acceleration is at max torque. The data shows it clearly. I don’t think that is at cross purpose.

I disagree that a CVT would win the race IF kept at peak HP. Now kept at peak torque…yes the CVT would win. More gear steps beats less. The multiplication of torque would provide greater acceleration from lowest to highest with no steps because the CVT can keep the engine at max torque, i.e. max acceleration, across both a wider range and with no steps in between.

Consider this a calculus problem. The plot shows a 4 speed. In each gear the max acceleration occurs in the very start of each gear and drops off as the RPM rises to max HP. If we increase the number of gears from 4 to 6 to 8 to 10 to infinity (our CVT), the steps get progressively smaller but the peak accel still occurs in the earliest portion (max torque) of each gear step. Once we reach infinity, our CVT again, the max acceleration would occur as the RPM is held at max torque rather than max HP.

The trace for 1st gear is different. There’s a pronounced peak in the middle range.
Since the car starts from a standstill the engine speed rises from below the max torque speed.
Then acceleration hits that peak as the engine reaches max torque.

Consider the car at some particular speed.
You’re given the choice of two gear ratios, one that puts the engine at max torque, and one that puts the engine at max power.
The max power gear is a shorter gear with higher multiplication of torque to the wheels.
I don’t have time to write out the equations, but do you see how the shorter gear can make up for the lower engine torque at max power?

From my first post:

Try this:

Pull-quote:

The classic mistake is to conclude that the fastest way down, let’s say, a 1/4 mile drag strip is to keep the engine RPM at the torque peak (or as close as possible).

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I know You are a clever guy, but to get the real facit of that plot, a lot more is needed. To interpret the plot, You need speed - in increments - and rpm’s in increments to deduct anything. I could get ten different results from it as it is shown. If You could provide those data, I’ll be all eyes.

In Europe, back in the seventies, Volvo bought DAF - a Dutch car manufacturer, from whom they adapted their multi million gear transmissions (DAF actually invented the cvt transmission back in the late fifthies, used in the DAF Dafodil) into a smallish car, Volvo 343, which you could either get with a 4-speed manual transmission or a belt driven transmission with two drive belts, working each on to conical disc’s. That’s when the cvt-trans got attention. Many test’s was made by car magasine’s because this was the first time such a trans was being introduced to a such a high volume of potential carbuyers.
The car was with a 63 hp 1,4 liter Renault engine with both transmissions, the weight was the same, everything was the same.
Only difference was, in a race, that the manual, you drove as a manual, the other you just flattened the accellerator whereafter the engine raced up to 5300 rpm (max hp) and stayed there. In no test was the manual trans less than 1,2 sec slower than the “automatic” from 0-100 km/h. And mpg was quite a bit better in daily driving to top it off. Would I like such a transmission in my car? No, I like to do my own shifting.

There’s more to a CVT’s advantage than always having the perfect gear ratio, there’s also the uninterrupted power delivery.

Studying Mustangman’s graph, it seems that his quarter mile run began at T=457.9 seconds and ended at T=472.9 seconds. I estimated these times by measuring the graph with a dial caliper. In other words, the quarter mile run lasted about 15 seconds according to this graph.
However, it seems that the first to second shift required 0.48 seconds, the 2nd to third shift required 0.72 seconds, and the 3rd to 4th shift required 0.59 seconds. That’s a total of 1.79 seconds spent coasting.
You need one of these gear boxes.

That is because there is slippage of the tire and the clutch. Once “hooked up” the peak occurs.

I do see how that can happen. The actual result depends on what RPM the max torque and max HP occur. If the torque is broad and flat, I think that may be true. Note the chart I posted however. Once the gear change has occurred, the acceleration Gs at the beginning of the gear is lower than the the very end of the previous gear. The ratio and HP did not make up for the torque.

I agree, the shift time is huge. The car was a road race car with a the rather weak mandated full syncromesh transmission and I was trying not to break it again. Road racing concentrates the effort in the 3-4 shift where the torque multiplication is a bit less.

Here is the RPM data. Note shift points around 6400 RPM and the fact that the run ended before hitting peak HP in 4th. The gear ratio is not optimum for 1/4 racing. Also note there is a drop off of acceleration after the shift point. Since the acceleration, in G’s is plotted to the left axis, by definition you have the acceleration data for the entire run. Speed does not matter to the argument nor does RPM matter to the specific comments about max accel. The engine is kept between max torque and max HP in each of the first 3 gears.