Energy need to Accelerate 4300 lbs from 55 mph to 84mph within a quarter mile?

In summary, the conversation is about the energy needed to accelerate a 4300 lbs vehicle from 55 mph to 84 mph within a quarter mile. The speaker is trying to find the additional energy needed to achieve this acceleration, taking into account the vehicle's weight, horsepower, and gear ratio. They also mention that they have tested this on a closed runway and found it impossible. However, they are asking for assistance in calculating the energy needed and state that an approximate result would be sufficient. The conversation ends with a humorous remark about needing a lawyer to dispute a ticket for allegedly achieving this acceleration.
  • #1
dorcom
2
0
This might be an amusing one for you.

I need help how do I calculate the energy needed to accelerate a 4300 lbs mass traveling at 55 mph to 84mph within a quarter mile.

In real life terms: (No exact science needed)
I'm trying to find out, roughly how much, theoretically, approximate additional energy (ideally in horsepower) is needed to accelerate a vehicle, in this case a 1986 Toyota Landcruiser with a 125 hp 6 cylinder carburetor engine... with a vehicle curb weight (mass) of 4,300 lbs, traveling in direct (4th) gear accelerating from 55mph (V1) up to 84mph (V2) within about a quarter mile. Ignoring all frictions, wind, etc.

Btw. the engine of the vehicle produces 125 net hp at max. 3,600 rpm. the axle (wheel to engine) gear ratio is 3.7 (At 55mph the engine revs around 2,500 rpm. Given by actual tire size, 27" diameter, the engine would be deep into the redline range at almost 4,000 rpm at 84mph!

Thank you very much in advance for your helpful consideration of an answer and solution helping me solve this!

An approximate result or approach would be fine.
Q

PS Why I ask this?
Because a nut-job knucklehead of a CHP officer claims I did that!
Thereafter I tried it a couple of times on a closed old airport runway, but it is impossible to achieve. In part simply because the vehicle mechanically by design can not go that fast neither does it have the horsepower to move it's 4300 lbs so quickly within a 1/4 mile.
 
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  • #2
The energy needed is called "Work" - the work is the difference in kinetic energies.

[itex]W=\frac{1}{2}m(v_f^2 - v_i^2)[/itex]

Of course, you also have extra energy losses due to drag and internal frictions and so on - this is just the energy that goes into the motion.

To do it in a quarter-mile limits the time the energy needs to be supplied in - the work divided by the time will give you the mean power required.

[itex]P=W/T[/itex]

The time depends on the acceleration - if this is constant acceleration, then it is easy.

[itex]d=\frac{1}{2}(v_i + v_f)T[/itex]
 
  • #3
Ignoring air resistance, for continuous acceleration the maximum power expenditure occurs when the car is still accelerating and is reaching its final velocity (since instantaneous power is given by F*V, and Force is constant for constant acceleration). In this case the acceleration required is about 2.2 mph/second or 1.00 m/s2, the final velocity being 84 mph or 37.5 m/s. So

Pmax = M*a*V = 7.3 x 104 W = 98.3 hp.

That's less than the engine is rated for (120 hp) by a pretty good margin.
 
  • #4
For you to go from 55-84mph at constant acceleration over 0.25mi would take about 0.003579hr ... about 13secs.

Ages since I've worked in non-SI units... let's see:

Work = (1.0143e+6 - 4.3483e+5) = 579470 ft.lbs
Power = 44575 ft.lb/s or about 80hp.

(using 1hp = 550 ft.lb/sec)

So if I got the numbers right - you probably deserve that ticket.
[edit]ah - got overtaken.

<head slap>
Of course, W=Fd and F=ma so W=mad :) so P=mad/T ... vave=d/T so P=mavave
Which is why gneill has a bigger number than me - his is peak power.
 
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  • #5
Drag coefficient
Fd = 1/2 rho v^2 Cd A

Cd =0.3 assumption
A = Frontal area = 0.6m^2 assumption
rho = density = 1.3 kg/m^3
v = 37.5m/s ie 84nph

Fd = 1/2 ( 1.3 ) ( 0.3 ) ( 37.5 * 37.5 ) ( 0.6 ) = 164 nt

P = F v = 164 * 37.5 = 6150 watts =~ 10hp

Assume rolling resitance is also 10 HP

total HP to go 84 mph steady velocity = 20HP

Also , assumptions are low, double the HP ( wild guess )
So, total HP to go 84 mph = 40HP

Acceleration HP + Drag HP = 98 + 40 = 138 HP

Vehicle max HP at 2500rpm = 125 HP
Drop off HP at 4000 rpm = 50HP ( wild assumption )

Available HP at 84mph = 125 - 50 = 75HP

HP deficit at 86 mph = available - needed = 75HP- 138HP = -63HP

You are short 63 HP to go 84 mph.
You need a bigger and more powerful engine.

This is a gross calculation, but it looks like your vehicle will not reach 84mph starting from 55mph in a one quarter mile stretch.
Your ticket seems unjustified by this estimated analysis.
 
  • #6
256bits said:
<snip>

Also , assumptions are low, double the HP ( wild guess )

<snip>

Drop off HP at 4000 rpm = 50HP ( wild assumption )

<snip>

Your ticket seems unjustified by this estimated analysis.

I suspect that the policeman's wild assertions will trump your wild assumptions in court :-p :biggrin: :smile:
 
  • #7
gneill said:
I suspect that the policeman's wild assertions will trump your wild assumptions in court :-p :biggrin: :smile:

I imagine so.
PS 5 minute engineering analysis at its finest :)

And i am glad you replied, as you did catch the gross assumtion part which I should have spelled out explicitly as to how difficult the OP case in court would be.

Unless the OP has more exact figures than mine of drag force (Cd and frontal area) and horsepower drop off, any sane judge would not accept. To get valid certifiable figures for a particular car would cost more than the ticket.


What the OP needs is:
Lawyer Vincent Gambini (Joe Pesci - one of my favourites ) from the movie My Cousin Vinny , with a mechanic girlfriend Mona Lisa ( Marisa Tomei )
 
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  • #8
The assertion is that the vehicle is incapable of the alleged performance.

The way to support this is with empirical testing not theoretical calculation - remember, the officer is claiming to have measured his speeds somehow. Also - how was the accelerating distance measured? After all, the officers presumably only claiming the final speed, not how you got there.

So you need to get someone to attempt to repeat the feat in your car - preferably some kind of expert. You also need to be able to show that your car was not tampered with before the test ... or use another car the same make and model/year.

The complimentary calculation is to work out the minimum runup this car would need to hit 84mph.

We used to say:
A physical theory is a set of published statements that are believed by nobody but the author. An experimental result is a set of published statements that are believed by everyone except the experimenter.
 
  • #9
Add:Energy need to Accelerate 4300 lbs from 55 mph to 84mph within a quarter mile?

First of all thank you to all for your replies.

Simon Bridge wrote"
"The assertion is that the vehicle is incapable of the alleged performance."

Right! Some thoughts: (I'm aware I'm drifting away from the forums core interests)

In 1986 the national speed limit was 55 mph. Vehicles from that period typically have a speedo that ends at 85mph like mine with 55 mph being marked red and/or larger.
Emission control was still in it's infancy and the typical trade off was low hp performance.
Like mine, 4.2 liter 6 cyl, pushrod (OHV) cast iron engine with laughable 125hp max at 3,600rpm. Compare that to a today's 2 liter 250 hp 7,000rpm "green" emission engine!
Since I am required in California to pass emission (smog) test all is and must be original.

Note:
A) Pushrod engine, a 1920 engineering concept: 1928, Arthur Chevrolet filed with the US Patent Office for an 'Overhead Valve Engine'. Patent #1,744,526
B) Some specific problems that remain with overhead valve (OHV) engines: Limited engine speeds or RPM. Over rpm can easily cause valve "floating" i.e. oscillation and pushrods can flex, bend or snap.

A Proof: Given the tire diameter, rear axle ratio 3.7 and direct (4th) gear the engine would have to rev at over 3900 rpm! Given the age and wear, it barely makes it to 3,200 rpm these days and if, only in first or second gear.

Tire diameter = 27" x pi = 84 3/4" ≈7ft circumference
1 mile = 5280 ft
84 miles (p/hr) = 443,520 ft divided by 7ft = 63,360 tire rotations p/hr divide by 60 = 1056 tire rpm p/min.

1056 x 3.7 (rear axle gear ratio) = 3,907 engine rpm...
That's 10% over the mfg max rpm for the engine; well into the redline.
An engine that's 25 years old and has over 315,000 miles on it and age related lower compression... (Originally 7.8:1)

...I'd be picking up part and pieces all over the highway...

I'll be Vincent Gambini and Mona Lisa in one! Cousin Vinny here I come!PS I love 256bits last line, VERY TRUE! Thanks! :-D
 
  • #10
I've driven a car from that era - it was a Caddy and cost NZ$5 just to start it. It is kinda hard for a modern driver not to notice they have passed the double-nickle though people used to do it all the time back in the day. At least it explains the power to weight ratio.

Even so - what was the officer basing his allegation on?
If he's just guessing then you may be able to overturn it - but if he got you on radar then you'll have a harder time.

Did he really clock you at 55 a quarter-mile earlier?

If you want to be sure, then you'll need empirical evidence backed by an expert-ish witness. Take your car to a track and see how close anyone can get to that 85mph. I've seen a Buick resurrected from some field go on to win dirt-track races (briefly) against modern cars. Maybe you can get it tested by the AA. But the others are right: none of these calculations will stand up in court by themselves.

See what you are up against?
 

FAQ: Energy need to Accelerate 4300 lbs from 55 mph to 84mph within a quarter mile?

How much energy is needed to accelerate 4300 lbs from 55 mph to 84mph within a quarter mile?

The amount of energy needed to accelerate an object depends on its mass and the change in velocity. In this case, the 4300 lbs car will need more energy to accelerate from 55 mph to 84 mph than a lighter car. The exact amount of energy can be calculated using the formula E = 1/2 * m * (v2^2 - v1^2), where E is the energy, m is the mass, and v1 and v2 are the initial and final velocities, respectively.

What type of energy is needed for the acceleration?

The energy needed for acceleration is kinetic energy, which is the energy an object possesses due to its motion. In this case, the car will need an increase in kinetic energy to accelerate from 55 mph to 84 mph within a quarter mile.

How does the weight of the car affect the energy needed for acceleration?

The weight of the car plays a significant role in the amount of energy needed for acceleration. As the weight of the car increases, so does the amount of energy needed to accelerate it. This is because heavier objects have more inertia and require more force to accelerate.

Can the energy needed for acceleration be reduced?

Yes, the energy needed for acceleration can be reduced by making changes to the car's design, such as reducing its weight or improving its aerodynamics. This can also be achieved by using more efficient engines or alternative fuels.

What are the potential consequences of using a large amount of energy for acceleration?

Using a large amount of energy for acceleration can have several consequences, such as increased fuel consumption and emissions, which can contribute to air pollution and climate change. It can also put a strain on the car's engine and other components, potentially leading to maintenance issues. Additionally, using excessive energy for acceleration can also be a safety hazard if not done properly.

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