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Greetings:

Your proposed experiment is very interesting and

I’ll present the steps that you will need to

calculate horsepower from you measurements.

However, inertial (accelerometer)measurements are

great for rapid changes but they do not measure

power in steady state conditions. I’ll go through

an example experiment to explain the procedure and

it’s limitations.

I will use English system for engineering units (lbs,

ft, sec)in my example because you are probably more

familiar with these units than the metric system most

engineers now use. To demonstrate steady state

conditions I’ll simulate a Jaguar’s acceleration of 0

to 60 mph in 4 seconds and then back off when it

reaches 114 mph rather than continuing up to the Jag’s

200 + mph. I don’t know the Jag’s weight so I’ll use

your cars weight of 3000 pounds.

One g of acceleration at sea level is 32.174 ft/sec^2

(feet per second squared). One half g = 16.082

ft/sec^2, one tenth g = 3.217 ft/sec^2 etc.

The mass in English engineering units is named “slugs”

and is the weight divided by g (m= w/g).

The mass of your car in the English system of units is:

m = 3000 pounds divided by 32.174 = 93.24 slugs

For this simulation I will use the time between samples

as one second (Delta t= 1 sec). Your sample time will be

Delta t= 0.05 sec. My simulation will accelerate to 1 g

in 3 seconds and drop back to zero g at 10 seconds

There are 2 steps in the process:

Step 1. From the acceleration data plot a curve to

graphically determine the velocity (v) as a function of

time (t). The area under the acceleration (a) curve

from t=0 to a given time is the velocity at the

measurement time. My acceleration curve is in Gs which

is converted to feet per second squared (ft/sec^2)by

multiplying the g force by 32.174.

To obtain the velocity we average the acceleration

before and after the sample and divide it by the sample

time and add the previous velocity to the result.

For example the velocity (v1) at one second (t1) is;

v1=((a0+a1)/2 *Delta t)+a0=((0.00+12.868)/2) *1)+0 = 6.434

v2=((a1+a2)/2 *Delta t)+a1=((12.868+25.736)/2)*1)+6.434=

25.736

etc.

If we plot the velocity as a function of time we obtain

the distance (d)traveled.

For example the distance d1 at t1 is

d1=((v0+v1)/2 *Delta t)+d0=((0.00+6.434)/2)*1)+0 = 3.22

d2=((v1+v2)/2 *Delta t)+d1=((6.434+25.736)/2)*1)+3.22=

19.30

etc.

Time__acceleration____velocity-MPH_distance

(sec__Gs_(ft/sec^2)___(ft/sec)_MPH_____(ft)

t0___0.0____0.00____0.000___0.00____0.00

t1___0.4___12.87____6.434___4.39____3.22

t2___0.8___25.74___25.736__17.55___19.30

t3___1.0___32.17___54.689__37.29___59.51

t4___1.0___32.17___86.859__59.22__130.29

t5___0.7___22.52__114.204__77.87__230.82

t6___0.6___19.30__135.114__92.12__355.48

t7___0.4___12.87__151.199_103.09__498.64

t8___0.2____6.43__160.850_109.67__654.66

t9___0.1____3.22__165.676_112.96__817.92

t10__0.0____0.00__167.284_114.06__984.40

t11__0.0____0.00__167.284_114.06_1151.69

Step 2. Using the velocity and fixed mass (m) of the

car, you then calculate the kinetic energy of the car

as a function of time.

Kinetic energy (KE)equals the car’s mass (m)divided by

2, times the velocity (v)squared (^2))

KE = (m/2)*v^2

t0 KE0 = 0

t1 KE1 =(93.24/2)* 6.434^2 = 1929.90

t2 KE2 =(93.24/2)* 25.736^2 = 30878.37

t3 KE3 =(93.24/2)* 54.689^2 =139435.14

etc.

The instantaneous power (P) equals the change in work

(Delta W) done, divided by the change Delta t the time between

samples.

However, the change in work is (Delta W) =(Delta KE)

Therefore the Power (ft lb/sec)= the change in kinetic

energy over the sample time (Delta t).

P = (Delta KE)/(Delta t)

There are 550 ft lb/sec per horsepower therefore

HP = P/550

P1 = (1929.9-0)/1 sec =1929.9 ft lb/sec

or 1929.9/550 = 3.51 HP

P2 =(30878.37-1929.9)/1 sec = 28948.47ftlb/sec 52.63HP

etc.

Time_____KE_____Power_______HP

(sec) (ftlb/sec)

t0_________0.0_______0.0______0.0

t1______1929.9____1929.9______3.51

t2____30878.37___28948.47____52.63

t3___139435.14__108556.77___197.38

t4___351723.93__212288.79___385.98

t5___608038.53__256314.59___466.03

t6___851085.07__243046.54___441.90

t7__1065786.23__214701.17___390.37

t8__1206186.32__140400.09___255.27

t9__1279643.07___73456.75___133.56

t10_1304611.13___24968.06____45.40

t11_1304611.13_______0.00_____0.00

Now here is the problem that I mentioned up front.

After 10 seconds the velocity does not change. Thus the

kinetic energy does not change and stays constant. Thus

the work is zero (W = 0) and the power is zero (P =0).

How can a car traveling at 114 mph use zero horse
power?

The answer is that it does use a great amount of power.

Your inertial measurements only measures the power and

the change in power needed to get the car’s mass to 87

mph. At a steady velocity of 114 mph, power is also

used to overcome aerodynamic drag and friction. Power

must overcome friction in the transmission and drive

train, is needed to drive the power steering and to

overcome the rolling friction of the wheels and tires.

Also, the alternator is very power hungry and it turns

out that at highway velocities many of today’s smaller

cars use one half of the available horsepower to run

the electrical accessories! Soon electrical power

control is not lost when the engine fails adding to the

power load.

Your inertial measurements do not include these steady

state power requirements. However, they were included

in your acceleration data before the car reached a

steady velocity.

Inertial navigation systems such as you are using are

calibrated by using true ground distance compared to

calculated distance. Therefore you should measure the

actual time to the distance of max g, about 130 ft in

my example, and also measure the time to reach steady

velocity.

The actual and the calculated distances should be close

during the high g acceleration and perhaps at steady

state. Inertial measurements are not effected by wheel

slippage and in the near future all cars will use

accelerometers in 3 D to operate , air bags, anti-skid,

positive traction and anti-roll over control systems in

place of todays sensors that measure wheel rotation

which is slip sensitive.

If your automobile has a manual transmission you might

put the transmission in neutral at high velocity and

use your accelerometer to measure the deceleration

caused by aerodynamic drag and friction with the engine

and transmission disconnected from the system. Plot the

deceleration data and calculate the breaking horsepower

and velocity reduction and distance in the same way as

the acceleration curve was used. It would be very

interesting to compare the drag and friction horsepower

at each velocity with the acceleration horsepower.

So how are engines measured in a steady state manner?

We measure the engines “break horsepower” over it’s

operating RPM with varying loads. In simple terms the

engine is placed in a test stand and the engine output

shaft is loaded with precise amounts of frictional

force. Something like large break drums rubbing on

large break pads. Depending on the loading (breaking)

force, the friction system converts all of the engine

output power to heat. One horse power is also equal to

746 watts of heat. The rate of heat generated under

loading gives a true measure of the engines output

power without the power losses to external sources such

as the power train and aerodynamic drag. Also, there

are dynamometer test stands that load a cars drive

wheels with a resisting force. These measure the power

delivered to the wheels including all of the drive

train losses.

Good luck with your experiments. After a number of

checks by myself and John Link the Mad Science

administrator, I believe my spreadsheet calculations

are correct. However, you might find a error or two

caused by round off error.

Your Mad Scientist

Adrian Popa

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