MadSci Network: Physics

Re: how to calculate horsepower from acceleration rate and weight?

Date: Mon Jan 22 22:25:09 2001
Posted By: Adrian Popa, Director Emeritus, Hughes Research Laboratories
Area of science: Physics
ID: 979271150.Ph


Your proposed experiment is very interesting and
Iíll 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. Iíll go through
an example experiment to explain the procedure and
itís 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 Iíll simulate a Jaguarís acceleration of 0
to 60 mph in 4 seconds and then back off when it
reaches 114 mph rather than continuing up to the Jagís
200 + mph. I donít know the Jagís weight so Iíll 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 ďslugsĒ
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=
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=


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 carís 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

The instantaneous power (P) equals the change in work
(Delta W) done, divided by the change Delta t the time between
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

(sec) (ftlb/sec)

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 carís 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 todayís 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

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 ďbreak horsepowerĒ over itís
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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