Efficiencies are the quotient of used to taken power (energy). What sounds very simple isn't that easy to measure sometimes. Within the
model-rr this value is referenced without manual how to get it. More: often real values are missing! The following text describes a simple
way how to get efficiencies, results can be ssen in Wirkungsgrade. Please note,
the resulting values neglecting the "power-factor" cos (phi) this will be treated later.
The secondary power P2 is easily to be estimated as product of pulling weight (in kg) * 9.81 * speed in m/s. The speed is the
real speed of the model train, to be measured by distance and time for this distance. The use of digital speed measuring models is
futile due to their inexactness!. The pulling weight correspond to the weight force neccessary to move the train. It can be measured by
special balances or digital force meters. Another way is to calculate it for any single car/coach from it's mechanical values, a possible
with the online-calculator at http://www.sheyn.de/Modellbahn/FAQ/Zuggewicht_e.php.
Keeping the voltage constant (e.g. at 12V for H0) and stopping the time for the given distance for each single appended car, You can get
the P2. It is useful simulateously taking the current draw.
The taken power is the product of voltage * netto current through the motor (if the so called "power-factor" can be neglected;
this is true for small frequencies). Don't forget to measure the current of any electrical device without the motor, since only motor
consumed current is of interest!.
Thus the taken power P1 is:
and the secondary power P2:
Resulting in the efficiency h (eta):
Ambiguous values are obtained in cases where any electronic device controls the motor in the model. In this case the reference-value
of the voltage has to be named for the efficiency! E.g.: a model pulling a train with the pulling weight of 112g (a pretty heavy train!)
with 0,3 m/s real speed. This equals to a P2 of about 0,33W. The voltage of the tracks is 12V, the net-current draw of the motor
is 195,9mA. The conncetor-voltage of the motor equals to 0,6 Utrack - 2,8V (in this case thus Uk = 4,4V).
In regard to Uk we have P1 = 0,86W and therefore h = 38,3%. With Utracks as
reference we have but P1 = 2,35W and then only h = 14%. Since the motor can not operated
without electronic, it isn't useful to use Uk as reference!
Additionally another effect has to be regarded when using impuls-control of the motor: the duty-cycles which is in out case 2/3. The
effective current is now 2/3 of the instantaneous current drawn during the impulses, whereas during the pauses no current flows and
thus no power is consumed. Therefore we have to include the duty cycle to the primary power (regarding impulses only, not the pauses!).
Our example results in P1 = 3,52W and so h = 9,4%. These details must be kept in mind!
Further informations are found in the german text at
Wie funktioniert die Modellbahn