Calculating average power of a set of data

Click For Summary

Discussion Overview

The discussion revolves around the calculation of average power for a group of individuals running up a slope, exploring two different methods of calculation and their resulting discrepancies. Participants analyze the implications of their approaches and the mathematical principles involved, including the harmonic mean.

Discussion Character

  • Debate/contested
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant proposes two methods for calculating average power: one based on individual power calculations and the other on average mass and average time.
  • Another participant argues that the second method is invalid, highlighting that it can lead to misleading results due to the influence of larger values on the mean.
  • This participant introduces the concept of harmonic means, suggesting that the average power calculation should consider the reciprocal of time.
  • A later reply reflects on the relationship between average velocity and the calculation of average power, drawing parallels with the harmonic mean and suggesting a similar approach for power calculations.
  • There is acknowledgment of the complexity involved in averaging quantities that are not directly additive, such as power and speed.

Areas of Agreement / Disagreement

Participants express differing views on the validity of the second method for calculating average power, with no consensus reached on which method is correct. The discussion remains unresolved regarding the best approach to take.

Contextual Notes

Participants note the importance of considering the nature of the quantities being averaged, particularly in relation to time and mass, and the potential for misleading results when using simple averages in these contexts.

kurt
Messages
8
Reaction score
0
There are n people with masses mn . Each run up a given slope of height h in times tn. Gravitational accelartion, g, is constant.
Now there are 2 ways to calculate the average power of this group of people with 2 different results. The question is which way of calculating the average power (hence which result) is correct.
1st way: I calculate the power of each individual: Pn =mn g h/tn and take the average of these Pn. This is the average power of the group calculated using the first way: Pavg.1 = (P1 + P2 + P3 +…….+ Pn ) /n
2nd way: I take the average of n masses, mavg and the average of n times, tavg, then use the power formula and obtain the average power of the group using the second way: Pavg.2 =mavg g h/tavg .
When I do the two ways on the Excel I get slightly different results. Here is an example:
g (m/s2) h(m) m (kg) t (s) P=mgh/t (W)
9,81 2,24 51 7 7844,861
9,81 2,24 48 7 7383,398
9,81 2,24 46 4 4043,290
9,81 2,24 47 8 8262,374
9,81 2,24 51 5 5603,472
9,81 2,24 49 4 4306,982
9,81 2,24 50 6 6592,320
9,81 2,24 52 4 4570,675
9,81 2,24 54 7 8306,323
9,81 2,24 48 5 5273,856
9,81 2,24 47 8 8262,374
9,81 2,24 51 4 4482,778
6244,392 ←average of the individual powers
9,81 2,24 49,5 5,75 6254,464
average m average t power ↑ of averages

So which way (result) would be correct? By the way, plotting t against m and using the slope of the best fit line to calculate power is not appropriate here, as the slope comes out to be negative (-0.087) and the correlation is very weak (R2 = 0.017). I greatly appreciate your comments.
 
Physics news on Phys.org
Your second method is not really valid. Just take a simple example of two runners who do it once, each. (nonsense values to make it obvious)
If one runner had mass 100kg and ran up 10m in 10s - power is mgh/t = 100*10*g/10 =100g
The next runner weighed 11kg and ran up 10m in 1s - power = 11*10*g/1 = 110g
Mean of those two values is 105g
taking both together, the power would be (by your calculation) 111*10 *g/11 =101g
The data is dominated by the larger value.

What you are dealing with is Harmonic Means (you are dealing with 1/t) and you need to be careful when they come into it.

There is a similar problem when working out average speeds around a circuit. If you move round a square course (side length d) with the speeds on each side being 1m/s, 2m/s, 3m/s and 4m/s and want to work out the average speed, the 'obvious' choice is to say (1+2+3+4)/4 = 2.5 but the real answer involves finding how long it takes actually to cover each side
t=d+d/2+d/3+d/4 =2.25d
then the speed is
4d/t = 1.78

In this case (and we know from experience) the distances traveled at slow speed will dominate the average speed. There is a limit to just how much you can 'make up', on the fast bits for the slow bits.
That's the Harmonic Mean for you! Watch out for it.
 
Hi sophiecentaur, thank you much for your comments..
 
Does it make sense now?
 
Thank you for your follow up sophiecentaur. Rethinking about it, I got another reasoning out of your reminding me taking the average of velocities (or simply the average velocity). In calculating the average velocity of several known paths of d1 to dn with known times t1 to tn, but with unkown velocities on these paths, we divide the total displacement by the total time. v = Σd/Σt. This is the most general case of finding the average velocity for the whole path. Yes, with the unknown but equal distances taken with known velocities and known times, we get the harmonic mean of these known velocities. Similarly, as you know, if the times are equal on different length of paths then it will yield to the arithmetic mean of these given velocities.
It is exactly the same procedure with the densities of mixtures. Since both have the same type of formula: v = Σd/Σt for velocity and d = Σm/ΣV for density.
Now I see the same structure of formula for power: P = gh m/t. since gh is only a constant factor, it can be taken to unity, or to the left side. Then we would have P’ = m/t, which has the same form with m and t known parameters, and P (orP’) as unknown parameter, of which we seek the average. Since this looks like the most general case of velocities above, this leads me to apply the same procedure. i.e. Pavg = gh Σm /Σt .
I had not thought of this before your response. Thank you much for your time and thoughts.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
9
Views
2K
  • · Replies 169 ·
6
Replies
169
Views
16K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 25 ·
Replies
25
Views
12K
  • · Replies 8 ·
Replies
8
Views
11K
Replies
8
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
5
Views
4K