How do you calculate the average force on a piston?

[KitsonSteam22]

1. How do you calculate the average force on a piston in one stroke from TDC to BDC given the following Kg forces and their distances in mm from TDC?

1 mm 2 mm 3 mm 4 mm 5 mm 6 mm 7 mm 8 mm 9 mm 10 mm 12 mm 13 mm
9.9 Kg 7.8 Kg 6.3 Kg 5.3 Kg 4.4 Kg 3.8 Kg 3.2 Kg 2.8 Kg 2.4 Kg 2.1 Kg 1.6 Kg 1.4 Kg14 mm 15 mm 16 mm
1.3 Kg 1.1 Kg 1 Kg
2. I am absolutely blank.3. Tried a sample average, but as the sample increases so the average drops.

 

[BvU]

Hello Kitson,

Surely you can't let the sample increase further than the BDC ?
TDC is at 0, but what is the stroke length ?
You could let excel fit a polynomial and integrate, but you still need the BDC position.

 

[KitsonSteam22]

Hey BVU
Correct, any force still around at BDC is waste.
BDC is 30 mm and the force is 0.2 Kg so insignificant. I did not put the last figures in because they are easy to include once I understand the solution.
I'm old school, pensioner, specialized in rotor dynamics and balancing so very grey in this area.
Total stroke is 31 mm.

Regards

 

[KitsonSteam22]

Here are the rest of the distances and forces:
17 mm 18 mm 19 mm 20 mm 21 mm 22 mm 23 mm 24 mm 25 mm 26 mm 27 mm 28 mm 28 mm 29 mm 30 mm
0.9 Kg 0.8 Kg 0.7 Kg 0.6 0.6 Kg 0.5 Kg 0.5 Kg 0.4 Kg 0.4 Kg 0.3 Kg 0.3 Kg 0.3 Kg 0.3 Kg 0.2 Kg 0.2 Kg

 

[Irene Kaminkowa]

Try statistical approach.

Suppose x ∈ [a, b], and the interval [a, b] is partitioned into n subintervals, each of length ∆x = (b – a)/n, so that n = (b – a)/∆x. Then, the mean of f(x) is reasonably well approximated as
$$\mu = \frac{\sum_{i=1}^{n}f(x_i)}{n}$$

 

[BvU]

I had fun going through this:

A 6th order (bit much..) polynomial (the red line) gave
y = -0.00000408x⁵ + 0.00038121x⁴ - 0.01390910x³ + 0.25371572x² - 2.47577020x + 11.96556136

Integrating from 0 to 31 gives the area under the red curve
A = -0.00000408x⁶/6 + 0.00038121x⁵/5 - 0.01390910x⁴/4 + 0.25371572x³/3 - 2.47577020x²/2 + 11.96556136x with x = 31 that gives an average of 68.7/31 = 2.22 (if I didn't make any typos...)

A simple average gives 2.05

If you add an extra measurement at TDC of 12 kg, then a direct average gives 2.29.

However, this gives too much weight to that TDC point (it is an estimate for 0-0.5 mm, the measured 10 kg is for 0.5 to 1.5 mm) so a better estimate is to give that point half the weight of the others:

(sum of measurements + 12/2)/30.5 = (61.4+6)/30.5 = 2.21

(close to the integral value)

 

[KitsonSteam22]

Try statistical approach.

Suppose x ∈ [a, b], and the interval [a, b] is partitioned into n subintervals, each of length ∆x = (b – a)/n, so that n = (b – a)/∆x. Then, the mean of f(x) is reasonably well approximated as
$$\mu = \frac{\sum_{i=1}^{n}f(x_i)}{n}$$

Hi Irene

I like your suggestion, the last time I did stats was in the 1980s. I would be interested to see the result if you ran the figures. Con't stop thinking that an effective average would give a more accurate result. A bit like the rifle bullet accelerating down a barrel, versus a shotgun coasting down the barrel. The rifle having the slower burning propellant??

 

[Irene Kaminkowa]

KitsonSteam22, if you chose statistics, do not forget to calculate f(11), since it's omitted. Use linear approximation: f(11) = (f(10) + f(12) )/ 2

There are also methods to find the integral by the points given: method of left(right) rectangles, method of trapezoids, Simpson's method and many others.
https://en.wikipedia.org/wiki/Rectangle_method
https://en.wikipedia.org/wiki/Trapezoidal_rule
https://en.wikipedia.org/wiki/Simpson's_rule
http://pages.cs.wisc.edu/~amos/412/lecture-notes/lecture18.pdf

 

[KitsonSteam22]

I had fun going through this:
View attachment 103129

A 6th order (bit much..) polynomial (the red line) gave
y = -0.00000408x⁵ + 0.00038121x⁴ - 0.01390910x³ + 0.25371572x² - 2.47577020x + 11.96556136

Integrating from 0 to 31 gives the area under the red curve
A = -0.00000408x⁶/6 + 0.00038121x⁵/5 - 0.01390910x⁴/4 + 0.25371572x³/3 - 2.47577020x²/2 + 11.96556136x with x = 31 that gives an average of 68.7/31 = 2.22 (if I didn't make any typos...)

A simple average gives 2.05

If you add an extra measurement at TDC of 12 kg, then a direct average gives 2.29.

However, this gives too much weight to that TDC point (it is an estimate for 0-0.5 mm, the measured 10 kg is for 0.5 to 1.5 mm) so a better estimate is to give that point half the weight of the others:

(sum of measurements + 12/2)/30.5 = (61.4+6)/30.5 = 2.21

(close to the integral value)

Hey BvU

Thanks, (Have a look at what I said to Irene's reply). There's defiantly weight in what you said about adding an extra measurement at TDC. In an engine, the firing occurs a couple of degrees before TDC (varies from manufacturer), a force is applied to the piston which can't go any where during the dwell angle, the potential energy then becomes kinetic energy as the piston clears the dwell angle (crank momentum). Maybe this is for somebody with more than the two brain cells that I have. I think you onto something I haven't read about.
Back to the average, I think looking at an effective average might give a better result.

regards

 

[Nidum]

There are well established methods for analysing what happens in an engine .

Can you tell us more generally what you are trying to understand or calculate ?