Argthtjtr

Averaging destroys information. Do it as late as is practical in your analysis.

A commenter points out that, if you are making many position/velocity measurements of the same object as it moves once, a simple linear regression is more robust as a velocity estimator than an ensemble of point-by-point analyses.

Calculate velocity ten times and then average it. If you were looking for average distance, then average the distance, if you were looking for average times, then average the times.

What you are doing doesn't even give the same number as the average velocity:

Averaging rates is tricky.

To see why, let's use a different example

You're driving a car over a hill. For the first mile, you're going up the hill and you use 1/20th of a gallon of gas. On the second mile you coast down the hill using 1/100th of a gallon. What is your average gas mileage?

For the first mile, you were traveling at 20 mpg, and for the second mile, you were traveling at 100 mpg. So your average is 60 mpg, right?

Wrong!

Let's try a slightly different problem.

You're going up the same hill again. Getting to the top requires another 1/20th of a gallon, but this time you turn off your car at the top and glide down to the bottom without using any gas at all. What is your average mpg this time?

So let's see... add 1 mi / .05 gal + 1 mi / 0 gal, then divide by 2... good news! Your average gas mileage was infinite! You can travel anywhere in the world without ever running out of gas, and all you need to do is start on a hill!

Obviously, this result is ridiculous.

Summing first, and then dividing ( 2 mi / .05 gal = 40 mpg ) gets us a far more useful answer.

To understand why, I'll leave you with two final problems.

Bob went for a drive. He drove for 1 mile at 30 mph, then he drove 2 miles at 60 mph. What was his average speed?

Alice went for a drive. She drove for 1 hr at 30 mph, then she drove for 2 hrs at 60 mph. What was her average speed?

Now consider: if you have two different trials that took different lengths of time, which one will contribute the most to your result? Is that what you want?

There's no such thing as a "right" way to summarize data. The best you can do is to understand how your summarization formula affects the results.

My original intuition was this:

If you measured $x$ and $t$ once for each trial, then in your case $v$ is a function not a measurement. Every measurement generates its own pair of $x_i$ and $t_i$, which implies $v_i$. Thus you get $hat v$ by averaging $v$. $x$ and $t$ depend on each other, as opposed to $x_i$-$x_j$, so it doesn't make sense to average $x_i$.

If you doubted your ability to precisely measure $x$ and $t$ (as well you should) then in each trial $i$ you would make $n$ measurements of $x_{ij}$. Now before you can do $v_i=x_i/t_i$, you must first have an estimate of $x_i$, which is given by $frac{1}{n} sum_j{x_{ij}}$. Thus you average $x$ for that trial, ditto $t$, get a $v_i$, repeat for each trial, and then average $v_i$.

However, I tried to simulate it and was surprised to find that it seems like there is hardly any difference. Below are sections of my R notebook along with results. Maybe I have a bug?

pacman::p_load(tidyverse, ggplot2)

Function to simulate an imprecise measurement:

# Relative measurement error

em = 0.01



measure = function(x, n) {

  # Attempt to get the value of a quantity x, and n measurements

  x_measured = mean(x*rnorm(n, 1, em))

  return(x_measured)

}

Let's test it with some simulated measurements:

df = expand.grid(case=1:3, measurement=1:5)

df$result = Vectorize(measure, vectorize.args = 'x')(replicate(nrow(df), 1), 1)



p = ggplot(df) + 

  geom_point(aes(y=result, x=as.factor(case))) +

  theme_classic()

show(p)

We expect repeated measurements to converge, but with diminishing returns:

df = expand.grid(case=1:1000, m=1:50)

df$result = Vectorize(measure, vectorize.args = 'n')(1, df$m)



df_sum = df %>% group_by(m) %>% summarise(mu = mean(result), sigma = sd(result), measurements = first(m))



p = ggplot(df_sum) + 

  # geom_boxplot(aes(y=result, x=measurements, group=measurements)) +

  geom_pointrange(aes(x=measurements, y=mu, ymin=mu-sigma, ymax=mu+sigma)) +

  theme_classic()

show(p)

We set up the ball experiment: The ball thrower has a velocity setting (the exact value of which we don't know) and the camera has a timer setting (the value of which we also don't know). So both v and t converge to one value, but have some experiment error. We try doing several experiments:

# Set ball roller to a certain energy level (value unknown to you)

v_expected = 5



# And the camera to photograph after a certain delay (value unknown to you)

t_expected = 20



# Relative apparatus error

ea = 0.001



# Number of trials

n = 10



# Number of experiments

k = 1000



df = expand.grid(experiment = 1:k, trial=1:n)



# Roll the ball for each trial, but the machine is slightly faster or slower sometimes

df$v_actual = v_expected*rnorm(nrow(df), 1, ea)



# The camera's timer isn't very consistent either

df$t_actual = t_expected*rnorm(nrow(df), 1, ea)



# We don't know the true distance yet, but nature does

df$x_actual = df$v_actual * df$t_actual



# Visualize

p = ggplot(df) + 

  geom_point(aes(x=x_actual, y=t_actual, color=v_actual), size=0.5) +

  theme_classic()

show(p)

Now we try measuring our experimental outcomes:

# You try to measure the distance, but your ruler isn't very accurate

df$x_measured = Vectorize(measure, vectorize.args = 'x')(df$x_actual, 1)



# You also tried to measure the time, but your stopwatch is not the best

df$t_measured = Vectorize(measure, vectorize.args = 'x')(df$t_actual, 1)



# Number of repeat measurements

m = 20



# What if you measured x multiple times?

df$x_hat = Vectorize(measure, vectorize.args = 'x')(df$x_actual, m)



# And had multiple assistants, each with their stopwatches?

df$t_hat = Vectorize(measure, vectorize.args = 'x')(df$t_actual, m)

Of course multiple measurements are much better:

df_sum = df %>% gather(key='method', value='measurement', c(6,8)) %>%

  mutate(error=abs(measurement-x_actual)/x_actual)



# Visualize

p = ggplot(df_sum) + 

  geom_boxplot(aes(x=method, y=error)) +

  theme_classic()

show(p)

But does it matter when we average?

df_sum = df %>% mutate(v_measured = x_measured/t_measured) %>% group_by(experiment) %>% 

  summarise(v_bar = mean(v_actual),

            t_bar = mean(t_actual),

            x_bar = mean(x_actual),

            v_bad = mean(x_measured)/mean(t_measured),

            v_good = mean(v_measured),

            v_best = mean(x_hat/t_hat),

            early = abs(v_bad-v_expected)/v_expected,

            late = abs(v_good-v_expected)/v_expected,

            multiple_measurements = abs(v_best-v_expected)/v_expected) %>% 

  gather(key='method', value='rel_error', 8:10)



df_sum$method = fct_inorder(df_sum$method)



# Visualize

p = ggplot(df_sum) +

  geom_boxplot(aes(x=method, y=rel_error)) +

  theme_classic()

show(p)

Calculate your velocities and then report the following:

• Average (mean, median, or mode. Whatever makes sense for your application)


• Spread (Range, stdev, 95%CI, etc... Again which one depends on your application)


• Number of samples (in this case, n=10)

If you leave out any of the above, your number becomes less useful for the next person looking to build off of your work.

Since its only 10 measurements, it is easy enough to include them at the end in a supplement. That way if someone is unhappy with the values you reported, they can calculate their own charaters

I assume you're talking about a straight line (so that by velocity you just mean speed)?

In all honesty I'm open to being wrong about this, but to me your method is what makes sense.

My reason is that repeatedly flying a distance 10 times is the same as flying 10 times the length of that distance only once, and hence I'd expect you to get the same average speed for both. Otherwise you're saying that the average speed over a longer distance can be a different value depending on how you choose to interpret chopping it up, which doesn't really make sense to me.

You can try to figure out which one has more error by working it out by the way... I think Markov or Chebyshev's inequalities could be helpful if you assume bounded moments etc.

But if you're like me and this doesn't sound too fun to you, just simulate a bunch of measurements using Gaussian noise with code and see which estimate ends up being more accurate. I tried to quickly whip up some code and the first one seemed to be more accurate.

For each of the 10 measurements you should compute the target quantity - velocity in your case - and then average the resulting values.

The error estimate however should be derived from the uncertainty of the actually measured quantities - that is time and distance for your example. This is called propagation of uncertainty and for the velocity estimate it is computed as follows:

$$
sigma_v = lvertbar{v}rvertsqrt{frac{sigma_L^2}{bar{L}^2} + frac{sigma_T^2}{bar{T}^2} - 2frac{sigma_{LT}}{bar{L}bar{T}}}
$$

where $bar{x}$ denotes the average of $x$.

So for example you collected the following data points (velocity already computed from the distance and time pairs):

       Time   Distance  Velocity

0  5.158157  10.674957  2.069529

1  4.428671   9.234457  2.085153

2  4.967043   8.768574  1.765351

3  4.777810   9.754345  2.041593

4  5.237122   9.833166  1.877589

5  4.813971   9.861635  2.048545

6  5.108288  10.432516  2.042272

7  4.668738  10.344293  2.215651

8  4.642770   9.806191  2.112142

9  5.012954   9.992270  1.993290

Then the average velocity is computed as $bar{v} = 2.025112$.

For the error estimate we consider the standard deviation as the measurement uncertainty for distance and time respectively, $sigma_{x = L,T}^2 = mathbb{E}left[(x - bar{x})^2right]$: $sigma_T^2 = 0.066851, sigma_L^2 = 0.315579$. The covariance is computed as $sigma_{LT} = mathbb{E}left[(L - bar{L})(T - bar{T})right] = 0.052633$; $mathbb{E}$ denotes taking the average as well.

By using the above relation for the propagation of uncertainty we obtain a measurement uncertainty on the velocity:

$$
sigma_v = 2.025112cdotsqrt{frac{0.315579}{9.870240^{,2}} + frac{0.066851}{4.881553^{,2}} - 2frac{0.052633}{9.870240cdot 4.881553}} = 0.125817
$$

Note that the result differs only slightly from the standard deviation on the velocity itself but for the general case, depending on the uncertainties of the measured variables, a larger discrepancy might occur. In the end it is the measured data that is afflicted with an uncertainty and for any derived quantity also the measurement uncertainty must be derived.