Model railroaders frequently like to talk about scale speeds, and often quote the scale speeds at which their models run. However, it is very clear there are no physicists amongst them, because they way they scale speed is incorrect by most measures.

**How It's Done**

Let me back-up. I should first explain how modelers currently do it, and the reason why they do it the way they do.

To do this, I need to do a little math. Bare with me, because I don't need a lot of math, but for this article to be meaningful to anyone, a little bit of simple algebra needs to be invoked.

Speed is calculated by the equation

`v = d/t`,

where `d` is the distance travelled and `t` is the time it took.

Additionally, we will introuce an equation to take a real distance `d` and change it into a scale distance `d'` using the scale factor m

`d' = m*d`.

Now if I want to calculate the scale speed `v'`--according to the vast majority of modelers--I can do it by cleverly combining these too formulas. I must first observe that I can change out the scalable quantities in the velocity equation to produce

`v' = d'/t`,

and then combine that with the scaled distance equation to produce the relationship

`v' = m*v`.

For those of you that love examples, an HO scale railroad would give `m` a value of 1/87.1, or approximately 0.0115. Thus a track speed of 60 mph would be equivlent to very nearly 0.7 mph, which is 1 ft/sec.

**What About Time?**

Of course, the question we should ask ourselves, is why do we scale both the velocity and the distance in the above equation, but not time? Is there a fundamental reason why time should not be scaled?

To most, the reason is intuitive. It just seems absurd. And if you were to naively scale time in the same way we scale distance, you'd be right.

Going back to the ugly math, let us write the velocity equation with all the coordinates scaled, and then scale both distance *and* time by the factor `m`. Doing so produces the equation for scaled velocity as
`v' = d'/t'`,
and after relating it back to the original velocity through the scaling equations, `d'=m*d` and `t'=m*t`, we get

`v' = v`,

which is, of course, an absurd result!

Imagine having to run your model trains at an *actual* 60mph around the track! Your very expensive train, if it could even produce that much speed, would go flying off the track and probably injure someone in the process!

But this is not the end of the story for time scaling; Indeed, this is just the beginning.

**One Man's Scale Clock**

One of the justifications one hears for measuring scaled time the way it does, is because our concept of time is so thoroughly rooted in our own wrist watches. Indeed, when timing out the distance a model train goes, we run off to our wrist watches, happily counting off non-scaled seconds, and never even give it a second thought.

Human cognitive function is *so* rooted in this absolute concept of time, that the theory or relativity (which dictates that time runs at seemingly contradictorily different rates depending on where you are) is still impossible for most scientists to intuitively work with.

That being said, let's devise an interesting thought experiment that messes with our concept of a uniform clock.

In the real world, I set up a simple pendulum clock next to the train tracks, and seeing how far the next locomotive to pas the clock gets in one cycle of the pendulum.

Now, if we made a scale model of the simple pendulum clock, and put it next to our model train tracks, how fast (in the real world) would we need to run that model train to cover the scaled distance the real train traveled in one cycle of the pendulum?

In other words, given the scaled distance and the scaled concept of time from the ticking pendulum, how does the speed scale?

To solve this problem, one must (once again) invoke some math. More specifically, the first step is to find the equation that relates the length of one tick of the pendulum, `T`, with the length of the pendulum `L`. A little looking around the web will show you that the period of a pendulum's swing is given by

`T = 2*pi*sqrt(L/g)`,

where g is the acceleration of gravity, which is a constant at any position on the Earth. Combining that with the speed equation, we find that the speed of a the train is related to the pendulum's length via

`v = (d/(2*pi)) * sqrt(g/L)`.

If one scales the distance

`L` by the factor

`m`, and uses the techniques presented above, it is very easy to show the surprising result that the scaled velocity

`v'` is related to the real velocity

`v` by the relationship:

`v' = v * sqrt(m)`.

**Surprise!**

So what does that mean?

Think about it for a moment. Let's say, in the real world, we have a clock with a pendulum that swings a complete cycle once every second. This allows a train passing at 60mph to travel 88ft during that one second. Now if we made an HO scale model of this clock, and placed it next to our model train, we'd have to run the train at about 9.43 ft/sec in order to cover a scaled 88 ft distance (12 1/8 inches) in one tick of the pendulum! That is nearly 10 times faster than the speed of 1 ft/sec we calculated using the classic relationship.

Of course, the reason for this is that the clock itself runs faster. Indeed, a pendulum that is shorter by a factor of 100 will run at a rate that is 10 times faster than that of the longer pendulum.

Thus, in this example at least, it is clear that we have scaled not just distance, but also *time*.

**Superelevation**

Of course, I hear you say, this is just one example. One example that could have been carefully constructed to give the desired result. And I'm happy to hear that skepticism, because that is what real science is all about.

So let's try another situation entirely. Train curves are superelevated, just like roads, in order to make corners smoother. Ideally, this is done by finding the elevation angle such that the rails tilt the coach in a way that exactly balances the forces so that they pull straight down through the bottom of the coach, causing you to feel no lateral forces at all (and keeping the tea in your cup from spilling).

So the question is, if we built a scale model of a real world curve that is ideally superelevated, and we perfectly model that curve in our model train set, what speed would we have to run the model train at so that the forces are balanced for the model passengers with their model tea cups in the model coach?

Of course, to answer that, we need to know that the equation for relating the superelevation *angle* `theta` of the coach (and hence the rails it is on), to the curve's radius `R` and the design speed of the train around the curve `v`.

Fortunately, this is a problem that is solved in the first few months of class by every student of physics in the world, and the correct answer is always

`v^2 = g*R*tan(theta)`.

Applying the same techniques to this equations as the examples before it, we find something surprising: yet again, the relationship between the scale velocity `v'` and the real world velocity `v`, is given by

`v' = v * sqrt(m).`

In other words, if a real world curve has its superelevation perfectly designed to a 60mph train, then an HO scale model train would have go around that curve at 9.43 ft/sec to keep the tea in the model cup from spilling while going around the corner. (This, of course, being a good deal faster than the 1 ft/sec scale speed that modelers would calculate!)

**Welcome to the Movies**

It's hard to argue with physics, and so far I've given two entirely unrelated physical situations in which I have no choice but to produce the same, very surprising result.

But, being the stubborn kind of people we are, it is hard to accept radically different ideas (indeed cognitive dissonance is a very fascinating field of psychology), and so I fear I must produce one more example before you actually believe what it is I have to say.

Imagine you are a special effects artist working for Robert Zemeckis and Bob Gale, and are in charge of the crash sequence at the end of Back to the Future III, where the beautiful steam engine drives off the cliff at 88mph, and plummets into the Earth below.

Obviously, Sierra Railway will *not* let you drive their prized engine number 3 off a cliff, so you must resort to models to do it!

Being the sort of person you are, you want the shot to look realistic. This requires you to satisfy two criteria:

- You must find the correct speed to drive your model train off the cliff so that it follows the exact same parabolic path through the air as the real thing would, except scaled down according to the scaling equations we now know and love.
- You must run the film through the camera at a faster rate so that, when slowed back down to the standard 24 fps of a theater projector, it
*looks* like it took the same amount of time to hit the ground as the real thing, even though it actually hit much sooner due to the much shorter distance.

Now the math for this one is much too ugly to show here, but rest assured that over the last year I have calculated it several different times, several different ways, and always come to the same conclusions:

- Velociy scales by the same
`sqrt(m)` factor as we found in all of the previous examples.
- Time scales by the
`sqrt(m)` as well!

**Time Does Scale**

So we've determined that velocity scaled according to the square root of the scale factor, and that time also scales by the square root of the scale factor. These are not independent results! Indeed, following the lead of theories such as special relativity, the fundamental equations are best taken as atransformation of the basic coordinates; that is to say, the basic equations are:

`d' = m*d`

`t' = t*sqrt(m)`

One can then easily *derive* the velocity transormation equations by applying these transforms to the scaled velocity equations `v'=d'/t'` to get the velocity transform equation `v'=v*sqrt(m)`.

**But Why?**

But *why* does time scale in this seemingly absurd way? It is very counter intuitive, even to me, that this would be the case.

As it turns out, the common thread is gravity. The pendulum uses gravity to drive it. Superelevation is determined as balancing gravity with centrifugal aceleration. And, of course, a projectile motion is the quintessential problem of gravity.

More specifically, gravity is *scale invariant*. No matter how you scale your model railraod, the gravitational acceleration roots you down to a very specific scale of motion. In other words, you have no choice but to accelerate a real 9.8 meters per second per second, no matter how you scale!

This scale invariance, then, gives a way for both the brain, and for measuring devices, to determine the scale factors involved, *unless you change the rate of time*!

So why does the rate of time change the way it does? The easiest way is to use a bit of differential calculus. But that is beyond the scope of this article. Instead, I will note that the constraint on how to scale time is imposed by the quadratic relationship between distance and time under the effects of gravity. More technically, since `d` is proportional to `t^2`, and we know that `d` scales according to `d'=m*d`, it necessarily constrains `t` to scale by `t'=t*sqrt(m)`.

**So Now Then**

While modelers as a whole have determined what *feels* to a reasonable scaling methodology (space scales, but time does not). But to anyone serious about modeling the physics of trains correctly, it is important to realize that it isn't a matter of what *feels* right, but instead is determined by what *is* right.

And as much as everyone *feels* that time does not scale, the physics does not agree. Indeed, time must scale in order to replicate the balance of the physical forces involved.