The Science of Back To The Future [BTTF Week]

A delightful romp through the worlds of time travel debunking, high school geometry, and the Metric System.

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bttf_week_logoThere come times, in the course of human events, whereby an official-unofficial OTI Staff Scientist such as myself feels compelled to write about pop-cultural issues which he or she (statistically: he) finds irksome on a professional level.  Back to The Future is a nearly endless font of these issues for me—from the obligatory complications surrounding causality in time-travel paradoxes to hoverboards which don’t work above water to the continued reminder of the sheer mathematical improbability of  Huey Lewis and the News.  But as my esteemed colleagues have already touched on these with great aplomb, I’ve got some another problem to Overthink™ today.

And oh yes, there will be math involved.


Pop culture has envisioned a multitude of timetravel devices, and these give rise to the usual scientific musings (physical and metaphysical alike): questions of causality, internal paradoxes, and the “inconveniences” associated with the creation a billion alternate universes or self-referential circular time lines.  Also, there’s usually lightning and dramatic music.

However, there’s a much smaller–though fundamentally difficult–caveat to time travel that’s always kinda’ bothered me.  With the notable exceptions of Doctor Who and Bill and Ted’s Excellent Adventure, the spatial aspects of time travel are almost universally overlooked when characters decide to go gallivanting through temporal mayhem.  I was reminded when questioned by a fellow OTI™ writer in a recent email; so as  to protect his/her (statistically: his) anonymity,  I’ll keep the author’s identity a secret :


“Watthew Mrather” writes: The DeLorean can’t go anywhere in space, except in the conventional driving/flying sense.  Your [sic] always in the same place you left, but in a different time.


Well, for starters, that’s not a question, “Mr. Mrather,” but I’ll go along with it anyway.  Remember, though: without proper punctuation and grammar, we are no better than the animals.

In fact, in order to pull off the kind of time travel we see in the Back To The Future trilogy–the kind where the traveler is transposed in time, but remains stationary in the same relative position to where he/she left–the DeLorean would have to be an outstanding space ship, in addition to its already laudable work as a time-ship.  A major issue of freely traveling within time while limiting one’s self to a local reference frame–say, a California mall parking lot–is that the reference frame itself isn’t stationary.  As an illustration, let’s figure out how far the DeLorean would have to travel in order to stay in sync with the Earth over a relatively small time-jump.  We’ll look at the simplest example (and the first one, diagetically speaking) of the whole BTTF trilogy.  You all remember the scene, right?  (Spoiler alert: Professor Plum and Alex P. Keaton send a dog one minute into the future.)

(Why no one in sci-fi movies ever names their pet Smoluchowski, I’ll never know.)

As most of us know, the Earth both rotates about its axis (accounting for the regular cycle of day and night, and hence a good bit of poetry) and revolves around the sun (accounting for the seasons, and hence the rest of poetry).  However, the litany of additional factors affecting Earth’s absolute position in space makes it fantastically difficult to calculate where a given point on its surface would be after some interval of time has passed.  A little professional lingo, here: we in the science biz refer to this level of complexity as a “colossal mind-fuck.” So, to make my life easier,  I’ll first lay down the standard boilerplate warning we employ for all Overthought(™-pending) mathematical meta-transmutations:

For all calculations pertaining to a period of time less than one hour, the Overthinking It™ writers reserve the right to disregard effects due to the nonperpendicular axial-tilt of the Earth relative to its plane of revolution, the precession of the Earth about its axis, the gradual decay of Earth’s orbit into the sun, the rotation of the Milky Way, analogous rotational and linear velocities within the Virgo Supercluster, and the general expansion of the Univese.

Also, we won’t factor in Daylight Savings Time.

Got it?  Good.   Let’s get this party started.

SO, according to Doc Brown’s stopwatch, Einstein (the dog, not the Princetonian–or the grossly mispronounced Russian film director) travels precisely one minute into the future on this first jump, arriving, relative to their frame of reference, at the same location he left.  But how far has this reference frame itself traveled during that one minute?  Let’s calculate and see.

PART THE FIRST. In Which the System Is Set Up:

We know that BTTF takes place somewhere in California–and for the sake of argument, we’ll say that Hill Valley is near Los Angeles, hence at ~35o N latitude.  It’s October 26, meaning that it’s autumn in the northern hemisphere, and so for our calculation we’ll take the Earth’s distance from the sun to be roughly perihelion, and unchanged during the minute time jump.

According to Doc’s watch, the time-jump starts at 1:20A.M.  Now, the Earth rotates 360o every 24 hours, a rate of 15o per hour.  If we assume that Hill Valley was furthest away from the sun at midnight local time, it’s therefore rotated (counter-clockwise, as viewed peering down from the North Pole) exactly 20o since midnight.  Here’s a summary of what’s going on at the moment Einstein makes his temporal leap:

I tried to get more arrows on this figure, honest.

Figure 1. I tried to get more arrows on this figure, honest.

PART THE SECOND. In Which Rotation is Considered:

As stated above, the earth makes a full rotation (360o) every 24 hours.  This is (24hr)*(60 min/hr) = 1440 minutes in a day.  Hence, over the course of a  single minute, the earth will rotate (360o)/1440 = 0.25o.  The Earth’s radius at the equator is 6378.1km, but with increasing latitude the distance between a surface observer and the axis of rotation decreases (consider the case of a person standing on the north pole–he/she (statistically: he) wouldn’t move at all due to rotation. Note also: for subterranean Mole-Men, the calculation’s even harder.)  But a little trigonometry tells us that the distance between a surface observer this axis is (6378.1) * cos(latitude), or (6378.1)*cos(35) = 5287.1km.

So the point at which Einstein takes off/lands is at a distance of ~5300km from the Earth’s rotation axis, and rotates through a  0.25o angle during the minute.  Checkout figure two, which summarizes the effects due only to rotation:

Figure 2

Figure 2. Angles not to scale. Greenland ablated for clarity. And the sheer joy of it.

Now, in order to solve for D1, we’ll take the standard OTI approach and–just as Mr. Belinkie did for the Dark Bailout movie–employ the law of cosines.  To jog your memories:

It's not just a good idea... it's the law.

The Law of Cosines. It's not just a good idea... it's the law.

Plugging in a=b=5287.1km, and γ =0.25o, we arrive at…

D1 = 23.07km

Also, (though this isn’t labeled in figure 2), since the triangle in Figure 2 is isosceles, each of the non-0.25o angles are equal.  And since the sum of the angles of a triangle is always 180o,these therefore have a value of 89.875o.

PART THE THIRD. In Which Revolution is Considered:

What about the Earth’s revolution around the sun?  Here’s what we know.  Since the Earth is ~perihelion, we’ll call the distance between its center of mass (the Eternal Winter Palace of the High Czar of the Mole-Men) and the sun to be 149.5 million kilometers.  In truth, our orbit is an ellipse, but it’s not that elliptical (I mean, I’ve certainly seen more elliptical ellipses–like this: “…”) so we’ll estimate it as a circle.  To fully traverse this circle the Earth takes a full year, which is 365.25 days.  Hence, despite anything certain overtly melodramatic people might tell you, a year consists of:

(365.25 days/year) * (24 hours/day) * (60 minutes/hour) = 525960 minutes.

(How do  I measure, measure a year?  With some fucking precision.)

So as before, we calculate that in a minute’s time, the Earth revolves 360o/525960 = 0.0006844626o.  Check out figure three which takes into account the effects due only to revolution:

(Angles not to scale.  Earth and Sun ARE to scale, though.  You've been lied to all your lives...)

Figure 3. (Angles not to scale. Earth and Sun ARE to scale, though. You've been lied to all your lives...)

Again, we’ve got an isosceles triangle to which we can apply the law of cosines to yield:

D2 = 1785.88km

And as before, the two larger angles within the triangle are calculated to be 89.99966o.

PART THE FOURTH. In Which We Finally Get to Go Home:

Let’s look at the overall effect en toto.  Though revolution and rotation are of course simultaneous, we can parse the mathematical effects of both out into separate steps.  Hence each displacement becomes the leg of a triangle–one for rotation, one for revolution.  Check out figure four–which puts it all together:

Figure 4.  This is probably the first OTI image to have discrete A. and B. panels.

Figure 4. This is probably the first OTI image to have discrete A. and B. panels.

We already have D1 and  D2.  From the calculations in figure 3, we compute the smaller of the two interior angles of the central parallelogram to be 180o – (89.99966o + 20o) = 70.00034o, as is shown in Fig. 4A.  From Figure 2, we know that the small isosceles triangle on the right of 4A  has angles of 89.875o.  As in figure 4B, adding these two gives a total angle of 159.87534o.   Applying the law of cosines one last time yields:

D3= The total distance Einstein would have to travel in order to arrive back in the Twin Pines Mall parking lot one minute into the future…




That distance is ~4/5 the length of the eastern sea board of the United States!  Or–to put it more juvenile terms–it’s approximately the distance from Bangor, Maine to Morehead, Kentucky.  But less sexy.  And that’s how far the Earth would travel in just one minute!  I don’t even want to go near the calculation for the 1985-1955 jump…

Now, the machine is traveling at 88mph.  While that’s hardly fast enough to cover this distance, how fast would it have to go to make the jump in space as well as time?

Speed is just the distance traveled over an interval of time.  Doc’s stopwatches would have you believe that Einstein’s trip took no time at all, at least as far as the dog is concerned:

Courtesy of the good people at the BTTF Wikia page.  Thanks for saving me the trouble of taking a screen-cap!

Courtesy of the good people at the BTTF Wikia page. Thanks for saving me the trouble of taking a screen-cap!

But then, his watches aren’t perfect, and this allows us to try and estimate something we’re unable to measure.  Let’s be generous and assume that the stopwatches have an error of ~1 millisecond.  That is, Einstein’s watch could be .001s slower than we think it is, without a perceptible effect: we’d still see the two watches changing time in pretty much perfect sync with one another.  Hence an upper-limit to the time it took Einstein to jump 1807.56km is 0.001s.  This yields an average speed of:

(1807560m)/(0.001s) = 1.808 x 109 m/s

…roughly six times the speed of light in a vacuum.  Of course, an object moving faster than the speed of light would theoretically travel backward in time, so that kinda’ bones us right there.

* * *

So what can we take from this?  Well first off, were the DeLorean to really just pop-out and pop-in to time in the same absolute place, then even this first tiny future-leap would drop poor Einstein nearly two-thousand kilometers into the vacuum of space.  Longer leaps of time would be even more potentially disastrous: you might reappear inside solid rock, or within the fusion furnace of a sun, or near Steven Segal.  One shivers at the thought.

But that kind of speculation is moot.  We know that the DeLorean does pop back in pretty much the same relative place it popped out.  Maybe it’s a conscious choice–maybe it’s a nimble space ship that Doc could have engineered it to go anywhere, but kept it within frame for safety’s sake.  After-all, while I’m unable (and certainly unwilling) to do more elaborate calculations, that’s not to say that —given more sophisticated calculation equipment–a computer on board the DeLorean wouldn’t be able to handle them with aplomb.  I mean, check out all the buttons and toggle-switches inside that thing–maybe some of them control the distance-computation circuits, in addition to the flux capacitor et. al. This gets less and less plausible when one imagines how Doc Brown could have constructed a second time-ship out of a 19th century train–as is revealed in the closing scene of BTTF 3.  After-all, the retroengineering required in fashioning a flux capacitor from pre-period goods would be hard enough without having to fabricate silicon microprocessors as well.

No, my guess is that we’re supposed to assume the DeLorean is somehow coupled to the local reference frame  as it travels freely through time–much in the same way that you and I are coupled to it as we travel through time in a constrained fashion.  While the easiest way to achieve this coupling is through gravity, the big “G” tends to muck up time in whatever way it sees fit, regardless of your fancy-shmancy time travel plans.

Now that I think of it, one could always synthesize a worm-hole that encloses a timeless space, with its exterior coupled to the Earth’s timeline.  This is of course absurd: any object passing through it would be exposed temperatures approaching absolute zero and would hence instantly become incomparably cold.