How this instrument works
The barn-pole paradox (also told with a ladder and a garage) asks what happens when a pole longer than a barn is run through it at relativistic speed. At rest, a 10 m pole obviously cannot fit inside an 8 m barn. But length contraction shortens the pole in the barn's own rest frame, and past a certain speed the contracted figure drops below 8 m — long enough that, for an instant, the barn's front and back doors could both be shut with the whole pole enclosed. This calculator computes exactly that contracted figure: the pole's rest-frame length run through L = L₀√(1 − v²/c²) at the velocity you set, so you can check it against the barn's own rest length.
The catch is that velocity is relative, so the pole's own frame is just as entitled to call itself at rest and watch the barn go by. From the pole's point of view, the barn is the thing in motion, and it contracts by the identical factor — which makes the barn even shorter relative to the pole than the pole was relative to the barn. Naively that reads as two frames flatly disagreeing about a fact as simple as whether an object fits inside another: barn's-frame says yes, pole's-frame says no. Nothing in the arithmetic is wrong on either side; both apply the same formula correctly to the same 10 m and 8 m rest lengths.
What breaks the apparent contradiction is that 'the doors close at the same time' is not a frame-independent statement. Two events separated in space — the front door shutting and the back door shutting — can be simultaneous for the barn and strictly out of order for the pole, because special relativity ties simultaneity to the observer's own motion, not to some shared cosmic clock. Neither observer is wrong about what they see happen at each door; they are only wrong to assume the other observer must see the two door events in the same order they do.
- Enter the pole's own measured length, at rest, into Pole's rest-frame length, in metres.
- Enter the barn's own measured length, at rest, into Barn's rest-frame length — this is the yardstick, not part of the formula's arithmetic.
- Set the pole's speed relative to the barn into Pole's velocity, in m/s, staying below 299,792,458.
- Read Pole's length in the barn's frame and compare it to the barn length: shorter means the contracted pole momentarily fits with both doors closed.
Worked example — a 10 m pole through an 8 m barn at 0.8c
Set Pole's rest-frame length to 10 m, Barn's rest-frame length to 8 m, and Pole's velocity to 240,000,000 m/s — about 80.06% of light speed. Squaring the ratio v/c gives 0.640886, and 1 minus that is 0.359114; its square root is 0.599261. Multiplying by the 10 m rest length gives Pole's length in the barn's frame = 5.99260851143 m. That is almost exactly 6 m, comfortably under the 8 m barn, so in the barn's rest frame there is a stretch of time when the entire pole is between the doors and both could be shut at once.
Run the same 0.599261 contraction factor the other way and the paradox shows its teeth: in the pole's own frame it is the 8 m barn that is moving, so the barn contracts to 8 × 0.599261 ≈ 4.79 m — well under the pole's own 10 m rest length. From the pole's seat, the barn is never big enough to hold it at all. Both figures are correct arithmetic on the same L = L₀√(1 − v²/c²) formula; the resolution is that the barn's two door-closing events, simultaneous in the barn's frame, happen at different times in the pole's frame, so the pole's frame never needs both doors shut on it together.
Questions
Why does a 10 m pole fit inside an 8 m barn at all?
Because rest length is not what a relatively moving observer measures. At 240,000,000 m/s, about 80.06% of light speed, the factor √(1 − v²/c²) works out to 0.599261, so the barn's frame measures the pole at 10 × 0.599261 ≈ 5.9926 m — almost 2 m shorter than the 8 m barn, with room to spare while it passes through.
Doesn't the pole's own frame see a barn too short for it?
Yes, and that is the actual paradox, not a mistake in either frame. Riding with the pole, the 8 m barn is the thing in motion, so it contracts by the same factor to roughly 4.79 m, shorter than the pole's own 10 m rest length. Both frames apply the formula correctly; they only disagree about whether the two door-closing events happen at the same moment.
So which frame is right — does the pole fit or not?
Neither frame outranks the other; 'fits inside' stops being a frame-independent fact once 'at the same time' stops meaning the same thing to both observers. In the barn's frame both doors close together around a fully enclosed pole. In the pole's frame those same two door-closing events happen at different times, and the front door has reopened before the pole's back end ever reaches it.
Is this the same puzzle as the ladder paradox?
Yes. Barn-pole and ladder-and-garage are the identical scenario with different props, both commonly traced to physicist Wolfgang Rindler's 1961 discussion of what he called the length-contraction paradox. Swap pole for ladder and barn for garage, and the formula, the numbers, and the simultaneity resolution carry over unchanged.
At what speed does a 10 m pole first fit an 8 m barn?
Exactly 0.6c. Solving 10√(1 − v²/c²) = 8 gives v²/c² = 0.36, so v = 0.6c, or 179,875,474.8 m/s. Below that speed the contracted pole is still longer than 8 m and cannot fit; above it the margin grows, reaching the roughly 5.99 m figure of this calculator's own worked example at 0.8c.
Do the barn doors actually collide with the pole?
Not in the standard version of the puzzle. The doors are idealized as opening and closing instantly, so the thought experiment only asks whether the pole is momentarily enclosed, never whether anything crashes into anything. A version where the doors stay shut trades the paradox for an engineering problem: the pole then has to physically stop, which breaks the constant-velocity assumption the contraction formula depends on.