How this calculator works
Both lengths entered are rest lengths. The barn observer measures a contracted moving pole, while the pole observer measures a contracted barn. The apparent contradiction disappears when the same door events are transformed: doors that close together in the barn frame close at different times in the pole frame. The timeline sets the front of the pole at the entrance at t = x = 0. When fitting is possible, both ideal doors close briefly at the barn-frame rear-entry time. The exit door reopens immediately; a permanently closed door would cause a collision, not trap an unchanged rigid pole. Speed and size inverse modes calculate the just-fitting boundary.
Formula
gamma = 1/sqrt(1-v²/c²); moving pole length = l0/gamma; moving barn length = L0/gamma. t_prime = gamma(t-vx/c²); x_prime = gamma(x-vt). Minimum fit speed = c sqrt(1-(L0/l0)²) when l0 > L0.
Worked example
At 0.6c, a 10 m pole has barn-frame length 8 m and can briefly fit a 9 m barn. In the pole frame the barn is 7.2 m long, but the two barn-simultaneous door events are no longer simultaneous.
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Frequently asked questions
Which frame decides whether it fits?
Simultaneous spatial fit is frame-dependent. All frames agree on the individual crossings and door events.
Why can a pole longer than the barn fit?
Its barn-frame length is contracted and the endpoints are measured at one barn-frame time.
Can the doors remain shut?
Then collision and finite-speed material stresses matter. This timeline assumes doors close only momentarily and reopen without obstructing the pole.