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Equation 124 · Part 1 · The Bit Comes Back Before the Bearing

Symbol Δ t_LG

ΔtLG≈+8.9 ms\Delta t_{LG} \approx +8.9\,\mathrm{ms}
ΔtLG\Delta t_{LG}

What this part means

Δ tLt_LG is a part of this expression. Its role is fixed by the surrounding article and by the operations shown in the formula.

Its job in the formula

Δ tLt_LG is a part of this expression. Its role is fixed by the surrounding article and by the operations shown in the formula.

The passage around this formula

…above. Taking f1f_1 = 1 as an illustrative, explicitly assumed per-quantum information content, a modest tolerance δ\delta = 0.1\,rad\mathrm{rad} (about six degrees) gives tGt_G/β\beta = 100 , so tGt_G ≈\approx 12.4\,ms\mathrm{ms} and Δ\Delta tLGt_{LG} ≈\approx +8.9\,ms\mathrm{ms} : the compass is the slower debt, as the title claims. But a coarser tolerance, δ\delta = 0.5\,rad\mathrm{rad} (about twenty-nine degrees), gives tGt_G/β\beta = 4 , so tGt_G ≈\approx 0.50\,ms\mathrm{ms} and Δ\Delta tLGt_{LG} ≈\approx -3.0\,ms\mathrm{ms} : under this same model,…

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