Doubling Time from OD600 Calculator
Compute the doubling time and specific growth rate of a culture from two OD600 readings and the time between them.
How it works
µ = ln(OD₂ ÷ OD₁) ÷ t · doubling time = ln(2) ÷ µ · generations = log₂(OD₂ ÷ OD₁)
During exponential growth OD rises proportionally to cell number, so the natural log of the OD ratio divided by elapsed time gives the specific growth rate µ. The doubling time is ln(2) ÷ µ — the time for the culture to double. This holds only while the culture is in log phase and within the linear OD range; readings taken in lag or stationary phase, or above the instrument’s linear limit, give a misleadingly slow rate.
Worked example
From OD 0.1 to OD 0.8 the culture grows 8-fold. Over 3 h, µ = ln(8) ÷ 3 = 2.079 ÷ 3 = 0.693 per hour, so the doubling time is ln(2) ÷ 0.693 = 1.0 h and exactly 3 generations (log₂8) have passed.
Sources: Bacterial growth — Wikipedia
Frequently asked questions
Why use log of the OD ratio instead of the plain ratio?
Exponential growth is linear on a log scale. Taking the natural log converts the multiplicative growth into a constant rate µ, which is what defines the doubling time.
Do both readings need to be in log phase?
Yes. The formula assumes a constant growth rate. A reading taken during lag or stationary phase, or above the OD linear range, will distort the calculated doubling time.
How do doubling time and generation time relate?
They are the same quantity for a growing culture — the time to double the population equals the time for one generation. Both equal ln(2) ÷ µ.
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This calculator is for educational purposes. Double-check important results before acting on them.