As per the diagram below (and other graphs available online), why do unmyelinated fibres have a higher conduction velocity than myelinated fibres when the axon diameter is less than around 1 µm?

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2 Answers 2


Graphs like these are based on a complete cross-sectional area. Myelination adds thickness, so if you want to add myelin outside and keep the same diameter, you have to have a smaller lumen inside the axon.

As you can see in the "unmyelinated" plot versus diameter, smaller fibers have lower conduction velocity. Since the benefits of cross-sectional area are most important at small diameters (because the velocity changes with the squart root of diameter), for small diameter axons you end up losing more from losing internal area than you gain from adding external insulation.

If you keep the internal diameter the same and just add myelin, you will always increase the conduction velocity, but the total diameter including the myelin won't be the same anymore.

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    $\begingroup$ This explains it perfectly. So going from unmyelinated to myelinated is always beneficial in terms of conduction velocity, but it may not be in terms of space-saving. Thank you! $\endgroup$ Jul 23, 2020 at 22:41
  • $\begingroup$ @pincushion44 Exactly, and space is at a premium in nervous systems! $\endgroup$
    – Bryan Krause
    Jul 23, 2020 at 22:45

That's observed due to passive conduction. With more area of cross section resistance is decreased (ohm's law). Myelin has just function of insulation the graph showing conduction in neuron fibre varying with thickness of neuron (hereby also thickness of myelin sheath). Conduction is ruled by multiple factors not only just thickness of insulation. check this out https://www.ncbi.nlm.nih.gov/books/NBK10921/ When the area is so less than certain level the resistance is really high and insulation provided by myelin sheath works against conduction rather than supporting it. Conduction happens superficially instead in the bulk. Where as in non myelinated fibre the insulation of myelin is not barrier.

Straight line tells me that my reason is legible probably since current is proportional to (1/R and hence) A

  • $\begingroup$ Sorry, I am slightly confused as to why insulation from myelin works against conduction when the area is low (and so longitudinal resistance is high)? $\endgroup$ Jul 23, 2020 at 21:59
  • $\begingroup$ will try to make answer better after studying more about it 🙂 $\endgroup$
    – daemon
    Jul 23, 2020 at 22:01
  • $\begingroup$ try that citation if it helps tell me. $\endgroup$
    – daemon
    Jul 23, 2020 at 22:02
  • $\begingroup$ It might also be possible that at smaller diameter cohesive forces of myelin sheath material close the node of ranvier and reduce it in significant amount $\endgroup$
    – daemon
    Jul 23, 2020 at 22:05

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