A section has internal nodes and external nodes. The number of internal nodes is specified by the section's discretization parameter nseg. Internal nodes are located at the centers of segments. Membrane potential is calculated at internal nodes by numerical integration; at each internal node, membrane potential equals (the amount of charge stored on the corresponding segment's membrane capacitance)/(the segment's membrane capacitance).

A section has two external nodes, one at each end.

The axial resistance (i.e. path resistance) between an external node and its adjacent internal node is simply the path resistance between the external node and the center of the segment that contains the internal node. If a section's geometry is specified by the stylized method, i.e. specifying L and diam, then its segments are cylindrical. Example:

Code: Select all

```
create dend
dend diam = 1
dend L = 100
dend for (x) print x, ri(x)
```

returns

The function ri(x) returns the axial resistance, in megohms, between the center of the segment that contains location x on the currently accessed section and that node's parent node. In this case, dend is not the child of any other section, so the axial resistance between its node at 0 and the parent node of that node is infinitely large (well, 1e30 is approximately infinitely large). I think you will find that the axial resistance of a cylinder 50 um long x 1 um in diameter that contains an electrolyte with resistivity of 35.4 ohm cm is 22.53... megohms ( Ra*50/(PI*0.5^2), and toss in a factor to convert um to cm)

If you connect the 0 end of one section (the child section) to the 1 end of another section (the parent section), and nseg is 1 for both sections, the path resistance between the internal nodes of the parent and the child equals the path resistance between the parent's internal node and its external node at 1 added to the path resistance between the 0 end of the child and its internal node (which is at 0.5).

If a section's geometry is specified with the pt3d method, it is possible to specify tapering of diameter. In this case, axial resistance and surface area along each segment are computed using formulas that take tapering into account. For example,

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```
create axon
axon pt3dclear()
axon pt3dadd(0,0,0,1)
axon pt3dadd(100,0,0,10)
```

specifies that axon will be a frusta with length 100 um and diameters of 1 and 10 at its 0 and 1 ends, respectively, and that in between 0 and 1 diameter varies linearly with position. Now observe that axial resistance is much larger over the first half of axon's length than it is over the second half. In fact, almost 90 percent of the resistance along the entire length of axon is due to resistance between its nodes at 0 and 0.5.

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```
axon for (x) print x, ri(x)
0 1e+30
0.5 4.0975164
1 0.40975164
```