Question about edge of cylinder and axial resistance

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oren
Posts: 54
Joined: Fri Mar 22, 2013 1:03 am

Question about edge of cylinder and axial resistance

Post by oren » Tue Apr 28, 2015 4:49 am

Hello,
I am trying to understand how Neuron is building the section and I am not sure I understand correctly. Especially I have problem understanding the edges of the cylinders.

Questions:
1. there is no membrane resistance in the edges of the cylinders? If I want to calculate the total membrane resistance of a cylinder I need to calculate the surface of the cylinder without the cylinder ends surface?

Code: Select all

2*pi *(diam/2)*L
and not

Code: Select all

2*pi *(diam/2)*L + 2*pi*(diam/2)ˆ2

2. When I dived the cylinder to compartments what happens in the edges ( when the cylinder is isolated and not connected to other cylinders)
For Example:

If I build the following cylinder

Code: Select all

create soma
diam = 1 //um
L = 500 //um 
Ra = 100 
cm =1
insert pas
g_pas = 1/10000
e_pas = 0
nseg = 3

3. This will be the equivalent circuit?
( The values are Ra-end =106103295.39459689 , Ra-middle = 212206590.78919378
Rm = 1909859317.1027443
Cm = 5.235987755982987e-12
)

Image


4. the edges are not connected to nothing?
If for example I build 2 cylinders and connect them with Gap Junctions, as the following:

Code: Select all

create soma
diam = 1 //um
L = 500 //um 
Ra = 100 
cm =1
insert pas
g_pas = 1/10000
e_pas = 0
nseg = 3

create soma1
diam = 1 //um
L = 500 //um 
Ra = 100 
cm =1
insert pas
g_pas = 1/10000
e_pas = 0
nseg = 3

objref GJ1
objref GJ2
GJ1 = new Gap(0.5)
soma { GJ1.loc(0.9999)}
GJ2 = new Gap(0.5)
soma1 { GJ2.loc(0.9999)}

setpointer GJ1.vgap, soma1.v(0.9999)
setpointer GJ2.vgap, soma.v(0.9999)

5. This will be the corresponding circuit?
Image


6. Where will the gap junction be connected if I do ?

Code: Select all

...
objref GJ2
GJ1 = new Gap(0.5)
soma { GJ1.loc(1)}
GJ2 = new Gap(0.5)
soma1 { GJ2.loc(1)}

setpointer GJ1.vgap, soma1.v(1)
setpointer GJ2.vgap, soma.v(1)


Thank You.

hines
Site Admin
Posts: 1577
Joined: Wed May 18, 2005 3:32 pm

Re: Question about edge of cylinder and axial resistance

Post by hines » Tue Apr 28, 2015 5:15 pm

1. there is no end area. However one can use pt3dadd statements to obtain the effect of end area. ie. starting with your code below

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{
pt3dclear()
pt3dadd(0,0,0,0)
pt3dadd(1e-6,0,0,1)
pt3dadd(1000-1e-6,0,0,1)
pt3dadd(1000,0,0,0)

for (x) print x, ri(x), area(x)
}
prints

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0 1e+30 0 
0.16666667 1.2732395e+15 1047.9829 
0.5 424.41318 1047.1976 
0.83333333 424.41318 1047.983 
1 1.2732395e+15 0
but notice how physically nonsensical is the resistance from the distal zero area node to the center of the distal compartment. The 1-d approximation breaks down.

2 and 3. Your equivalent circuit is identical to the representation in NEURON

4. correct

5) correct, the gap connects the middle of the 3rd compartment

6) the gap resistance connects the distal zero area nodes.

oren
Posts: 54
Joined: Fri Mar 22, 2013 1:03 am

Re: Question about edge of cylinder and axial resistance

Post by oren » Wed Apr 29, 2015 5:34 am

Thank You,
Oren

youngJose

Re: Question about edge of cylinder and axial resistance

Post by youngJose » Thu Apr 28, 2016 6:10 am

oren wrote: If I build the following cylinder

Code: Select all

create soma
diam = 1 //um
L = 500 //um 
Ra = 100 
cm =1
insert pas
g_pas = 1/10000
e_pas = 0
nseg = 3

3. This will be the equivalent circuit?
( The values are Ra-end =106103295.39459689 , Ra-middle = 212206590.78919378
Rm = 1909859317.1027443
Cm = 5.235987755982987e-12
)

Image

I am just wondering what formula he used to calculate the Rm. I checked the neuron book and didn't find answer.
Because in my mind, I always think we can calculate Rm only if we have rm which represent the specific resistance of one unit area of membrane (in m2).
So Rm = rm * S. Am I wrong?
Thank you!
Wenbin

hines
Site Admin
Posts: 1577
Joined: Wed May 18, 2005 3:32 pm

Re: Question about edge of cylinder and axial resistance

Post by hines » Thu Apr 28, 2016 8:50 am

Considering...

Code: Select all

$ units
Currency exchange rates from 2013-07-11 
2430 units, 68 prefixes, 64 nonlinear units

You have: mho micron2 / cm2
You want: micromho
	* 0.01
	/ 100
You have: 
I see that after executing your statements
oc>1/( g_pas(.5)*area(.5)*(0.01) )
1909.8593
where the printed value is in megohms.

area of a uniform cable segment is PI*L/nseg*diam
It will be a bit different for 3-d points which define the shape as a sequence of truncated cones.

youngJose

Re: Question about edge of cylinder and axial resistance

Post by youngJose » Thu Apr 28, 2016 11:49 am

hines wrote:Considering...

Code: Select all

$ units
Currency exchange rates from 2013-07-11 
2430 units, 68 prefixes, 64 nonlinear units

You have: mho micron2 / cm2
You want: micromho
	* 0.01
	/ 100
You have: 
I see that after executing your statements
oc>1/( g_pas(.5)*area(.5)*(0.01) )
1909.8593
where the printed value is in megohms.

area of a uniform cable segment is PI*L/nseg*diam
It will be a bit different for 3-d points which define the shape as a sequence of truncated cones.
Thank you Hines.
I though g_pas was capacitance... Silly me.

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