Is there any way to calculate the cumulative width of synaptic clefts? It's my understanding there are 150 trillion synapses in the human brain (give or take). Is there a way to calculate the cumulative width of all synaptic clefts associated with the synapses? I would expect that some would be shared by several synapses. At 20-40nm per synaptic cleft, 150 trillion of them would add up to a sizeable distance, which seems implausible, given the size of the brain's surface.

Anyone have any suggestions on the best way to calculate this -- if there is one?
Mathematically, you're correct to think that linearly the total distance would be sizable, however don't forget the third axis, depth.

Given the following:


  The adult human brain is estimated to contain from 100–500 trillion synapses. (Drachman D (2005). "Do we have brain to spare?".)


Let's go with 150 from your example

... and ...


  Every cubic millimeter of cerebral cortex contains roughly a billion of them. (Alonso-Nanclares L, Gonzalez-Soriano J, Rodriguez JR, DeFelipe J (September 2008). "Gender differences in human cortical synaptic density")


Let's calculate the total volume these synapses occupy:

150,000,000,000,000 / 1,000,000,000 = 150,000 cubic mm => 150 cubic cm (
You can picture this as a cuboid 15cm wide by 10cm tall which is roughly the size of a butter bar )

So if all synapses can fit in a 15cm x 10cm cuboid then the clefts could all fit in a significantly less space. 

If you want to calculate linear distance for, let's say an average of 30nm clefts then:

150,000,000,000,000,000 * 30nm = 4,500,000,000,000,000nm => 4500km