I found a PDF online that says:

The diagram shows that one unit of power is required to achieve a given data rate over one unit of distance. But to maintain the same data rate over x the distance, a traditional wireless network requires not x but rather 16x the power.

It refers to this diagram:

RX power vs distance

Shouldn't it have been 9X the power (instead of 16X) or 4X the distance (instead of 3X)?

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    \$\begingroup\$ calculate the surface area of a sphere that has a radius of 1 meter, then calculate the same for a radius of 2m and 3m .... what is the ratio of areas? \$\endgroup\$ – jsotola Jun 1 '18 at 16:18
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    \$\begingroup\$ It all depends on what link loss formula they assumed. Does the pdf tell you that? \$\endgroup\$ – Andy aka Jun 1 '18 at 16:36
  • \$\begingroup\$ @Andyaka they didn't \$\endgroup\$ – Luís Marques Jun 1 '18 at 16:45
  • \$\begingroup\$ Can you link the pdf please \$\endgroup\$ – Andy aka Jun 1 '18 at 16:46
  • \$\begingroup\$ motorola.com/innovators/pdfs/mesh-ntwks-wp-7.24.06.pdf \$\endgroup\$ – Luís Marques Jun 1 '18 at 16:48

How they modelled the path loss is key to how they came up with 16x the power: -

enter image description here

Slideshow here.

In free space, the exponent is 2 and this means that for 3 times the transmit distance you will need 9x the power.

But, conceivably they could have used an exponent of 2.53 and \$3^{2.53}\$ = 16.1.

Personally I think the marketing guys have messed-up because this is known to happen.

| improve this answer | |
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    \$\begingroup\$ I like the understatement of the last sentence :) \$\endgroup\$ – Luís Marques Jun 1 '18 at 17:15
  • \$\begingroup\$ Marketing guys (and politicians) have a great talent for making things so simple that they are basically useless. So, 9m or 16m is just alternative facts ;-) \$\endgroup\$ – Stefan Wyss Jun 3 '18 at 14:07

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