Will the walkie talkies work if there are hundreds in a small area all transmitting data with each other? Besides, there's just not that much bandwidth there.

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rm30 2 hours ago | parent | next [-]

The smartphone is just an advanced walkie-talkie, currently limited only by the mobile operator, the law, the radio chipset, and the OS.

In a true emergency, who can stop you from modifying that architecture? Once you treat the device as an independent radio node (using its DSP power to run custom modems) you can establish a mesh network with a range of several kilometers.

We have a '4x4 car in our pockets; we’ve just been conditioned to treat it like a toy.

	▲	catlifeonmars 13 minutes ago \| parent [-]
		Not disagreeing with you, but you’re papering over a lot of complexity. Note that cellular radios are highly specialized and the filtering circuits are tuned to specific bands. It’s not exactly like having a software defined radio in your pocket. Next, at the modem level, you’ll need to implement and then sideload custom firmware. Finally, you’ll need to expose the right APDUs to the kernel to manage the whole thing. TBH it sounds like a fun side project, but my point is you need to repurpose a lot of different parts of the stack to accomplish what you want.

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modeless 3 hours ago | parent | prev [-]

Walkie talkies as licensed today wouldn't because they are required by law to use exclusively stone-age radio technology. But modern unlicensed radio technology is incredibly good at sharing scarce 2.4 Ghz spectrum. Sometimes devices do interfere with each other, but they remain useful and they are far better at sharing than any expert would have predicted years ago. Let the radio engineers try.

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kanbankaren 3 hours ago | parent | next [-]

It is not as easy as you think.

RF attenuation is proportional to frequency and at 2.4 GHz, it is very high. Also, the distance over which one could communicate depends on antenna height, so if both parties are at ground level, it is not feasible over a few hundred meters unless both are in wide open space.

Source: ham operator who has played with long distance device to device communication without using a repeater.

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lxgr 2 hours ago | parent [-]

> RF attenuation is proportional to frequency and at 2.4 GHz, it is very high.

Through building materials, foliage etc, but not in free space/line-of-sight.

> Also, the distance over which one could communicate depends on antenna height, so if both parties are at ground level, it is not feasible over a few hundred meters unless both are in wide open space.

Isn't it just the opposite? Antenna height is only the limiting factor with line-of-sight, otherwise NLOS considerations like attenuation by building materials, multipath propagation etc. start to matter much more. Modern radio standards are extremely good at that.

Of course line-of-sight usually remains the ceiling, since there usually isn't much in the sky to helpfully reflect signals back down, at least with mobile transmitter compatible transmission levels (i.e. excluding shortwave).

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kanbankaren 2 hours ago | parent [-]

> Through building materials, foliage etc, but not in free space/line-of-sight.

Yeah. Even in free space. For example, attenuation at 1 km for 144 MHz (ham VHF band) is about -76 dB while for 2.4 GHz, it is about -100 dB. That 24 dB drop could mean, the signal is below the noise floor of your receiver unless you increase the RF power output which means more battery drain.

For example, BT audio gets cut just moving to the next room despite the RF power of BT transmitters being ~ 5mW( 7 dBm ) and at 10m, the attenuation is -60 dB(just free space loss which is ideal condition), so 53 dBm (7-60) at the receiver is usually sufficient, yet they struggle.

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lxgr 2 hours ago | parent [-]

No, attenuation in vacuum is exactly the same, and the difference between humid air, dry air, and vacuum doesn't really matter at frequencies below a few GHz.

> For example, attenuation at 1 km for 144 MHz (ham VHF band) is about -76 dB while for 2.4 GHz, it is about -100 dB.

This is a common misunderstanding of the free-space path loss formula, which is expressed in terms of the idealized isotropic radiator, the length of which is frequency-dependent. In other words, this calculation is assuming a proportionally (much) smaller antenna for the 2.4 GHz case.

With the same antenna size, the path loss is exactly the same. After all, where else should the radiated energy go?

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kanbankaren an hour ago | parent [-]

> With the same antenna size, the path loss is exactly the same.

What do you mean? The size of the dipole or monopole antenna is dependent on the wavelength, so obviously the 2.4 GHz is just a few centimeters and not the same size as a VHF antenna.

> After all, where else should the radiated energy go?

Well, most of RF energy is wasted. There are software that can plot the radiation pattern, but even without knowing the exact pattern, very little RF energy is received at the target.

	▲	lxgr 35 minutes ago \| parent [-]
		> The size of the dipole or monopole antenna is dependent on the wavelength, so obviously the 2.4 GHz is just a few centimeters and not the same size as a VHF antenna. Sure, if you want to stay omnidirectional, but you don't have to. You can use one of several antennas based on feedback, beamforming etc.

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Workaccount2 2 hours ago | parent | prev [-]

This is great on paper until some jackass wants to access their home NAS over the public frequency range so they can watch anime all day at their desk, which only works when they use multiple channels at once.

There are tons of cool things society could enjoy if it wasn't for a small handful of shameless actors.