How is this pot connected?

Hi, along with my little static sensor project I am building with my grandson, we are also going to build a simple EMF sensor.

I have the schematic but there is one bit I am just not sure how it actually connects. I have copied that part of the schematic and attached here.
emf2

As I explained before I am not that great with some bits of electronics so you are looking for the pot off pin 8 of the LM3914.

Does it show that the pot wiper is actually needs connecting to pin 8 along with the fixed end of the pot (ie pins 1 and 2 of the pot connect to LM3914 pin 8)

I have tried asking the person who made the design by email last week but not heard anything, so asking the bright minds here if they think it is correct. I ask because my understand is that the having 3 pins. so pins 1 and 3 form the resistance range and pin 2 (wiper) is what chooses the variable resistance. So can you have pins 1 and 2 connected together?

Thanks all for reading and replying.

This changes the pot to a variable resistor!

Yes, that is what it shows

But does it make any difference, connecting the wiper to that other terminal? If the wiper alone was connected to the chip and the other terminal to ground, leaving that third terminal unconnected, wouldn't that be the same?

Those 1 Ohm series resistors for the LEDs must be an error.

In normal circumstances, it wouldn't make a difference. However, in the presence of oxidation, it prevents the potentiometer from (momentarily) going completely open-circuit.
I'm not sure how useful that is for this circuit, but it doesn't hurt to connect it like that anyway.

Indeed, the LM3914 has constant-current LED drivers, the current is usually set by the resistance between pins 7 and 8. Connecting pin 8 to ground through a potentiometer like that is not right. The datasheet contains many more details:

https://www.ti.com/lit/ds/symlink/lm3914.pdf#page=2

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you do realise that pot only controls the led intensity?

Your input voltage connects to pin 5

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