Maybe this will help your further understanding. Math is involved, but it is very simple:
Voltage = Current times resistance, ie. V = I * R. All but the most pedantic people will refer to this as "Ohm's Law".
With a little algebra, this equation can also be arranged as:
I = V / R
-or-
R = V / I
This means if you know any two of the variables (voltage, current, resistance) you can calculate and predict the remaining one. Science rules.
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OK, now let's consider the LM317 design goals, there are two main ones:
1) Vary the amount of current allowed to pass from Vin (pin 3) to Vout (pin 2) in order to maintain exactly 1.25V between the Vout and ADJ (pin 1). If the voltage is too low, allow more current, if the voltage is too high, pinch off some current.
2) Allow the least possible amount of current in or out of the ADJ terminal. So little in fact, as to allow us to disregard it completely in our calculations.
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Now look at the basic circuit for the device.
R1 is your 270 ohm resistor, and R2 is your 10K pot, (technically a rheostat in this case.) Call the voltage across R1 "V1", and call the current through it "I1". Similarly with R2. Now,
if the LM317 is doing it's thing, we know that V1 will be 1.25V, and the current in or out of the ADJ pin is too small to care about, so I2 is the same as I1.
Since we know R1 and V1, we can say that I1 = V1/R1 = 1.25/270 = 0.0046, or 4.6mA. And as long as that LM317 is operating within its limitations, this current stays exactly the same! It's what we call a "constant current source".
Now look at the output voltage Vout. Kirchoff tells us that it is the sum of V1 and V2, but that should be fairly intuitive I hope. We know what V1 is, it is always 1.25V, and V2 = I2 * R2. And we know what I2 is, it is always 4.6mA. So that allows us to know what the output voltage can be for any given value of R2. Consider the following conditions at various percentages of rotation for R2:
0%, R2 = 0 ohms: Vout = 1.25 + 0.0046 * 0 = 1.25V
20%, R2 = 2000 ohms: Vout = 1.25 + 0.0046 * 2000 = 10.45V
50%, R2 = 5000 ohms: Vout = 1.25 + 0.0046 * 5000 = 24.25V
100%, R2 = 10000 ohms: Vout = 1.25 + 0.0046 * 10000 = 47.25V
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I think you might see a problem above, the voltage out seems to be independent of the voltage input. Not really, as the LM317 cannot output more than the input voltage minus something called the "dropout voltage", which is the minimum voltage across Vout and Vin that needs to be maintained in order for the device to keep doing its thing. This dropout voltage changes with temperature and loading, but at room temp with minimal load it's roughly 1.5V. One of the graphs in the data sheet describes this.
Now all this should explain at least a couple things: 1) In normal operation the pot shouldn't be overloaded because it only has about 5mA going through it, and 2) Why you reach max voltage at low settings on the pot. In fact, 10K is a bit large for your project, it probably should be more like 5K to make more of the rotation range useable.
Hope this helps. It is Sunday and I have a little time to spare.