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What's the maximum precision (after the decimal point) of a float in Javascript

An algorithm I'm using needs to squeeze as many levels of precision as possible from a float number in Javascript. I don't mind whether the precision comes from a number that is very large or with a lot of numbers after the decimal point, I just literally need as many numerals in it as possible.
(If you care why, it is for a drag n' drop ranking algorithm which has to deal with a lot of halvings before rebalancing itself. I do also know there are better string-based algorithms but the numerical approach suits my purposes)
The MDN Docs say that:
The JavaScript Number type is a double-precision 64-bit binary format IEEE 754 value, like double in Java or C#. This means it can represent fractional values, but there are some limits to what it can store. A Number only keeps about 17 decimal places of precision; arithmetic is subject to rounding.
How should I best use the "17 decimal places of precision"?
Does the 17 decimal places mean "17 numerals in total, inclusive of those before and after the decimal place"
e.g. (adding underscores to represent thousand-separators for readability)
# 17 numerals: safe
111_222_333_444_555_66
# 17 numerals + decimal point: safe
111_222_333_444_555_6.6
1.11_222_333_444_555_66
# 18 numerals: unsafe
111_222_333_444_555_666
# 18 numerals + decimal point: unsafe
1.11_222_333_444_555_666
111_222_333_444_555_66.6
I assume that the precision of the number determines the number of numerals that you can use and that the position of the decimal point in those numerals is effectively academic.
Am I thinking about the problem correctly?
Does the presence of the decimal point have any bearing on the calculation or is it simply a matter of the number of numerals present
Should I assume that 17 numerals is safe / 18 is unsafe?
Does this vary by browser (not just today but over say, a 10 year window, should one assume that browser precision may increase)?

Short answer: you can probably squeeze out 15 "safe" digits, and it doesn't matter where you place your decimal point.
It's anyone's guess how the JavaScript standard is going to evolve and use other number representations.
Notice how the MDN doc says "about 17 decimals"? Right, it's because sometimes you can represent that many digits, and sometimes less. It's because the floating point representation doesn't map 1-to-1 to our decimal system.
Even numbers with seemingly less information will give rounding errors.
For example
0.1 + 0.2 => 0.30000000000000004
console.log(0.1 + 0.2);
However, in this case we have a lot of margin in the precision, so you can just ask for the precision you want to get rid of the rounding error
console.log((0.1 + 0.2).toPrecision(1));
For a larger illustration of this, consider the following snippet:
for(let i=0;i<22;i++) {
console.log(Number.MAX_SAFE_INTEGER / (10 ** i));
}
You will see a lot of rounding errors on digit 16. However, there would be cases where even the 16th decimal shows a rounding error. If you look here
https://en.wikipedia.org/wiki/IEEE_754
it states that binary 64 has 15.95 decimal digits. That's why I'd guess that 15 digits is the max precision you will get out of this.
You'd have to do your operations, and before you save back the number to any representational form, you'd have to do .toPrecision(15).
Finally this has some good explanations. https://floating-point-gui.de/formats/fp/
BTW, I got curious by reading this question so I read up as I wrote this answer. There are many people with better knowledge of this than me.

Does the presence of the decimal point have any bearing on the calculation or is it simply a matter of the number of numerals present
Kinda. To answer that, you'll need to look into how 64bit "double precision" floating point numbers are represented in memory. The "number of numerals" roughly translates into "length of the mantissa", which is indeed fixed and independent from the position of the point. However: it's binary digits and a binary point, not decimal digits and the decimal point. They do not correspond to each other directly. And then there's stuff like subnormal numbers.
Should I assume that 17 numerals is safe / 18 is unsafe?
No. In fact, only 15 decimal numerals would be "safe" if that's the representation you're starting with and want to exactly represent as a double.
Does this vary by browser (not just today but over say, a 10 year window, should one assume that browser precision may increase)?
No, it doesn't vary. The JavaScript number type will always be 64bit doubles.
Am I thinking about the problem correctly?
No.
You say you're considering this in the context of a drag'n'drop ranking algorithm, and you don't want do this string-based. However, thinking about decimal places in numbers is essentially thinking about string representation of numbers. Don't do that - either go all the way to strings, or treat numbers as binary.
Since you also mention "rebalancing", I assume you want to use numbers to encode the position of each item in a binary tree. That's a reasonable approach, but you really need to consider the binary representation of the number for that. And you really should use integers there, not floating-point numbers, as the logic would be much more complex otherwise. Start by deciding how many bits you want to use. There are some limitations for each, so choose wisely:
31/32 bit are what JS bitwise operators for numbers work on. Supported by all browsers easily.
53 bit are the range of integers you can exactly represent with floating-point numbers. Integer arithmetic will work as expected up to that size. Bitwise operations require extra code.
Fixed multiples of 8 (say, 64 bit) are what you can represent with typed arrays. Bitwise operations can be done part-wise, arithmetic operations require extra code. Or use a BigUint64Array that gives you 64 bits as a bigint to calculate with/operate on, but is not supported in old browsers.
Arbitrary precision can be achieved with bigint numbers, which support both bitwise and arithmetic operations, but again don't work in old browsers. Polyfills and bigint libraries are available though.

Description of numeric type in javascript

I am looking to describe how numbers are stored in javascript to a lay person. Would the following statement be accurate:
Very large numbers in javascript are often approximated.
However,precision should be guaranteed to 16 digits.
For example, 123455.373849 can always be represented accurately,
but the number 9,007,199,254,740,991,293 may not be.
Is there a better way to explain it, or any inaccuracies in the above statement?

16 digits? No, not really. Up to 53bit integers can be represented accurately and every number that can be represented as (53bit) * 2 ** (10bit).
Also, there are no 64bit integers in JavaScript, there are 64bit floating point numbers (and only 53bit of that hold the integer part), and BigInts that can have far more bits.
Very large numbers in javascript are often approximated.
Kind of, very large integers can only be approximated (or you use BigInts), however even small non integers, e.g. 0.1 can also not be represented exactly.
For example, 123455.373849 can always be represented accurately
No, probably not.
but the number 9,007,199,254,740,991,293 may not be.
Yup, thats far beyond 2 ** 53 - 1.

How to implement parseFloat

Wondering how a low-level implementation of parseFloat such as how it works in JavaScript would be implemented.
All the examples I've seen of typecasting resort to using it at some point, such as this, this, or this. On the other hand, there is this file which is quite large (from here).
Wondering if it is just a very complicated function or there is a straightforward implementation. Wondering just generally how it works if it is too complicated.
Perhaps this is closer to it.

The essential mathematics of parseFloat is very simple, requiring no more than elementary-school arithmetic. If we have a decimal numeral, we can easily convert it to binary by:
Divide the integer part by two. The remainder (zero or one) becomes a bit in a binary numeral we are building. The quotient replaces the integer part, and we repeat until the integer part is zero. For example, starting with 13, we divide to get a quotient of 6 and a remainder of 1. Then we divide 6 to get a quotient of 3 and a remainder of 0. Then 1 and 1, then 0 and 1, and we are done. The bits we produced, in reverse order, were 1101, and that is the binary numeral for 13.
Multiply the sub-integer part by two. The integer part becomes another bit in the binary numeral. Repeat with the sub-integer part until it is zero or we have enough bits to determine the result. For example, with .1875, we multiply by two to get .375, which has an integer part of 0. Doubling again produces .75, which again has an integer part of 0. Next we get 1.5, which has an integer part of 1. Now when the sub-integer part, .5, is doubled, we get 1 with a sub-integer part of 0. The new bits are .0011.
To determine a floating-point number, we need as many bits as fit in the significand (starting with the leading 1 bit from the binary numeral), and, for rounding purposes, we need to know the next bit and whether any bits after that are non-zero. (The information about the extra bits tells us whether the difference between the source value and the bits that fit in the significand is zero, not zero but less than 1/2 of the lowest bit that fits, exactly 1/2 of the lowest bit, or more than 1/2 of the lowest bit. This information is enough to decide whether to round up or down in any of the usual rounding modes.)
The information above tells you when to stop multiplying in the second part of the algorithm. As soon as you have all the significand bits, plus one more, plus you have either one non-zero bit or the sub-integer part is zero, you have all the information you need and can stop.
Then you construct a floating-point value by rounding the bits according to whatever rounding rule you are using (often round-to-nearest-ties-to-even), putting the bits into the significand of a floating-point object, and setting the exponent to record the position of the leading bit of the binary numeral.
There are some embellishments for checking for overflow or underflow or handling subnormal values. However, the basic arithmetic is simply elementary-school arithmetic.
Problems arise because the above uses arbitrary-size arrays and because it does not support scientific notation where an “e” is used to introduce a decimal exponent, as in “2.79e34”. The above algorithm requires that we maintain all the space needed to multiply and divide decimal numerals of any length given to us. Usually, we do not want to do that, and we also want faster algorithms. Note that supporting scientific notation with the above algorithm would also require arbitrary-size arrays. To fill out the decimal numeral for “2.79e34”, we have to fill an array with “27900000000000000000000000000000000”.
So algorithms are developed to do the conversion in smarter ways. Instead of doing exact calculations, we may do precise calculations but carefully analyze the errors produced to ensure they are too small to prevent us from getting the right answer. Also, data may be prepared in advance, such as tables with information about powers of ten, so that we have approximate values of powers of ten already in binary without having to compute them each time a conversion is performed.
The complications of converting decimal to binary floating-point arise out of this desire for algorithms that are fast and use limited resources. Allowing some errors causes a need for mathematical proofs to ensure the computations are correct, and trying to make the routines fast and resource-efficient lead people to think of clever techniques to use, which become tricky and require proof.

Why does JavaScript use base 10 floating point numbers (according to w3schools)?

I read this on W3Schools:
All numbers in JavaScript are stored as 64-bit (8-bytes) base 10,
floating point numbers.
This sounds quite strange. Now, it's either wrong or there should be a good reason not to use base 2 like the IEEE standard.
I tried to find a real JavaScript definition, but I couldn't find any. Either on the V8 or WebKit documentation, the two JavaScript implementation I could find on Wikipedia that sounded the most familiar to me, I could find how they stored the JavaScript Number type.
So, does JavaScript use base 10? If so, why? The only reason I could come up with was that maybe using base 10 has an advantage when you want to be able to accurately store integers as well as floating point numbers, but I don't know how using base 10 would have an advantage for that myself.

That's not the World Wide Web Consortium (W3C), that's w3schools, a website that isn't any authority for any web standards.
Numbers in Javascript are double precision floating point numbers, following the IEEE standards.
The site got the part about every number is a 64-bit floating point number right. The base 10 has nothing with the numerical representation to do, that probably comes from the fact that floating point numbers are always parsed and formatted using base 10.

Numbers in JavaScript are, according to the ECMA-262 Standard (ECMAScript 5.1) section 4.3.19:
Primitive values corresponding to a double-precision 64-bit binary format IEEE 754 value.
Thus, any implementation using base 10 floating point numbers is not ECMA-262 conformant.

JavaScript uses, like most modern languages, IEEE754. Which isn't at all stored in base 10.
The specificity of JavaScript is that there is only one number type, which is the double precision float. Which has the side effect that you're somewhat limited contrary to other languages if you want to deal with integers : you can't store any double precision integer, only the ones fitting in the size of the fraction (52 bits).

How many bits does JavaScript use to represent a number?

How many bits does JavaScript use to represent a number?

Generally JS implementations use 64-bit double-precision floating-point numbers. Bitwise operations are performed on 32-bit integers.

That depends on the specific implementation, not the language itself.
If you want to know what range of numbers is supported, then see section 8.5 (The Number Type) of the specification.

From the referenced spec:
The Number type has exactly 18437736874454810627 (that is, 264−253+3)
values, representing the doubleprecision 64-bit format IEEE 754 values
as specified in the IEEE Standard for Binary Floating-Point
Arithmetic, except that the 9007199254740990 (that is, 253−2) distinct
"Not-a-Number" values of the IEEE Standard are represented in
ECMAScript as a single special NaN value. (Note that the NaN value is
produced by the program expression NaN.) In some implementations,
external code might be able to detect a difference between various
Not-a-Number values, but such behaviour is implementation-dependent;
to ECMAScript code, all NaN values are indistinguishable from each
other.
That said be aware that when using the bit operators &, ^, >> << etc only the least significant 32 bits are used and the result is converted to a signed value.