# Can't fit 147 K32 Plots into a 16TB Drive

So i have a 16TB HDD.

Formatted like this:

`sudo mkfs.ext4 -m 0 -T largefile4 -L <drivename> /dev/sda`

I used this tool to calculate the perfect plot plan.

I have put 146 K32 Plots into this drive. Now the free space is 107,2 GB, which means, that i cant fit the 147. plot inside this HDD.

Is there a way to make it fit? Maybe some capacity is reserved, which i can delete safely without damaging existing plots?

Will this command fix the problem? Could it corrupt my existing plots on the drive or is it safe? Should i unmount the drive, before using this command? I am using Ubuntu btw.

``````tune2fs -m 1 /dev/sda
``````
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Not sure I understand. Is the objective to create a plot in the 107,2 GB of available space of the mentioned hard drive?

From what it ways in the Chia plotting basics. A single k32 plotting process never needs more than 239 GiB of space. One needs to be careful here as 239 gibibytes uses 1024 as its divisor where GB or gigabytes uses 1000 as the divisor. That means you will need 256.6 GB of temporary space and the final plot file will take 108.8 GB.

Update:
I think I understood your question. You want to know why the capacity of your hard drive is not actually what the manufacturer has advertised?
Here is an interesting article that Seagate talks about hard drive capacity

Hard drive manufacturers market drives in terms of decimal (base 10) capacity. In decimal notation, one megabyte (MB) is equal to 1,000,000 bytes, one gigabyte (GB) is equal to 1,000,000,000 bytes, and one terabyte (TB) is equal to 1,000,000,000,000 bytes.

Programs such as FDISK, system BIOS, Windows, and older versions of macOS use the binary (base 2) numbering system. In the binary numbering system, one megabyte is equal to 1,048,576 bytes, one gigabyte is equal to 1,073,741,824 bytes, and one terabyte is equal to 1,099,511,627,776 bytes.

Capacity Calculation Formula

Decimal capacity/1,048,576 = Binary MB capacity
Decimal capacity/1,073,741,824 = Binary GB capacity
Decimal capacity/1,099,511,627,776 = Binary TB capacity

Example:
A 500 GB hard drive is approximately 500,000,000,000 bytes (500 x 1,000,000,000).
When using the GB binary calculation, (500,000,000,000/1,073,741,824) that same 500 GB will show as 465 gigabytes.
This is why Windows will show a 500 GB drive as 465 GB.

A 5 TB hard drive is approximately 5,000,000,000,000 bytes (5 x 1,000,000,000,000).
When using the TB binary calculation, (5,000,000,000,000/1,099,511,627,776) that same 5 TB will show as 4.54 terabytes.
This is why Windows will show a 5 TB drive as 4.54 TB.

TL;DR: 16TB advertised capacity is actually 14,901.16 GB. If you create k32 plots you will be able to make 136 plots total

You can only fit 146 k32 plots on a 16 TB drive, so youāre good

If you want to use all your available space, then you can use the Chia Plot Plan to plot a mixture of k32s and k33s.

Also, as stated above, I used this tool to calculate the perfect plot plan. It says 147 plots are possible.

However, on my drive it is actually not possible right now and i want to know why.
Is this command

tune2fs -m 1 /dev/sda

going to help me remove some of that drive overhead to get a little bit more space for the final plot? Should i unmount before?

As stated above, i am already using this tool and it does say 147 plots are possible.

And Iām saying this tool is wrong

They say

This calculation is based on k32 plots being 108.837 GB, k33 plots being 224.227 GB and k34 plots being 461.535 GB

But a k32 plot is 101.4GiB which is ā 108.877 GB.

So a 16TB drive can only hold 16000 / 108.877 ā 146.95.

2 Likes

Okay i understand you now.thanks

I just realized i already used
-m 0
in the format command

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Do you know another tool, where the best filling combination of K32 and K33 on different hard drive sizes are calculated? Because, as you said, the numbers in https://plot-plan.chia.foxypool.io/ are wrong.

For example at 8 TB, 57 K32 and 8 K33 are suggested. However if i do your math:
57 * 108.877ā6205.989
8 * xāy
How much is x or how big is a K33 Plot again?

6205,989+y will probably be more than possible on a 8TB driveā¦

Check your plots and verify they are ok. You could have broken plots, they could be smaller then should be. Then you can fit 147 plots on 16TB.

Yes, this is not exact science since plots are compressed files their final size might be slightly different.

If you remember any matrix math you can do some algebra to solve it yourself.

In factā¦I will have to do thisā¦been awhileā¦

EDIT: Hereās what you do: Graph `108.837x + 224.227y =D`, where D is your drive size in GB. Your mission, should you choose to accept it is: Find the whole numbers x, y so that that point is closest to the line. Closer to the line you are, the closer you are to using all your storage on that drive.

Itās 5:06am, I havenāt slept yet. Iāll figure out a better way to do this later.

EDIT 2: Better way maybe is just asking the foxypool guy to update the numbers?

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True, I was more thinking about file copy was interrupted by something like a reboot, process been killed off by Linux due to lack of memory.

Iāve been using a spreadsheet for this, you can see it and maybe clone it, it kind of sucks but it is what I am using at the moment. You change the number under āoptimizeā with the gigabytes capacity of the drive using ādf -BGBā. Then you just follow the boundaries, left to right, between green and red until you see a point of local max utilization (green cell closest to zero) and the row/col gives you the ideal mix of 32s and 33s.

Note the first max may not be the global maximum, you might find a better one further to the right but I donāt want to plot hundreds of 33s if I can help it

At some point I will write a go cli app that does this for you with a list of the top 3 or so combinations, but I have not bothered to sit down and write it yet

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Darn, youāre right!

P.S. Proudly contributing nothing to the conversation because it is Tuesday.

P.P.S. Seagate Exos 16TB with k32 size.

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I spoke with the plot plan guy. He said his numbers are an average of plot sizes. Hence the discrepancy.

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That makes sense

My numbers are theoretical maximums (regarding file size). In practice, since plots are random and compressed, they can be a tiny bit smaller so you can probably fit 1 or 2 more per disk.