New Subnet - Why does 10.55.50.1 with subnet mask 22 belong to 10.55.48.0

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Love the work that you are doing and can not wait for the VM stuff.

Anyway, I work at a school district and I have broken down my network into about 6 vlans. We are building a new school and there is a lot of IP infrastructure that is going into it (IP cameras, clocks, projectors, etc.) So I need to create a new subnet. I understand most of what I am doing, I just want to know why this is and how to figure it out on the fly -

I went to ye old local subnet calculator and put in the following address

10.55.50.1 and a subnet of 255.255.252.0/20 I would like to have 1024 addresses. I thought Vlan 50 would be good to remember. But when I put this in the calculator I get
Address: 10.55.50.1 00001010.00110111.001100 10.00000001
Netmask: 255.255.252.0 = 22 11111111.11111111.111111 00.00000000
Wildcard: 0.0.3.255 00000000.00000000.000000 11.11111111
=>
Network: 10.55.48.0/22 00001010.00110111.001100 00.00000000 (Class A)
Broadcast: 10.55.51.255 00001010.00110111.001100 11.11111111
HostMin: 10.55.48.1 00001010.00110111.001100 00.00000001
HostMax: 10.55.51.254 00001010.00110111.001100 11.11111110
Hosts/Net: 1022 (Private Internet)

My Questionr: Why does it start at 10.55.48.0 as the network address? I understand the range etc. Just why can it not start at 50 and go 10.X.50-53.X instead of 10.X.48-51.x. I do not understand how that part of the deal is figured.

Thanks

damewoodr,
Glad you're enjoying ITProTV and you're asking a great question!
The answer lies in your subnet mask! If you were to write out your subnetmask of the /22 = 255.255.252.0 This last octet is our only real concern here.
If you were to begin with a 10.0.0.0/8
One network 10.0.0.0, 16,777,214 hosts
With your subnet mask of /22, this means that you can have 2^14 number of subnets= 16,384 possible subnets from that 10.0.0.0/8 network. (I will not do all of them but I will set the pattern of where this begins.
1st - 10.0.0.0/22 00001010.00000000.00000000.00000000/22
2nd - 10.0.4.0/22 00001010.00000000.00000100.00000000/22
3rd - 10.0.8.0/22 00001010.00000000.00001000.00000000/22
4th - 10.0.0.0/22 00001010.00000000.00001100.00000000/22
5th - 10.0.4.0/22 00001010.00000000.00010000.00000000/22
6th - 10.0.8.0/22 00001010.00000000.00010100.00000000/22
...
(notice the interval of 4 in between the subnets--this comes from the fact that we are not concerned with 10 hosts bits) but we are concered with the FIRST 22 bits.)
If you were to copy and paste the above binary into a word proccessor and continue to work it all the way through to 10.55.40.0/22...
Xst - 10.0.40.0/22 00001010.00000000.00101000.00000000/22
Xnd - 10.0.44.0/22 00001010.00000000.00101100.00000000/22

Xrd - 10.0.48.0/22 00001010.00000000.00110000.00000000/22
Xth - 10.0.52.0/22 00001010.00000000.00110100.00000000/22
The /22 give you an interval of 4 because the 22nd bit is in the 3rd octet with a place value of 4. So your network ID numbers will always be increments of 4.
Hope this explaination helped you.
Cordially,
Ronnie Wong
Host, ITProTV