How do I manually calculate the Firestore automatic index size?

Question

I'm relatively new to Firebase and I'm looking to calculate how much data I incur from automatically created single-field indexes on Firestore native mode. I understand that the rule of thumb is to go through and exempt fields and there are guidelines on this. However, I am more interested in how much data is actually calculated as an overhead if I don't exempt any fields.

I am looking for someone to verify my math. I used the rules outlined here https://firebase.google.com/docs/firestore/storage-size. This is what I have

Document size is the sum of:

The document name size = 48 bytes

The sum of the string size of each field name = Average of 9 bytes * 97 attributes=873

The sum of the size of each field value = Average of 70 bytes * 97 attributes=6790

32 additional bytes

Document size 48+873+6790+32 =7,743 bytes

Collection size = (doc size) * (# of docs) 7,743*130 = 1MB

Single-field index with collection scope is the sum of:

The document name size of the indexed document = 48 bytes

The document name size of the indexed document's parent document = 0

The string size of the indexed field name = 9 bytes

The size of the indexed field value = 70 bytes

32 additional bytes

Single-field index size on the average attribute = (48+9+70+32) = 159 bytes

Single field index with 130 documents= 159*130=20,670 bytes

Automatically created ascending and descending indexes 20,670*2 = 41,340 bytes

Single field index for 97 attributes 41,340*97=4MB