Minimum Frame Size

[sivam siva]

Can anyone give me a trusted source or calculation to find the minimum frame size required to detect the collision?

I'm having confusion as I'm having a discrepancy with multiple sites and their various formulas.

from my knowledge:

Transmit Delay of interface(Td)= Size / BW 

Propagation Delay in the Wire(Pd)= Distance / Speed

Total Delay=Td+Pd

RTT=2 x Pd

please take 10Mbps for the example calculation 

The minimum frame size for 10Mbps is 64Bytes, and I want to find out this value by using the above formula if that is correct.

also please tell me what is the correct maximum distance for 10Mbps that can be used in the formula.

Thanks 

Siva

[Joseph W. Doherty]

"correct maximum distance for 10Mbps that can be used in the formula."

2,500 meters (10Base5, five segments, each 500m, 4 repeaters)

collision time limit, for 10 Mbps Ethernet is 51.2 us

[sivam siva]

But why many people are telling 100m is the maximum cable length?

[Joseph W. Doherty]

That's for twisted copper, #Base-T, 10Base2 (200m) and 10Base5 (500m) can have longer segment lengths. (Actually so can twisted copper, 100m is for certain "cat" categories at certain speeds. Transmitting at a lower "speed", with better/higher grade "cat" cable, you can usually go further than 100m.)

Also see: https://customcable.ca/cat5-vs-cat6/

[sivam siva]

Hi Thanks for the reply
Let's say I'm using a 10BaseT standard cat cable with 10mbps ethernet for the calculation.
Could you please help me to find the below value
If the max distance is 100m then what is the delay & RTT
if Delay & RTT known means what is the minimum frame size for it (with formula & calculation).
I hope you know what is Transmit Delay of the interface(Td), Propagation Delay in the Wire(Pd), if yes please tell me " RTT= 2xPd " or " RTT= 2xPd + Td ".

[Joseph W. Doherty]

First, understand, I believe the standard was defined assuming 10Base-5 (for timings). So, in theory, twisted copper could go the same length, but the 100m limitation is based on other physical properties of twisted copper. I.e. the calculation won't appear to make sense for twisted copper.

Most of the delay would be based on the PD of the medium. (I think [?] 10Base-5 and 10Bast-T would have very similar PD for same cable length.)

I also believe (?) the standard allow a certain amount of time for each repeater (max of four) - which would at least be, I think (?) 1 bit time.

[sivam siva]

@Joseph W. Doherty 

Very Thanks for your continuous reply 

finally could please tell me only how 64 bytes calculated using formulas (or please refer me any trusted source )

Thanks

Siva

[Joseph W. Doherty]

See if what you seek might be found in the two referenced links:

https://stackoverflow.com/questions/33039394/why-is-the-minimum-ethernet-frame-64-bytes

https://serverfault.com/questions/154558/why-does-ethernet-have-a-minimum-frame-size-specified