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USB 2.0 fast enough for 40x burner?

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USB 2.0 fast enough for 40x burner?

Postby articulate on Wed Mar 26, 2003 12:56 pm

I want to connect my Teac 540E to a new PC (2.6 Ghz) externally via a USB 2.0 IDE box, since there are no internal slots available. Question is: will USB 2.0 allow the full writing speed, or do I have to look at a Firewire unit?
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Postby cfitz on Wed Mar 26, 2003 12:58 pm

It's fast enough.

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Postby hoxlund on Wed Mar 26, 2003 1:42 pm

yeah usb2.0 specs are 480Mbps, which even still, tons of stuff has a hard time hitting, i have a Nomad Jukebox 3 that uses firewire for transfering music to the 20GB hard drive, and that thing like never hits half the spec speed of firewire (400Mbps)

so your better off with usb 2.0, cause in many cases its more widely used and compatible, and it offers higher transfer speeds, theoritcally
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Postby burninfool on Wed Mar 26, 2003 2:53 pm

I have my Sony DRU-500A in an external case connected by firewire.The compatibility issue between firewire and USB2.0 is negligable but for video editing firewire is a must.As to speed yes USB2.0 is faster than 1394 but there is now firewire800 1394b(800mbps) which is ideal for networking.
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Postby hoxlund on Wed Mar 26, 2003 5:43 pm

right now, apple is the only computer company that has anything out for 1394B, and besides, you don't need 800Mbps for burning, or even hard drive use, it barely uses all of the 400Mbps or 480Mbps for either usb 2, or firewire 1
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Postby brillman on Thu Mar 27, 2003 12:05 am

Da Math:
153.6KBytes/sec(1X speed) * 40X = 6144KBytes/sec
6144KBytes * 8 = 49152Kbits(or 49.152Mbits)

Will 480Mbits/sec capacity(USB2.0) cover 49Mbits/sec required?
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Postby hoxlund on Thu Mar 27, 2003 12:08 am

im just curious, whats this 6144KBytes * 8 = 49152Kbits mean, whats that *8 doing there?
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Postby Ian on Thu Mar 27, 2003 12:17 am

There are 8 bits in a byte.
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Postby hoxlund on Thu Mar 27, 2003 12:54 am

ok, just checking, i thought it was going to be something like that
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Postby hoxlund on Thu Mar 27, 2003 12:58 am

so you need 63Mbps to burn at 52x, wow never thought about it that way, so you still have 417Mbps bandwidth left over for the burn
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Postby dodecahedron on Thu Mar 27, 2003 5:18 am

brillman wrote:Da Math:
153.6KBytes/sec(1X speed) * 40X = 6144KBytes/sec
6144KBytes * 8 = 49152Kbits(or 49.152Mbits)

Will 480Mbits/sec capacity(USB2.0) cover 49Mbits/sec required?
You be the judge

i, on the other hand, was curious as to the 153.6KBytes/sec figures.
i thought 1x was 150KBytes/sec ???
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Postby MediumRare on Thu Mar 27, 2003 6:09 am

dodecahedron wrote:
brillman wrote:Da Math:
153.6KBytes/sec(1X speed) * 40X = 6144KBytes/sec
6144KBytes * 8 = 49152Kbits(or 49.152Mbits)

Will 480Mbits/sec capacity(USB2.0) cover 49Mbits/sec required?
You be the judge

i, on the other hand, was curious as to the 153.6KBytes/sec figures.
i thought 1x was 150KBytes/sec ???

This results from the general (sloppy) use of decimal prefixes for binary multipliers:
150KiBytes = 153,600 Bytes = 153.6 KBytes.

1 KiByte = 2^10 Byte = 1024 Bytes = 1.024 KByte.

See the discussion starting roughly here: http://www.cdrlabs.com/phpBB/viewtopic.php?p=55313#55313

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Postby articulate on Thu Mar 27, 2003 8:39 am

brillman wrote:Da Math:
153.6KBytes/sec(1X speed) * 40X = 6144KBytes/sec
6144KBytes * 8 = 49152Kbits(or 49.152Mbits)

Will 480Mbits/sec capacity(USB2.0) cover 49Mbits/sec required?
You be the judge


You've convinced me. :wink:

Thanks all !
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Postby hoxlund on Thu Mar 27, 2003 11:54 am

and in the end, thats all that matters, your welcome
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Postby technut on Fri Mar 28, 2003 4:00 am

brillman wrote:Da Math:
153.6KBytes/sec(1X speed) * 40X = 6144KBytes/sec
6144KBytes * 8 = 49152Kbits(or 49.152Mbits)


I largely agree with your math, but I wonder if it tells the whole story. The figures you cite only include the raw data being transferred. Most interface protocols include quite a bit of overhead besides the actual data, which uses up bandwidth. They may also include error correction and detection bits that make the actual bandwidth used greater as well.

Overall, of course, in this case there is no reasonable way there would be so much overhead to cause a problem (would require 3x overhead), but it is something to think about when discussing issues like this. I wish I had data as to the volume of the overhead in an IDE and USB interfaces, but unfortunately, I do not.
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Postby blakerwry on Fri Mar 28, 2003 4:59 am

I think you can test the overhead of certain IDE specs through the use of empirical data.

You would need a hard drive capable of speeds beyond that of the bus you are testing, but it wouldn't be too hard.

Take said HDD (any newer HDD based on 60 or 80gb platters) and benchmark the 1st 8GB's using HDTach or winbench99's disk inspection tool.

using IBM's drive feature tool set the max speed of the drive from UDMA 4 to 3 to 2 to 1 to 0... PIO 4, 3, 2, 1, etc.. Benchmark at each speed... Since you know the drive can perform better than the bus, the difference between the bus's theoretical transfer rate and the benchmarked results is approx. the overhead involved.

Since the data throughput of a drive with 80gb platters (like the 7200.7 or 40GB DM+8 or some DM+9 models) is about 60MB/sec at the beginning of the platter you might be able to get an idea of the overhead involved with ATA66... but otherwise you'll have to settle on ata44 (UDMA mode 3) or below.
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Postby dodecahedron on Fri Mar 28, 2003 11:54 am

MediumRare wrote:150KiBytes = 153,600 Bytes = 153.6 KBytes.
yeah, but how do you know what is the correct figure:
150 KiBytes (=153.6 KBytes) or 150 KBytes?

anyway, for me K is always 1024...
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Postby cfitz on Fri Mar 28, 2003 12:13 pm

dodecahedron wrote:yeah, but how do you know what is the correct figure:
150 KiBytes (=153.6 KBytes) or 150 KBytes?

You just have to know by the context, unfortunately. Meaning you really have to know independently. In general, hard disc capacities, flash memory, and many networking numbers are reported using the SI multipliers (powers of 10). RAM and CD-R/RW capacities are reported using the binary multipliers (powers of 2^10). In other circumstances, you are on your own... :( :wink:

All this confusion is why some advocate using the binary multipliers whenever appropriate, as MediumRare has done. If everyone did so, then there would be no confusion.

dodecahedron wrote:anyway, for me K is always 1024...

But it isn't always 1024 for the rest of the world (in fact many argue that by all rights 1K should mean only 1000), so you have to be aware of the differences. Nothing is ever easy... :evil: Oh well, if this was the worst of our problems we would all be doing quite well indeed...

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Postby MediumRare on Fri Mar 28, 2003 1:43 pm

dodecahedron wrote:anyway, for me K is always 1024...

so a km is 1024 meters ?

The thing is- if you talk about numbers, you have to agree what they mean. Ask someone in North America what a Billion is and they'll say a thousand million. Europeans call this a Milliard and a Billion is a million million. That's why we have SI multipliers like Giga and Tera to avoid the confusion. And why we should use the nonconfusing Gibi and Tebi etc. for the binary multipliers.

Unfortunately these have not yet been officially adopted, but are well on the way (see KCK's post). They were formalized at a rather late stage and have to fight with old habits.

[rant]Maybe we'll have to wait for a generation to die out- I don't know how many Britons or Canadians (probably more because of cultural diffusion from the south) still think in terms of miles and pounds and british thermal units, slugs, rods, gallons (imperial or US) or that whole zoo.[/rant] :wink: :P :o

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Last edited by MediumRare on Fri Mar 28, 2003 4:52 pm, edited 1 time in total.
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Postby KCK on Fri Mar 28, 2003 2:05 pm

Concerning dodecahedron's context, 1x means reading 75 sectors per second.

For a data CD with 2048 bytes per sector, this is 150 KiB/s = 153.6 KB/s.

For an audio CD with 2352 bytes per sector, this is 176.4 KB/s = 172.265625 KiB/s.
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Postby dodecahedron on Fri Mar 28, 2003 2:28 pm

KCK wrote:Concerning dodecahedron's context, 1x means reading 75 sectors per second.

For a data CD with 2048 bytes per sector, this is 150 KiB/s = 153.6 KB/s.

For an audio CD with 2352 bytes per sector, this is 176.4 KB/s = 172.265625 KiB/s.

that's the figures i was looking for!
thanks KCK!
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