No one actually knows how to lculate what speaker cable they need

Mr. stevecham

I would like to clear this subject:

1. This thread is about speaker cables.

2. Your (all of you) system is a given. I have nothing to do with your choice of equipment.

3. The idea of speaker cables to fit the DF has no significance when tube amplification with very low DF is involved. So is your case. There were other guys who asked me, with tubes, I've told them the same and they accepted. Please do too.

4. If such case occur (low DF), it is not me to say it is not working for that, it is the math behind the idea. As much as it works for high DF, it is not working for the low DF.  

5. The speakers are not much of a player in this game. Unless, they are not conventional. Even thou, in the case of Mr. keppertup it worked very well.

I think that the name: "speaker cables" are a mistake. They should be called amplifier cables!

 

I do not want to get personal in this thread. Except a few guys who need their daily dose of contemning my thread, for quite a while, all other are welcome.

 

Speakers sensitivity or efficiency: The range goes from the highest (106dB/w/m SPL) to the lowest (85dB/w/m SPL). There might be exceptions, but this is the majority.

The combination of 88dB/w/m SPL sensitivity with low impedance of 2.7 Ohms, makes the Thile a taught cocky to drive.

 

Understanding Damping Factor and its role in the chain of amplification, cables and speakers:

If I would reflect the audio world into the car's world, the amplifier would be the engine. The speakers would be the wheels. For the power (watts) would be the engines HP. The DF is the engines torque.

The speaker cables would be the driveshaft.  

We know that an engine with lots of HP but low torque will not deliver the sporty drive experience. A powerful engine must have both: the high HP and a high torque.

The fact, that tube amplification has a problem with DF is an old issue, that been tackled by SS. It is even better with class D but in a different way, not as with class A/B amplifiers.

Speaker cables as drive shaft, would present a serial device with some flexibility to be the extension of the amp's DF. A thin cable would be more flexible, and a thick one more rigid. Why would someone like to get a car with a strong engine with fantastic torque and than ruin it with a thin (flexible) driveshaft? To deliver the power from the engine to the wheels with a high torque engine is like delivering the power of an amplifier to the speakers from a high DF amplifier. It is about control! The high DF is more control. A coil loaded speaker requires such control, to overcome the reactive nature of a moving coil in a magnetic field (speaker structure).  

That's the analogy. I hope it helped some to better understand it.

This relation is not an empiric figure to guess of find by trial. It is a figure that has values and can be calculated. The calculation is telling with accuracy, what is the required cable for the task.

Regarding your system, did you try some other combinations like matching a very efficient and easy to drive, horn loaded speaker with your Manelly?

Did you try a powerful SS amp with high DF and thick cables attached to your Thile?

Even if you did, and you like your sound, that's O.k. My remarks were pure technical.

To OP: Aside from all the posturing on this thread, I am curious about a couple of your ideas but have a couple of questions.

1. Where is this "formula"?  I went back through two weeks of posts and still could find no formula, just multiple comments about basing the wire size on the amplifier and how "0 gauge" seems to usually work well, and also a couple of comments about needing thicker than 0 awg in cases where the SCs are longer.  When I started to see other posts asking "where is the formula" I assumed going back further would be of no help.  I would like to apply this formula to my situation and calculate what gauge wire I would need to meet your formula based on my SS amplifier.

2. I read your opinion about separating wires and assume your DIY results in running separate pos and neg wires from the amplifier to the speaker.  Your point seems to be that this should resolve mutual inductance and that resistance (i.e., inverse of wire size) is the governing factor.  Some manufacturers have done this (separated pos and neg wires) in the past but typically that approach has not seen long-term commercial success (low WAF?).

One way to get pretty close to your 0 awg would be to purchase 8/3 50A NEMA welding extension cord and then connect the three internal wires together and use a single run separately to each pos and neg terminal.  Each post would then see just under 3 awg (i.e.,  3 x 8 awg ~ 3 awg) 

To actually achieve 0 awg wire, I would probably purchase 0 awg battery jumper cable wire in the length I need and crimp a short, large gauge, lead-out wire to the jumper cable wire that would attach to my speakers and amp.   I would run the two battery jumper cables separately with one going to each of the pos and neg posts.

Where is the formula?

Mr. milpai

1. The formula: I’ll keep it to myself. It is not to be given away. If you would read the thread, you would find more than one to be answered about the formula.

2. Your ideas of implementation of a 0 AWG DIY is up to you. You can see some solutions in this thread.

   For some reason, you think that 0 AWG is good for all. Some need less (like 4 AWG or even less) and some need more, if the cable gets long or the DF goes really high.

   Please be kind to pay attention that the AWG table is not double incrementing by 2 on each number of gauge. If a cable would require a 4 AWG for 2m length, it will be 1 AWG on a 4m length.

   The part of what cable you need, I’ll calculate it for you, if you be kind to provide:

1. Amplifier’s model and brand.

2. Speaker cable length.

3. Speaker’s model and brand.

If you will read my answer to Mr. stevecham, there are some exceptions, regarding speaker technology and tube amplification. I’ll take care of that.

How do you take that requirement and turn it into a cable, is up to you. If so many cable makers from the industry used to do such cables, by your say, you might find one to buy. It must cost nickels as those are now left over’s. The last time I checked, those who may offer such a thick cable asked something north of $40k. But money should be no issue for a real audiophile that follows its ears. Common sense might be more difficult commodity to find!

The other way is to get a DIY project. There are plenty of options on eBay for cables. The problem is getting the ends fit a binding post. Some creative solutions were shared here, by Mr. keppertup and Mr. conradnash as well as with the sonic results. There might be more ways to do it. For anything thicker than 0 AWG, like 2x AWG or 4x AWG, a more difficult solutions will be required. It might also get more expensive…

b4icu OP"The formula: I’ll keep it to myself. It is not to be given away"

Well I don't think you can give it away anyway as has been demonstrated in this thread and you would not be able to sell, distribute or license your "formula" because it is just so much "mumbo jumbo" as Americans say!

Maybe a fun DIY project in my spare time would be to order this;

https://www.amazon.com/dp/B0167K786A/ref=sspa_dk_detail_3?psc=1&pd_rd_i=B0167K786A&pf_rd_m=ATVPDKIKX0DER&pf_rd_p=f52e26da-1287-4616-824b-efc564ff75a4&pf_rd_r=83DVKJH6N8G9QDV816AJ&pd_rd_wg=BsaKN&pf_rd_s=desktop-dp-sims&pf_rd_t=40701&pd_rd_w=kUgYS&pf_rd_i=desktop-dp-sims&pd_rd_r=f21cc8f0-dd1e-11e8-bee5-11f85c3f70ca

and these;

https://www.amazon.com/SELTERM-Gauge-Copper-Welding-Eyelet/dp/B073Y9RYFM/ref=pd_sim_328_4?_encoding=UTF8&pd_rd_i=B073Y9RYFM&pd_rd_r=fe7ca96e-dd1e-11e8-8073-9598c706805b&pd_rd_w=oz9zG&pd_rd_wg=xgHbu&pf_rd_i=desktop-dp-sims&pf_rd_m=ATVPDKIKX0DER&pf_rd_p=18bb0b78-4200-49b9-ac91-f141d61a1780&pf_rd_r=3QTYW38P4053NS2A9H5E&pf_rd_s=desktop-dp-sims&pf_rd_t=40701&psc=1&refRID=3QTYW38P4053NS2A9H5E

and then I could run 3.75 ft. lengths of two red and two black of each to my bi-wire speakers. I have a hydraulic crimp tool that has a die to accommodate 2 awg connectors so cutting and stripping the wire will be the most difficult part of the operation.  Those copper ring connectors might actually fit over my binding posts by removing the nut, installing the copper ring over the post and then replacing the nut.  About $70 to see what all the fuss is about.  Oh, BTW, I ran the calculation and 2 awg is what I need.