Is it good to drive the magnepan 3.6?

It sees half as much impedance because it has to supply twice as much current for a given voltage. With most amps the red speaker terminal swings back and forth from plus to minus while the black is fixed at ground potential. So lets say at some instant your red terminal is at 8V across 4 ohms and you'll have 2 amps of current flowing. If bridged to mono you'll be hooked to the other channel's red terminal which will be at -8V while the other is at +8V for a total of 16V across the 4 ohms which causes 4 amps to flow. If you ask 8V to supply 4 amps it is seeing a 2 ohm load.

You missed the point. You no longer have 8V across 4 ohms when you are bridged.

When it was not bridged it was 8V across 4 ohms for 2 amps. As you say, Ohm's law. Bridged you have 16V across 4 ohms so 4 amps. The difference between +8 and -8 is 16.

plus 8V------- 4 ohms ------- minus 8V so 4 amps will flow.

The 4 amps flows out of one terminal and into the other. If each terminal is developing 8V but supplying 4 amps then it is effectively a 2 ohm load. 8/4 = 2 = ohm's law.

Back to the original question, will this stress the amps? I think not. Your amp is capable of delivering a large amount of current as evidenced by the 1,600W rating into 4 ohms. Your ears would start bleeding before you got close to that limit. Try it and see how it goes.

rmlfour, you clearly don't understand basic electronics any more than Spectron does and clearly didn't comprehend the original questions, both of which I answered contrary to what you stated.

Vinhd asked if the amp would see a 2 ohm load. It does. Spectron said it did not not. Since Spectron gave the wrong answer I fail to see why it was wrong to correct him and therefore correctly answer the question. Your smart ass comment about the amp not wearing glasses demonstrates your lack of understanding. The expression "the amp sees" is a widely used one and understood by those who understand basic electronics which you and Spectron do not or we would not be having this conversation. When we say an amp sees a certain load we mean it is equivalent to that load. For instance; if you hook up two 8 ohm speakers in parallel to an amp it is effectively a 4 ohm load so it is common to say that the amp sees a 4 ohm load.

The fact that I corrected Spectron and used basic math to do so does not make it "kindergarten." How else can you try and explain a basic math problem without resorting to basic math? Yes, voltage is a difference of potential. After teaching basic electronics for almost 10 years I have a pretty firm grasp on that concept. That concept is exactly where you and Spectron are missing the point. You are hung up on 8V and 4 ohms when in reality we now have a difference of 16V which is the DIFFERENCE between +8 amd -8. In my example the difference of potential was 16V across a 4 ohm load causing 4 amps of current to flow. Each terminal is developing 8 of those 16 volts. An amplifier that is developing 8 V and has 4 amps of current flowing "sees" a 2 ohm load. I'm sorry you can't grasp that concept but that is as simple as I know how to put it.

If you go back and read my last post I also answered the question posted "Does that stress the amplifier's current capacity and will the amp be overheated?"

It is humorous that you chose to scold me because you believe I did "not add something to the questions posed in this thread" when I clearly addressed both parts of his question and your post added nothing but incorrect information and condescending remarks.

Since Spectron gave the wrong answer and after he was corrected insisted he was right with a condescending comment implying I didn't understand basic electronics, and someone else jumped in with a smart ass response defending his wrong answer, how else would you have replied other than to give a detailed response? Just let it go so Vinhd was left with the wrong answer?

I always required my students to show their work. Formula sheets will be provided so you will not be able to use your notes.

I have also been entertained by it. You would think that someone who represents Spectron amplifiers would take the time to get the right answer on a basic electronics question before sticking their foot in their mouth twice. His comments about reducing all types of distortion as well as bandwidth and slew rate doubling were also incorrect but I figured if he didn't understand the simple stuff there was no reason to go into anything else.