So that's my dream. That and some stuff about cigars and a tunnel.

Faith ,'Get It Done'


Buffistechnology 3: "Press Some Buttons, See What Happens."

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Liese S. - Mar 17, 2015 9:16:25 am PDT #24408 of 25496
"Faded like the lilac, he thought."

Great, Jon, thanks. I can install openSSH, which is probably what I should have gone with in the first place.


sumi - Mar 19, 2015 4:21:22 am PDT #24409 of 25496
Art Crawl!!!

I'm having an Instagram problem: when I try to take a picture it sends me to settings to enable access to my camera - but there is no spot in settings where I can enable access to my camera.

Anyone know what to do?


meara - Mar 19, 2015 6:20:54 am PDT #24410 of 25496

What kind of phone, sumi?


meara - Mar 19, 2015 6:23:00 am PDT #24411 of 25496

If an iPhone its under privacy, then camera--there should be a list of apps that have requested access and you can turn that access on or off.


Liese S. - Mar 19, 2015 10:32:40 am PDT #24412 of 25496
"Faded like the lilac, he thought."

So here's my progress: openSSH totally worked, so I got it installed on the NAS, port forwarding working, everything talking. Only I can't actually log onto the Crashplan software to control it over there, because apparently I still don't have internet connectivity from the Drobo. Which makes sense, because I couldn't get the email notifications up and running either, but had just deferred that problem to deal with later.

I'm thinking maybe I have my default gateway, etc. set up incorrectly. That is to say, I have everything set up as zeros, because I thought that set of routing was all about the LAN, not about getting to the internet. If I change the settings to what my modem uses, then I can't access the drobo through the LAN at all. So I am at another impasse.


sumi - Mar 24, 2015 10:21:14 am PDT #24413 of 25496
Art Crawl!!!

I realized that what I needed to do with my iphone is update the IOS. (Finally)


Typo Boy - Apr 05, 2015 12:50:36 pm PDT #24414 of 25496
Calli: My people have a saying. A man who trusts can never be betrayed, only mistaken.Avon: Life expectancy among your people must be extremely short.

I'm trying to figure out how much of the chemical energy in methane (CH4) is due to the carbon atom and how much due to 4 hydrogen atoms. (I'm looking at process to make methane out of hydrogen. Because there are good reasons why hydrogen gas is far less usable for certain purposes than methane (natural gas) I'm trying to figure out how much more energy is in the methane than in the carbon alone. According to my handing chemical table of elements the energy of ionization for hydrogen 131 kJ/mole whereas the the same figure for carbon is 153. So if "energy of ionization" is what I'm looking for, then the methane molecule has 4.4 time the energy that if the carbon atom was just used in a coal substitute. But am I figuring this right? Is "energy of ionization" really what I'm looking for?


Gris - Apr 05, 2015 2:27:50 pm PDT #24415 of 25496
Hey. New board.

Hmm. I am not as up on my chemistry as I once was, but I think that your approach doesn't really work because it doesn't look at the actual combustion reaction itself. You need to compare the energy change of burning methane vs. the energy of burning pure hydrogen gas or pure carbon gas. I think. The issue is that plenty of energy will be locked on the products in either case, so it is the energy released by the process that is actually interesting.

I don't really remember how to do that though.


Gris - Apr 05, 2015 2:31:29 pm PDT #24416 of 25496
Hey. New board.

This might help: [link]


DXMachina - Apr 05, 2015 7:10:21 pm PDT #24417 of 25496
You always do this. We get tipsy, and you take advantage of my love of the scientific method.

Gar, as Gris said, you have to look at the total energy of the reaction being carried out. If you burn methane, you also need oxygen as a reactant and you'll get carbon dioxide and water as products. The energy released from the reaction will be the difference between the total chemical potential energies of the molar quantities (the number of molecules of each compound required by the balanced reaction) of methane and oxygen, and the potential energies of water and carbon dioxide. The difference between the reactants and the products is the energy released (in this case) or absorbed.

If you want to look at the molecules themselves, it's not the atoms, but the breaking of the C-H and O=O bonds versus the formation of C=O and H-O bonds and the differences in bond energies among them.

If you need them, I can send you some of the powerpoint slides on the subject that I use in my introductory chemistry class.

Edit: I looked a Gris's link, and there's good basic information there. If you need more, feel free to email me.