Is there a Canon equivalent for the Nikon AF-ON focusing system?

This would be Canon's AI Servo autofocus, which is your choice of continuous servo approach. It is available on all Canon DSLR cameras, to my knowledge. The other two options are Single Shot and AI Focus. In high-end Canon bodies, you have the option of setting one of the rear body buttons to focus. In the newer bodies of Canon, there is usually a dedicated AF-ON button on the back of the camera already set up for this purpose. If you press and hold the AF-ON button (or the button of your choice) when using AI Servo drive mode, you can take continuous shots with focus between frames, subject tracking, etc.

When it comes to subject tracking, the camera can do some of the work, but you must make sure to keep the subject in the frame. You must also ensure that the subject is within an adequate range of active AF points, otherwise the camera will not be able to correctly determine what to focus on or track the subject. The proper technique of continuous / servo AF with subject tracking and focus adjustment between frames requires skill, and like any other skill, must be learned. You will have to put yourself into practice to be able to effectively use any AF system, and the more hours, the better you will get.

As for the highest AF points, it is actually Canon that has 61 points, in its two newest cameras of professional level: the 1D X and the 5D III. The Canon 61pt AF system is, specifically, the best in the world at the moment, with 41 cross-type sensors and 5 cross-type sensors, sensitive up to f / 5.6. The Nikon AF system is 51 points, with 15 types of crosses, whose central group is sensitive up to f / 8, while the rest is f / 5.6. Regarding whether one brand "exceeds" the other, that is truly irrelevant in the grand scheme of things. Both brands regularly outperform each other. At this time, Canon is the king of ISO and AF (many of the first users of the 5D III have been surprised with the capabilities and accuracy of the new AF 61pt system), where Nikon is currently the dynamic range and the Megapixel king. It is unlikely that these factors will remain the same for long, and after the next round of camera launches by both companies, the statistics will undoubtedly change again.

Do not use "who is better" as a factor to buy … that will never end well. Find out what you you want to do, what is your budget, if you want to be able to share glasses with friends that have the same brand as you, etc. The two Nikon cameras used by his friend, the D3s and the D700, are very advanced. Cameras, and they cost a lot of money. Canon has similar cameras, such as 1D X, 5D III, 1D IV, etc. that also cost a fortune. If you are just starting with the picture, it looks too close to the totem, and you might not need a camera like the one that appears at this time. Look at the lower range if you are starting. For Canon, that would be the Rebel series (numbers xxxD), and for Nikon, look at the D3xxx series. Buy a CHEAP camera body, since bodies come and go, and change every two years. The true long-term value in photography is in lenses, and in that area, Canon has a bit of an advantage with a larger selection and some unique entries in its line, as well as some unique designs that no other company uses. In any case, Nikon also manufactures excellent glasses, and you can not go wrong with any of the two brands.

Web Services – Equivalent for Authentication.asmx in SharePoint 2013

Everyone knows that SharePoint has a web service (ASMX), called Authentication.asmx, which is a good web service to do authentication. It has a method called Log in, which takes Username Y password and returns a cookie, if the user is valid. This is a great web service to use. But unfortunately, Microsoft recommends not using ASMX-based web services in SharePoint 2013 (probably due to performance reasons?) And mentions that it is present only for compatibility with previous versions. I was wondering if there is an equivalent (for example, WCF) to Authentication.asmx. Please let me know, if anyone has any ideas.

Unity Graphics.DrawMesh equivalent for physics?

I'm working on something where the same graphics grid is rendered several times in the scene. The problem with this is that each of these graphic elements has a physical mesh that is not found in the manually drawn meshes …

I have searched for something in the Physics and Collider documentation, but I have not found anything that does what DrawMesh does for the meshes of physics. Is there something for this? Or will I have to create a lot of physics objects manually?

router: the Linux equivalent of "no ICMP Unreachable" and "no ICMP Mask Reply"?

I have a Linux machine that is being used as a router. You have quagga installed and you have ip forwarding enabled. An OSPF adjacency is being formed with a Cisco 3925 router.

They ask me to make sure that this Linux machine (running Redhat 6.8) will not send any unreachable ICMP notification. I know that in a Cisco router it would simply establish "no unreachable" in any interface that needs it. But how can I be sure that they are disabled on a Linux machine?

They also ask me to make sure that "ICMP mask response" messages are not sent from the router. Similarly, I know that the "no ip mask-reply" command would be used on a Cisco router. Is there a type of equivalent configuration in Redhat?

Thanks for any help.

equivalent of tramadol morphine | Black Hat Seo and Affiliate Marketing Forum

Link —> equivalent of tramadol morphine

[​IMG]

Opioid dose calculator ← Back to AMDG start. Instructions: Fill in the mg per day * for the opioids your patient is taking. The website will automatically calculate the total morphine equivalents per day. Opioid conversion calculator for the morphine equivalents of the State of Washington and the bull; www.oregonpainguidance.org/opioidmedcalculator CAUTION: This calculator is designed to calculate the dose of equivalent doses of morphine (MED) for a patient taking one or more opioid medications. It should not be used to determine doses when converting a patient from one opioid to another.
This online opioid dose calculator is used to find the equivalent dose of total morphine. Example Calculate the equivalent dose of morphine, if the dose of codeine is 12 mg / day, tapentadol is 18 mg / day and morphine is 34 mg / day.
*** manufacturers' guidelines with a 2: 1 ratio of oxycodone: morphine (note that other conversions use a 1.5: 1 ratio for oxycodone: morphine) (3) Reference: (1) West Midlands Palliative Care Physicians ( 2003).
Conversions dependent on the dose: the conversion ratio of certain opioids may depend on the dose of the original opioid. In the case of converting morphine into methadone, methadone has a relative potency of 4: 1 at lower doses of morphine, but it becomes much more potent (12: 1) in patients who pass very high doses of morphine. 5, 7
9 mg / day of oral morphine equivalent to milligram. In other words, the conversion factor that does not take into account the days of use would be 9/5 or 1.8. However, since the buprenorphine patch remains in place for 7 days, we have multiplied the conversion factor by 7 (1.8 X 7 = 12.6).
Patient of tramadol to morphine. It has been reported that a potency ratio of oral morphine to oral tramadol is between 1: 4 and 1:10. Therefore, a total daily dose of 400 mg of oral tramadol is equivalent to 40 mg-100 mg of oral morphine. The potency ratio of parenteral morphine to parenteral tramadol is generally
DRUG NAME DOSE OF ORAL EQUIVALENT DRUGS MORPHINE CODEINE 30mg 4.5mg DIHYROCODEINE 10mg 1mg TRAMADOL 50mg 5mg to 10mg BUPECTIVE Accepts competition for buprenorphine Expiration of competition competition (buprenorphine)

Database theory: there are two ways to define the third normal form (3NF). How are these two definitions equivalent?

It is in BCNF, that is, for all dependencies of the form A -> B in F +, either:

i) is a trivial dependency, or

ii) A is a superclave of R

and if it is not in BCNF, at least:

iii) Each attribute in B-A is a main attribute

Discrete Mathematics: Why $ ∀x (P (x) → Q) $ is equivalent to $ ∃XP (X) → Q $

Yes $ X $ it does not happen free in $ Q $, so $ ∀x (P (x) → Q) $ is semantically equivalent with $ ∃XP (X) → Q $. How to understand this affirmation. And also, you need an example to show that the formula will not necessarily be equivalent if $ X $ it happens free in $ Q $.

I guess what worries me is that I do not understand what "free" means in this case. But it would be better to have an answer to the previous question.

Automata – What changes must be made in a Turing machine to be equivalent to a PDA, a DFA?

I think that to make a Turing machine have the same power as a DFA (by power I mean all the languages ​​that a DFA can decide, so can the Turing machine), we just do not allow any use of the rewind or rewrite. You can only move to the right and read the string in the order in which it is presented.

To simulate a PDA, however, I'm not sure what restrictions we might apply. I was thinking about allowing the backlash, but not rewriting the tape can make a Turing machine have a power equivalent to a PDA, but I'm not sure it's a correct statement or how it would prove the equivalence of the two.

Is $ mathbb {C} ^ n setminus V (f) $ equivalent homotopy with a "big ball add-on"?

Leave $ f in mathbb {C}[x_1,dots,x_n]$, and let $ V (f) $ denotes the escape locus. Is it true that for quite some time? $ N $, there is a homotopy equivalence
$$ mathbb {C} ^ n setminus V (f) simeq B (0, N) setminus V (f), $$
where $ B (0, N) = {| x | <N } $.

test verification: show that any divisor of order 0 in a non-singular projective curve of the genre $ g $ is equivalent to another

Could you please check if the solution below is fine?
There is a basic algebraic geometry exercise by Shafarevich, vol. 1, ex. 7.7.21.

Leave $ o $ Be a point of a smooth algebraic curve. $ X $ of the genre $ g $. Using the Riemann-Roch theorem, prove that any divisor $ D $ with $ deg D = 0 $ It is equivalent to a divisor of the form. $ D_0-go $, where $ D_0> 0, deg D_0 = g $.

It is equivalent to showing that space. $ mathcal {L} (go + D) $ has dimension $ geqslant $ 1. Suppose you have dimension $ 0 $. By the Riemann-Roch theorem we have
$ ell (go + D) – ell (K-go-D) = 1 $. Thus, $ ell (K-go-D) = – $ 1. As $ K-go + D $ Y $ K-go $ we are of the same grade, we have
$ ell (K-go) leqslant ell (K-go + D) $. Now apply Riemann-Roch to $ ir $:
$$ ell (go) – ell (K-go) = 1 $$
But the space $ mathcal {L} (go) $ It is the space of all rational functions having polo of order. $ leqslant g $ to $ o $, so of dimension $ g + 1 $,
Y $ ell (K-go) $ it's dimension $ g $ – the contradiction.

I do not like the last argument and we do not use the structure of a canonical divisor.

Thanks in advance.