College maths help.

Discussion in 'Science' started by Brett Nortje, Jan 21, 2017.

  1. Polydectes

    Polydectes Banned

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    Interesting thread. I learn and use mathematics as I need too I work or school but I quickly forget it because there isn't enough usage of it I'm my daily life. I think most people view it that way. Its always fascinating to come across someone who simply enjoys it. (I'm not assuming you don't use it for work.)
     
  2. Brett Nortje

    Brett Nortje Well-Known Member

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    Heck no, I don't use maths, I didn't even do it at school. I just like to speculate new techniques.
     
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  3. Polydectes

    Polydectes Banned

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    I'm trying to get into a field where I actually have to use some of this type of math.
     
  4. Brett Nortje

    Brett Nortje Well-Known Member

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    Clairaut's equation;

    This equation is about finding the ever changing system value, or, [f]. This would be for calculating the amount of energy used to get to [f], or, more precisely, how much [f] has to change to stay stable with the engine or building or such device - I am not sure, it does not explain that. However, to keep things brief, you will see this in your textbooks, what it is for, and, I suggest you understand what it is used for before you try to calculate it.

    We find [y] all over the place, and, it is in [y] we find by [x], yes? [1y] / [1x] * [x] + [1y] + [1x] * [f]? This is rather easy now, yes?

    So, we take, for the left hand side, [1y] * [1x] = [1]? Then, to find [f], we take [1f] + [1x] equals, oh, what is missing, [y]! This means [1y] = [2]!
     
  5. Brett Nortje

    Brett Nortje Well-Known Member

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    With trigonometry, the idea that the stress can be worked out easier with this idea of mine amuses me.

    If you were to take the two points of the graph, you could put a circle in the half way point of the two points. This could be scaled to allow for the circle to have the point you are looking for, scaled to the point that it should be, directly bisecting the two points as if they carried on. This would result in the new circle being able to predict the four sections of influence, or, three sections of influence for the circle. These would be 270 degrees between points, with the first or related one being where 'the 135 degrees to either side of the circle,' where the circle will gather stress. This would be where the stress bleeds of, in three sections, as if it were 'a circular triangle.'
     
  6. Brett Nortje

    Brett Nortje Well-Known Member

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    The Fredholm Integral equation is about finding constants in the engine to guide the engineer to find the stress values that the engine can withstand and operate at a constant speed under acceleration and top speed.

    [G / k] / [-f^w] would be the answer, as, that is where all the ones are crossed out. The people that put this theorem together had a lot of constants that they thought should be included, but, they all cancel each other out.
     
  7. Brett Nortje

    Brett Nortje Well-Known Member

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    The Hill differential equation is used to find the differences in the degrees of stress that the engine can handle over various distances.

    [​IMG]

    So, it would be [d4t] + 3 * n * [2nt] + [2m] + [2mt]. Then, we can simplify this by saying everything besides [d4t] + 0, as all that bracket junk is times by zero, the answer is [distance] by [four time].

    The answer is distance divided by four!
     
  8. Brett Nortje

    Brett Nortje Well-Known Member

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    The Ishimori equation is about finding the speed things may turn at without losing power, or, becoming too stressed. This is like where a piston will turn on a train's wheels, and, need to be kept in line with stress bearing materials - the materials need to be strong enough to keep the parts in the vortex, or, cylinder, or area of operation.

    [​IMG] [​IMG]

    As we can see, it comes in two halves, separated by the coma and [1] notes.

    If we were to be totally honest, it is just a lot of multiplication and division, yes? Let's take it on!

    [S / t] = [4s 4d 2x 2y] = [4sd + 2xy] + [duxys] = [4s 14d 2u 4x 4y],

    [2d - 2d =] 1d 1u 4x - [a2] = 8a * [s * x ^ y],

    This means, of course, that we will be taking Speed divided by time times by two, or so, with the 'graph score' being multiplied by itself. In short, we need to take the speed divided by time times two, so it is an average - you know you add two things together and then divide them by two, giving you an average? - then to the power of one. Putting it to the power of one will be where we find the two dimensions of it, but, if we want three dimensions, we merely put it to the power of three, of course, giving direct three dimensional measurements and values for 'the vortex.'
     
  9. Brett Nortje

    Brett Nortje Well-Known Member

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    In Laplace's equation, the objective is to put thee sums into three fields or vectors, so as to find the three dimensional values of the object. This would be where we calculate, maybe stress bearing or power outputs potential of the shape or engine, for example, by 'the three lines' of the diagram being given values based on their strength - they will hold the power under certain conditions before they break.

    Now, as we can see, he objective is to find the value of the system or triangle thing next to the [f]. If we were to ignore the zero on the right hand side, it would eliminate some of the sum, as anything equal to zero is zero, and that is a mistake!

    So, upon reading the whole point of this, we need to find the answer of the three dimensional shape. in the end it is [x] + [y] + [z], yes? So, we can say [1f] all the way along the top, with the squiggle d thing to the left of the [f] being equal to [4d] as to the power of two is two then four, yes? This means we are half way!

    Then, we need to observe that the [d4] / (or divided by) [d1] = [4d]. Seeing as how it is a three dimenaionl object, and, [f] is constant throughout, it is obviously already [1/3f], so, would come to [1d] * [4x], [4y] and [4z]. this means the answer to his whole sum, in total, is to simply [4xyz].

    In conclusion, the shapes can handle four times their value numerically than their lengths. This would also come down to materials used, and, I suspect it is because they are squared, and they compliment each other. This means, as there are three 'lines or vectors,' they will would handle more if there were more lines, yes? Just speculating...
     

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