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aivan3 [116]
2 years ago
14

How is an orthographic drawing similar to or different from an isometric drawing?

Engineering
2 answers:
evablogger [386]2 years ago
4 0
An isometrical drawing is a nearly 3d drawing showing the object's width and depth in a complete image, from each curved plane of the orthhographic view, the viewpoint is at a 45 degree angle. From an observations point of view, isometric differs, since all longitudes are true.
Oxana [17]2 years ago
4 0
An orthographic drawing is one that provides consistently measured non-perspective views from above, front and right side. Using Orthographic projection one can then develop a measured pseudo- perspective drawing:


An isometric drawing is a form of mechanical perspective drawing in which receding 90º angles are drawn at a consistent 30º. There is no convergence toward a vanishing point so it is not a true vanishing point perspective.
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olchik [2.2K]

Numerical weather prediction (NWP) uses mathematical models of the atmosphere and oceans to predict the weather based on current weather conditions. Though first attempted in the 1920s, it was not until the advent of computer simulation in the 1950s that numerical weather predictions produced realistic results. A number of global and regional forecast models are run in different countries worldwide, using current weather observations relayed from radiosondes, weather satellites and other observing systems as inputs.
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2 years ago
Find the differential and evaluate for the given x and dx: y=sin2xx,x=π,dx=0.25
Sedaia [141]

By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

<h3>How to determine the differential of a one-variable function</h3>

Differentials represent the <em>instantaneous</em> change of a variable. As the given function has only one variable, the differential can be found by using <em>ordinary</em> derivatives. It follows:

dy = y'(x) · dx     (1)

If we know that y = (1/x) · sin 2x, x = π and dx = 0.25, then the differential to be evaluated is:

y' = -\frac{1}{x^{2}}\cdot \sin 2x + \frac{2}{x}\cdot \cos 2x

y' = \frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}}

dy = \left(\frac{2\cdot x \cdot \cos 2x - \sin 2x}{x^{2}} \right)\cdot dx

dy = \left(\frac{2\pi \cdot \cos 2\pi -\sin 2\pi}{\pi^{2}} \right)\cdot (0.25)

dy = \frac{1}{2\pi}

By applying the concepts of differential and derivative, the differential for y = (1/x) · sin 2x and evaluated at x = π and dx = 0.25 is equal to 1/2π.

To learn more on differentials: brainly.com/question/24062595

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