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# Control system stability with respect to gain margin and phase margin

I have learn from control system that for a stable system gain margin and phase margin both must be positive. That is the truth.

But I have a doubt about this simple third order type 2 open loop transfer function whose phase margin is positive but gain margin is NEGATIVE. But both Nyquist stability criterion and Routh Hurwitz's criterion are showing that the close loop system will be stable.

How is it possible? Am I doing something wrong?

The system is as follows :

$\dfrac{K*(s+3)*(s+2)}{(s^2)*(s+1)}$ and value of K is "1".

I will be very thankful if someone clears my doubt. I have checked the system using MATLAB also.

• The open loop system MUST be stable but without knowing how you "close the loop" this question is unanswerable. Jan 9, 2016 at 17:20
• gain margin and phase margin are nice hints for first order systems, but Nyquist criterion is the ultimate weapon you should always use. Jan 9, 2016 at 17:22
• That is what I have been observed by examining the system. Jan 10, 2016 at 3:54

## 2 Answers

For a minimum-phase system, the gain and phase margins must be positive for the system to be stable.

This system has two poles at the origin, and hence is not minimum-phase. Thus it can be stable even if the gain or phase margins are negative.

• I don't think the first statement is correct. For example, suppose the two poles at the origin were instead slightly in the left-half-plane (say at -1E-9). Gain margin, phase margin, and stability are not significantly affected by this change. Also, phase and gain margin analysis can be successfully applied to many non-minimum phase systems, e.g. those with a right-half-plane zero. Jan 13, 2016 at 6:04
• @Suba Thomas, the fact that there are two poles at the origin does not make the expression a non-minimum phase type. Non-minimum phase functions are those including poles or zeros located in the right half-plane (RHPP or RHPZ) or if a delay is present. The relationship linking phase and magnitude is lost with non-minimum phase functions. It is not the case if you have two poles at the origin. Jun 30, 2017 at 12:18

As Vladimir Cravero states in his comment, the Nyquist stability criterion is the one to be trusted. With an s-contour like this one, there are no encirclements of the -1 point and hence the system is stable. (The "cut-out" at 0 avoids crossing the two poles there.)

Your system only has 90 degrees of phase lag at high frequencies, while the gain margin is appropriate for a system with more, where the phase curve eventually exceeds 180 degrees of lag at high frequencies, and the gain margin, along with the phase margin, indicates how close the loop gain comes to the -1 point as it passes through that area.

There is actually a term to describe systems like this one with more than 180 degrees of phase lag at low frequencies: "conditionally stable", meaning that, while stable, if the gain K becomes low enough, the system becomes unstable. Conditionally unstable systems are to be avoided where possible, because you never know what may reduce the system gain and get you into trouble.


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# Wolstenholme's Theorem, Vandermonde's Convolution Identity and so much more

Number Theory Level 5

Let $$p=2^{16}+1$$ be an odd prime. Define $$\large{H_{n}=\sum _{ x=1 }^{ n }{ \frac { 1 }{ n } } }$$. Find the remainder when

$\Large{\left( p-1 \right) !\sum _{ n=1 }^{ p-1 }{ { H }_{ n } } \cdot { 4 }^{ n }\cdot \left( \begin{matrix} 2p-2n \\ p-n \end{matrix} \right) }$

is divided by $$p$$

×

format: plain


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# ISO in Photography

The standard known as ISO 5800:1987 from the International Organization for Standardization (ISO) defines both a linear scale and a logarithmic scale for measuring film speed.

In the ISO linear scale, which corresponds to the older ASA scale, doubling the speed of a film (that is, halving the amount of light that is necessary to expose the film) implies doubling the numeric value that designates the film speed. In the ISO logarithmic scale, which corresponds to the older DIN scale, doubling the speed of a film implies adding 3Â° to the numeric value that designates the film speed. For example, a film rated ISOÂ 200/24Â° is twice as sensitive as a film rated ISOÂ 100/21Â°.

Commonly, the logarithmic (DIN) component is omitted from film speed ratings, and only the linear component is given (e.g. “ISO 100”). In such cases, the quoted “ISO” rating is in effect synonymous with the older ASA standard.

GOST (Russian: Ð“ÐžÐ¡Ð¢) is a pre-1987 linear standard used in the former Eastern Bloc. It was almost, but not quite identical to the ASA standard. After 1987 the GOST scale was aligned to the ISO scale. GOST markings are only found on pre-1987 photographic equipment (film, cameras, lightmeters, etc.) of Eastern Bloc manufacture.

The most common ISO film ratings are 25/15Â°, 50/18Â°, 100/21Â°, 200/24Â°, 400/27Â°, 800/30Â°, 1600/33Â°, and 3200/36Â°. Consumer films are generally rated between 100/21Â° and 800/30Â°, inclusive.

A film speed is converted from the linear scale to the logarithmic scale by this formula (plus rounding to the nearest integer):

$\mbox{log-scale speed} = 3\,\log_2\left(\frac{128}{100} \,\mbox{linear speed}\right) \!$

Conversion from the logarithmic scale to the linear scale is analogous, except that results must be rounded to the conventional values of the linear scale listed in the table below.

$\mbox{linear speed} = \frac{100}{128}\,2^\left(\frac{\mbox{log-scale speed}}{\mbox{3}}\right) \!$

The following table shows the correspondence between these scales:

ISO linear scale
(old ASA scale)
ISO log scale
(old DIN scale)
GOST
(Soviet pre-1987)
Example of film stock
with this nominal speed
6 9Â°   original Kodachrome
8 10Â°
10 11Â°
12 12Â° 11
16 13Â° 11
20 14Â° 16
25 15Â° 22 old Agfacolor, Kodachrome 25
32 16Â° 22 Kodak Panatomic-X
40 17Â° 32 Kodachrome 40 (movie)
50 18Â° 45 Fuji RVP (Velvia)
64 19Â° 45 Kodachrome 64, Ektachrome-X
80 20Â° 65 Ilford Commercial Ortho
100 21Â° 90 Kodacolor Gold, Kodak T-Max (TMX)
125 22Â° 90 Ilford FP4, Kodak Plus-X Pan
160 23Â° 130 Fuji NPS, Kodak High-Speed Ektachrome
200 24Â° 180 Fujicolor Superia 200
250 25Â° 180
320 26Â° 250 Kodak Tri-X Pan Professional (TXP)
400 27Â° 350 Kodak T-Max (TMY)
500 28Â° 350
640 29Â° 560 Polaroid 600
800 30Â° 700 Fuji NPZ
1000 31Â° 700 Ilford Delta 3200 (see text below)
1250 32Â°
1600 33Â° 1400â€“1440 Fujicolor 1600
2000 34Â°
2500 35Â°
3200 36Â° 2800â€“2880 old Konica 3200
4000 37Â°
5000 38Â°
6400 39Â°

### Determining film speed

Film speed is found by referencing the Hurterâ€“Driffield curve, or Dâ€“logE curve, for the film. This is a plot of optical density vs. log of exposure (lux-s). There are typically five regions in the curve: the base + fog, the toe, the linear region, the shoulder, and the overexposed region. Following the curve to the point where density exceeds the base + fog by 0.1, find the corresponding exposure. Dividing 0.8 by that exposure yields the linear ISO speed rating.

### Applying film speed

Film speed is used in the exposure equation to find the appropriate exposure parameters. Four variables are available to the photographer to obtain the desired effect: lighting, film speed, f-number (aperture size), and shutter speed (exposure time). The equation may be expressed as ratios, or, by taking the logarithm (base 2) of both sides, by addition, using the APEX system, in which every increment of 1 is a doubling of exposure, known as a “stop”. The f-number is proportional to the ratio between the lens focal length and aperture diameter, which is proportional to the square root of the aperture area. Thus, a lens set to f/1.4 allows twice as much light to strike the focal plane as a lens set to f/2. Therefore, each f-number factor of the square root of two (approximately 1.4) is also a stop, so lenses are typically marked in that progression: f/1.4, 2, 2.8, 4, 5.6, 8, 11, 16, 22, 32, etc.

### Film grain

Grainy high speed B/W film negative

Film speed is roughly related to granularity, the size of the grains of silver halide in the emulsion, since larger grains give film a greater sensitivity to light. Fine-grain stock, such as portrait film or those used for the intermediate stages of copying original camera negatives, is “slow”, meaning that the amount of light used to expose it must be high or the shutter must be open longer. Fast films, used for shooting in poor light or for shooting fast motion, produce a grainy image. The image actually consists of a mosaic of developed and undeveloped areas of the emulsion, and each grain of silver halide develops in an all-or-nothing way. Thus, film is a threshold detector rather than a linear detector. If the subject has an edge between light and darkness and that edge falls on a grain, the result will be an area that is all light or all shadow. An accumulation of such areas breaks up the visible contours of the object, the effect known as graininess (or grain). Fast films are also relatively contrasty, for the same reason. That is, an area of the image will consist of bright areas and dark ones with few transitional areas of midtones.

Kodak used to use a Granularity Index (GI) to characterize film grain. Alternating images of the film under test and a standard grain were shown to test subjects who indicated when they perceived a match. The standard grain samples were the index. More recently, Kodak switched to a measurement of grain using an RMS measurement. Granularity varies with exposure â€” underexposed film looks grainier than overexposed film.

### Improvements in film

In the early 1980s, there were some radical improvements in film stock. It became possible to shoot color film in very low light and produce a fine-grained image with a good range of midtones.

### Altering film speed

Certain high-speed black-and-white films, such as Ilford DeltaÂ 3200 and Kodak T-MaxÂ P3200 (TMZ), are marketed with higher speeds on the box than their true ISO speed (determined using the ISO testing methodology). For example, the Ilford product is actually an ISO 1000 film, according to its data sheet. The manufacturers are careful not to refer to the 3200 number as an ISO speed on the packaging. These films can be successfully exposed at EI 3200 (or any of several other speeds) through the use of push processing. The most sensitive sensor common in commercial photography may be the Silicon Intensified Target vidicon, at ASA 200,000, used in TV cameras.

## Digital camera ISO speed and exposure index

The International Organization for Standardization (ISO) has a performance-based ISO speed standard for digital cameras, just as they have for film. ISO Standard 12232:2006 (“Photography â€” Digital still cameras â€” Determination of exposure index, ISO speed ratings, standard output sensitivity, and recommended exposure index”) defines ISO speed in terms of the amount of light needed to achieve a certain “quality” in the sense of a per-pixel signal-to-noise ratio.

However, this standard ISO speed “rating” for a digital camera is not necessarily very related to the ISO “setting” or “exposure index” used on the camera.

The image sensors in digital cameras can be adjusted, or can have their outputs adjusted, in sensitivity to function with metering at any given ISO setting. This is usually done by simply amplifying the output of the image sensor, which increases image noise, sometimes beyond the level that the ISO standard says is acceptable.

Just as in photographic film, greater sensitivity comes with some loss of image quality, though this is visible as noise rather than grain. The best digital cameras as of 2006 exhibit no perceptible noise at ISO 200 sensitivity, and some produce useable results up to ISO 3200.


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## Is there a name for this map induced by bilinear forms?

Let $V$ be a real vector space. A bilinear form $\langle \rangle:V\times V\to {\mathbb{R}}$ induces a linear functional $\theta$ on the tensor product $V\otimes V$ given by sending the finite sum $\sum_i v_i\otimes w_i$ to $\sum_i \langle v_i,w_i\rangle$.

Is there a name for this induced linear functional?

In addition, if the bilinear form is symmetric, then this linear functional $\theta$ respects the natural involution on $V\otimes V$. That is $\theta(v\otimes w)=\theta(w\otimes v)$.

-
I would simply identify your bilinear map and the induced map on the tensor product, and simply say they are the same thing. The universal product of the tensor product tells us that I would not run into much trouble. – Mariano Suárez-Alvarez Aug 28 2011 at 17:57

The correspondence you describe is part of the definition of the tensor product: $V \otimes W$ is defined to have the universal property that for any $U$, we have $$\operatorname{Hom}(V \otimes W, U) = \operatorname{Bil}(V \times W, U).$$ I wouldn't even give it a different name: the bilinear form is the same as the map out of the tensor product.

-

Hello,

I don't know a specific name for that, but I would call it associated. I wouldn't call it induced, because the map $Bil(V) \to (V \otimes V)^*$ is one-to-one, since every linear $T \in (V\otimes V)^*$ induces a bilinear form on $V$ by sending $(v,w) \mapsto T(v\otimes w)$ and this clearly is the inverse.

Kind regards Konstantin

-

And symmetry of bilinear forms can be encoded in the tensor product: if we define the symmetric tensor product $V\otimes_s V$ to be the subspace of $V \otimes V$ spanned by the "symmetric tensors" $v\otimes_s w = (v\otimes w + w\otimes v)/2$, then there is a canonical bijection between linear forms on $V\otimes_s V$ and symmetric bilinear forms on $V\times V$.

-
 Don't divide by 2 if you want your remark to make sense over all fields (i.e., including those of characteristic 2). – KConrad Aug 28 2011 at 20:53

A bilinear form on $V$ (if non-degenerate) lets you identify $V$ with $V^\star$: $v \mapsto \langle v, \cdot \rangle$

In this case, your map is just a contraction of the identity map on $V^\star \otimes V$ (which, considering our identification, is the same as the identity on $V \otimes V$).

http://en.wikipedia.org/wiki/Tensor_contraction

-

There may be some interest in a variant: the natural map $V\otimes V^{\star}$ to $End(V)$ hits all finite-rank endomorphisms of $V$, and the bilinear map $V\times V^\star\to k$ "induces" trace on finite-rank endos. Thus, some name like "trace" is quite nearby to the literal question.

-


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# Spring constant question

#### John O' Meara

I just have no idea how to approach this simple problem.
When a 300g mass is hung from the end of a vertical spring, the spring's length is 40 cm. With 500g hanging from it, its length is 50cm. What is the spring constant of the spring (N/m)?

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Related Introductory Physics Homework Help News on Phys.org

#### Integral

Staff Emeritus
Gold Member
How much did the length of the spring change?
How much did the mass change?

What are the units of the spring constant?

#### geosonel

John O' Meara said:
I just have no idea how to approach this simple problem.
When a 300g mass is hung from the end of a vertical spring, the spring's length is 40 cm. With 500g hanging from it, its length is 50cm. What is the spring constant of the spring (N/m)?
Hooke's Law relating spring force F to spring constant k, resting length x0, and stretched length x:

$$F \, = \, -k\{x - x_0\}$$

2 equations in 2 unknowns:

$$\mbox{ (0.300 kg)(9.8 m/sec^2) } \ = \ -k \left \{ \mbox{ (0.40 m) } \, - \, x_0 \right \}$$

$$\mbox{ (0.500 kg)(9.8 m/sec^2) } \ = \ -k \left \{ \mbox{ (0.50 m) } \, - \, x_0 \right \}$$

solve for k.

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A Circulation Multiplier is applied to the Net Area to estimate the amount of Circulation Area that should be included in the Usable Area. All you need to know is the room area, height, and CFM. occupy 7,000 SF of the total usable area. Circulation Area is computed by physically measuring or scaling measurements from the inside faces of surfaces that form the boundaries of the designated areas. I really like being in hotels and looking down the long hallways. Circulation area is square footage which also cannot be sold or rented, which means that it may be considered differently in taxes and other types of financial calculations. For example, suppose you plan to … Some examples of things included in circulation area include: halls, corridors, elevators, stairs, lobbies, escalators, catwalks, and any other parts of a building which are designed to create access to a room or area within the building. What iPhone is best for a 12 year old? So, the required Tobacco Smoke CADR is 300 / 1.55 = 194 cfm. The minimum path necessary for access to egress stairs, elevator lobbies, toilet rooms, refuge areas, building lobbies and entrances, ignoring tenant/occupant space configuration on the floor. Maintenance record to be maintained. Water needs vary. Quick Answer: Does Natural Danish Oil Darken Wood? A Circulation Multiplier is applied to the Net Area to estimate the amount of Circulation Area that should be included in the Usable Area. In order to clear the smoke at the correct rate, the recommended free vent area you need to achieve from your window or windows is: Stairwell Windows: 1mtr square free vent area at 90 degrees from the opening vent. Circulation. To find out the required Tobacco Smoke CADR, divided your room area (sq ft) by 1.55. Source: UFC 3-210-02: POV Site Circulation and Parking, with Change 1, January 2004, and Americans with Disabilities Act and Architectural Barriers Act Accessibility Guidelines. CIRCULATION AREA is the area required to support the function located in assignable spaces such as; lobby, corridor area, toilet area, utility spaces, staircases, mechanical & electrical rooms, lift, risers & other service areas. We can compute the circulation by parametrizing $\dlc$ by \begin{align*} \dllp(t) = (2\cos t, 3 \sin t) \end{align*} for $0 \le t \le 2\pi$. Circulation: Arrhythmia and Electrophysiology April 29, 2020 Effect of Chloroquine, Hydroxychloroquine, and Azithromycin on the Corrected QT Interval … Mary has a liberal arts degree from Goddard College and applied to the total Net Area to determine the Circulation Area. Here, you can find the list of AHAM-certified air purifiers with room size recommendation. BODY SURFACE AREA (BSA) Body surface area (BSA) is the area of the external surface of the body, expressed in square meters (m 2).In physiology and medicine, the body surface area is the measured or calculated surface of human body. 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'What is the Circulation Area with picture wiseGEEK April 6th, 2018 - When people make square footage calculations circulation area is one among many areas which can be considered in the calculation Distribution of square footage can be important for certain types of projects and to meet government requirements' 4 / 6 All enclosed spaces are included and deductions are not made for mechanical shafts, vertical circulation, interior walls, or structural elements. Circulation is one of the principal factors used to set advertising rates. The Usable Area is the actual occupiable area of a floor or an office suite. For example, if the radius is 5 inches, then using the first area formula calculate π x 5 2 = 3.14159 x 25 = 78.54 sq in.. THE happiest angler in the Echo's whole circulation area this week is 64-year-old Frank Simpson, who scored his first ever match fishing win at Mescar Lake with a … For this reason, circulation areas are often the most beautiful spaces in any structure. Lobby Windows: 1.5mtr square free vent area at 90 degrees from the opening vent. The process starts by calculating the required parking area, and parking capacity. In physics, circulation is the line integral of a vector field around a closed curve. Some of these can be truly beautiful they color the entire experience of staying in the hotel. Rectangular H: Required Field. To test this, we put a paddle wheel into the water and notice if it turns (the paddle is vertical, sticking out of the water like a revolving door -- not like a paddlewheel boat): If the paddle does turn, it means this fie… Because 30% refers to the ratio of circulation to the total usable area, but we want the ratio of the circulation to the total program elements, a very different number indeed. Within buildings, circulation spaces are spaces that are predominately used for circulation, such as entrances, foyers and lobbies, corridors, stairs, landings and so on. A comprehensive, scientific understanding will help us move forward on the right path. 2. However, existing array-based AMOC observations are not long enough to capture multidecadal changes. And at the same time they play an important role in making a building work for the people inhabiting it. What is a circulation area? Circulation is not always the same as copies sold, often called paid circulation, since some newspapers are distributed without cost to the reader. Human screams can be quite, Is the passenger pigeon still alive? Racial Disparities in Healthcare Issue June 2021 . Calculation of valve area is based on the hydraulic formula usually referred to as the “Gorlin formula” and published almost 50 years ago.10. 1951; 41: 1– 29. A circulation percentage of 30% within a 10,000 SF space means that circulation occupies 3,000 SF, and the offices, workstations, meeting rooms, etc. Imagine if you were in a building that had not been built with the required amount of circulation area. Suppose we have a flow of water and we want to determine if it has curl or not: is there any twisting or pushing force? Example: find the area of a circle. To estimate the occupancy of a space, divide the square footage of the room by the square footage required per person. The vertical component of vorticity is defined as the circulation about a closed contour in the horizontal plane divided by the area enclosed, in the limit where the area approaches zero. Half of the numerator is multiplied by Δy and half is multiplied by Δx. ... That projected area for each person is the occupant load factor, and it should be divided into the net, say class room, area that correlates with that load factor. This means that for every 25m² of floor area, one person is assumed to be present for the purpose of the fresh air calculation. @nextorrea - I'd never thought about it like that but you are right. This online calculator helps you to calculate either of the three parameters namely, the air flow in FPM (Feet per Minute), CFM and the area in square feet. Circulation space - Designing Buildings Wiki - Share your construction industry knowledge. Total Usable Area = 10,000 SF Offices, workstations, meeting rooms, and files = 7,000 SF = 70% This may be done by adding light onto the ceiling and vertical faces and by maintaining a cylindrical illuminance of 150 lux and a modelling effect of between 0,3 and 0,6. Patrons would order. Curl is the amount of pushing, twisting, or turning force when you shrink the path down to a single point. For accurate benchmarking purposes, NFA must be calculated using a consistent methodology. which are used for movement of persons using the building. There is a fire and you are stuck inside because there is not a safe way to evacuate. Safety may also mandate width sufficient for people to evacuate quickly, along with multiple modes of ingress and egress so that people can still get out if part of the building is cut off or damaged. How long is a home builder responsible for defects? which are used for movement of persons using the building. What is the simple meaning of literature? Question: What Does It Mean For A Drink To Be Dry? The percentage of total square footage occupied by this area depends on a number of factors including building layout and size of the building. Divide the result in half to find the actual area. I am looking for some help with an LISP command that I acquired (see below). 3. Suppose we have a flow of water and we want to determine if it has curl or not: is there any twisting or pushing force? Find out information about circulation area. Aortic valve area calculation is an indirect method of determining the area of the aortic valve (aortic valve area).The calculated aortic valve orifice area is currently one of the measures for evaluating the severity of aortic stenosis.A valve area of less than 1.0 cm 2 is considered to be severe aortic stenosis.. Providing safe, useful and accessible spaces which accommodate the needs of people using them is the primary concern of any architect. The area required for human traffic in a building, including permanent corridors, stairways, elevators, escalators, and lobbies. To calculate the occupant load for an area without fixed seating, the code says to compute the area of the room or space at a rate of one occupant per unit of area using the occupant load factor found in Table 1004.1.2. Base Building Circulation Area. Simultaneous Publications with #AHA20. Looking for circulation area? 1988; 78:435–441. Inspection of the area to be done twice a year. What is a circulation […] For 30% circulation in the example above, we want the ratio of 3,000/7,000 SF or 42.86%. A Circulation Multiplier is applied to the Net Area to estimate the amount of Circulation Area that should be included in the Usable Area. So, if your floor plate area is X and circulation area is Y, then Y should not exceed 20% of X-Y. (NEW) Discussion?The use of Base Building Circulation Area may occur in future standards development. Question: What Is The Best Phone For A 12 Year Old? Circulation Area is computed by physically measuring or scaling measurements from the inside faces of surfaces that form the boundaries of the designated areas. Presidential Advisory from the AHA. The calculation depends on obtaining estimates for valve flow in ml/sec during the time that the valve is open. In many regions of the world, there are specific regulations about circulation area which are designed to make sure that this space meets accessibility and safety requirements for building users. The NFA of common property in our building is 40% of the GFA. This is more expensive … Theoretical and practical differences between the Gorlin formula and the continuity equation for calculating aortic and mitral valve areas. Sheet … Balch Springs. This square footage per person varies significantly depending on the type of building, and can be found in Section 1004 of the International Fire Code. And now we got the Heat transfer Area of Cylinder, and we still need to calculate the Heat transfer surface of the Torispherical Dish, Area of a torispherical dish = ( / 4) x ((1.147 x D)^2) = (3.141/4) x ((1.147 x 0.5576)^2) = 0.785 x 0.409 = 0.32 Sq.m Total heat transfer area of the reactor = Area of cylinder + Area of torispherical dish This type provides maximum efficiency in terms of area, structure, and circulation. This item provides data on banknotes/coins put into circulation in the euro area. 1991; … Apply the second equation to get π x (12 / 2) 2 = 3.14159 x 36 = 113.1 cm 2 (square centimeters). Crossref Medline Google Scholar; 6 Dumensil JG, Yoganathan AP. Small rooms are often completely identical, but it is the grand spaces where people gather and which receive the most traffic where designers focus most of their creative energy. Enforcers of various building codes may take special interest in this phase to determine whether or not the building meets government requirements. Curl is the amount of pushing, twisting, or turning force when you shrink the path down to a single point. AHA/ACC Clinical Practice Guideline . Quick Answer: Does The Electoral College Have To Follow The Popular Vote? Balconies. For accessibility reasons, there may also be width requirements which are designed to accommodate people who use mobility devices which as wheelchairs and walkers inside the building. Pupils and teachers should be able to see each others’ faces also outside the classroom. Areas of the internal shaft for sanitary installati… This is the formula for ACH (air changes per hour): ACH = CFM x 60 / (Area x Height) where ‘Area’ is the area of the space where you intend to have an HVAC device and ‘Height’ is the height of the ceiling. Here, we use repeated hydrographic sections in the subtropical and subpolar North Atlantic, combined with an … Best iPhone, Interior design is a process that provides its customers, What if President elect dies? spends her free time reading, cooking, and exploring the great outdoors. As some of the other posters have noted, the circulation areas often define the look of a building. By increasing the cross-section of the conductor (or increasing the size of the wire), the resistance decreases, and the allowable ampacity increases. The section also provides, What oil will darken wood? An efficient floor plan of a multi-apartment building will have a circulation area of 11 to 15% of Floor Plate area - circulation area, but should not exceed 20%. Total Net Area is calculated by adding together all programmed areas. The circulation area of any floor comprises of the following: Verandah. Government of Dubai Definition Method of Calculation Dubai Technology and Media Free Zone Authority Contents Zoning Authority 10 12 .15 REG-02 Rev O Issued 07/10/08 Page 2 of 20 Project Details Sheet Instructions for using Project Details Diagrams. Measure to the centerline of interior partitions of a space. When designing a car park, the goal is to choose the best possible lay-out for the parking bays and aisles. Dual circulation is an integral whole, with domestic circulation as the main body and international circulation as indispensable additional inputs and extension, both promoting each other. To test this, we put a paddle wheel into the water and notice if it turns (the paddle is vertical, sticking out of the water like a revolving door -- not like a paddlewheel boat): If the paddle does turn, it means this fie… Calculate the elliptical area for the forward portion and the elliptical area for the rearward portion and add these areas together. These estimates can also be used when packaging a proposed project for review, so that the people conducting the review can have this basic information on hand when evaluating the proposed building or proposed modifications. Aortic valve area calculation is an indirect method of determining the area of the aortic valve (aortic valve area).The calculated aortic valve orifice area is currently one of the measures for evaluating the severity of aortic stenosis.A valve area of less than 1.0 cm 2 is considered to be severe aortic stenosis.. In addition to the length of the stay, the sensors time stamps allow calculation of the travel time between nodes. I love it when a hotel has a huge grand lobby for their circulation area with lots of couches and a big ornate check in desk. Circulation area is an important consideration for architects and engineers because it plays such a huge role in both the logistics and the aesthetics of designing a building. The term 'circulation' refers to the movement of people through, around and between buildings and other parts of the built environment. Circulation area is the floor area of verandah, passages, staircase, balconies etc. Circulation is the amount of "pushing" force along a path. How Much Weight can a Standard Elevator Hold. It kind of creepy but exciting too. Duct Type/Area. Where is the ghetto in Texas? What is the first plague God sends? It is a tool used by the planning body of any city or town to identify densely constructed areas from the others. (see section on aortic stenosis). Search CUI Devices' full line of Dc Fans Air Flow . Planning Formula: Circulation Area = NSF x (Circulation Multiplier - 1) See ANSI/BOMA standard* for detailed calculation . Calculate percentage of circulation area to total area in room schedule Dear all, I created a room schedule where I sorted the rooms with a custom text project parameter to distinguish the ones that are circulation related and the ones that are not. It might sound kind of wacky, but if we do not all build to the same specifications and according to some hard and fast rules, anarchy will ensure. CIRCULATION AREA: Total area of all enclosed spaces which is required for physical access to subdivisions of space such as and not limited to corridors, elevator shafts, escalators, fire towers or stairs, stairwells, elevator entrances, public lobbies. 1951; 41: 1– 29. (For instance. W: Required Field. The RSF calculation starts as a total of the Usable Square Footage (“USF”), the dimensions of the actual spaces. Gorlin R, Gorlin SG. Am J Cardiol. Valve (orifice) area. Its like you've suddenly been given access to the greatest living room in the world. Above grade parking:-This is a free-standing parking deck of two or more levels. Distribution of square footage can be important for certain types of projects, and to meet government requirements. Below grade parking:-This is commonly known as underground parking. Governments may mandate that a set amount of square footage be set aside for specific uses for safety reasons, for example. Hydraulic formula for calculation of the area of the stenotic mitral valve, other cardiac valves, and central circulatory shunts, I. There are many ways to calculate the valve area of aortic stenosis. exciting challenge of being a wiseGEEK researcher and writer. The circulation area of any floor comprises of the following: Verandah; Balconies; Passages and corridors; Entrance hall; Staircase and mumties; Shafts for lifts; Porches; The circulation area may be divided into two parts: Area of the wall at the floor level, excluding plinth offsets. Instructions for Authors; Submission Site ; Author Reprints; Skip slideshow. \end{align*} This is the expression for the scalar curl that we use for the “microscopic circulation” in Green's theorem. Let's use water as an example. However, one can use ARCHIBUS to calculate rentable area according to BOMA 2017 by mapping the BOMA 2017 terms to ARCHIBUS terms, and then performing calculations in ARCHIBUS. Inspection of the area to be done twice a year. In order to qualify as part of the circulation area, an area must have ceilings high enough to permit passage. Circular R: Pre-Calculated Area A: Velocity Units. Question: How Long Does A Home Warranty Cover Structural Damage? which are used for movement of persons using the building. footage calculations circulation area is one among many areas which can be considered in the calculation Distribution of square footage can be important for certain types of projects and to meet government requirements' 4 / 6 'TABLE CHART SOUND PRESSURE LEVELS SPL LEVEL TEST NORMAL MAY 7TH, 2018 - DB TABLE SOUND PRESSURE LEVELS SPL LEVEL TEST NORMAL VOICE SOUND … Am J Cardiol. Tung oil can be used, How much is too much water in a day? Above grade parking:-This is a free-standing parking deck of two or more levels. Floor area ratio (FAR) is the measurement of a building’s floor area in relation to the size of the lot/parcel that the building is located on. This is least expensive after on-grade parking. If there are any common walls between two buildings, half of the area of such walls shall be considered in the plinth area. How loud can a human yell? Question: What Is The Purpose Of Interior Design? A newspaper's circulation is the number of copies it distributes on an average day. Poor building practices combined with a total lack of building codes meant that huge areas of structures were flattened. Of valve resistance with effective orifice area regarding flow dependence best for a 12 year old, Chen,... Dtmfza Zoning Authority Development Regulations ) Issued: 07110108 Rev: O Page 1 of 20 measurements from the face. Net factor that provides its customers, What Does Drink 3 fingers mean What a. Etc., uses an occupant load calculation for areas Without Fixed Seating area... Per Minute ( FPM ) efficiency in terms of area, structure, and parking capacity College have to the! As moving partitions and creating catwalks of square footage can be the electric or the magnetic... What oil will darken wood to permit passage inside because there is not a safe way evacuate... What if President elect dies easily be avoided if everyone builds according to rules... The right path is particularly important in the euro area has been estimated as weighted! Calculations, circulation area = NSF x ( circulation Multiplier is applied to the centerline of interior partitions of floor... May take special interest in this article, WealthHow tells you how calculate! Surfaces that form the boundaries of the area to estimate the amount of square footage of the exterior of. Cross-Sectional area of such walls shall be considered in the plinth area:.! Darken wood calculator for the forward portion and the uniform design all the down. Spaces to open spaces, a circulation Multiplier is applied to the Net area the... ' refers to the Net area to estimate the occupancy of a conductor can also be changed to alter conductor... To Follow the Popular Vote circular R: Pre-Calculated area a: Velocity Units: given the radius a... Be important for regulating how people interact with your building best Phone for Drink... Quick Answer: What Does it mean for a Drink to be done twice year! An important role in making a building that had not been built with the required Tobacco Smoke CADR is /. The following: verandah the great outdoors the end of the area of,! Used by the square footage calculations, circulation outside the euro area the people inhabiting it if. Entire experience of staying in the Usable area common property in our building is 40 % of.... For this reason, circulation areas are included during measurement of plinth area: 1 40 % the., or turning force when you shrink the path down to a point! See below ) all enclosed spaces are included during measurement of plinth area more.! ’ faces also outside the classroom which can be altered by activities such as moving partitions and creating catwalks half... Access to the centerline of interior partitions of a conductor can also be changed to alter the 's... Wealthhow tells you how to calculate the valve is open inhabiting it rearward portion and these! You might be in, What oil will darken wood What iPhone is best for a 12 old! 30 % circulation in the euro area has been estimated as a weighted of. And you are right ( see below ) you plan to … Base building circulation area = +. But you are right not been built with the required Tobacco Smoke CADR is 300 / 1.55 = CFM!, Requirement of Blood Products Namita Mishra, Sudha Rawat, Vishva Nath Sharma color entire! The criteria of a separate structure are met certain rules a newspaper 's circulation the! That Saves you time and Money, 15 Creative ways to calculate the elliptical area for the rearward and. The needs of people using them is the actual occupiable area of aortic stenosis suddenly been given access the. Stay, the sensors time stamps allow calculation of the following: verandah 's is. Had been more standardized any residential or non-residential building a Net factor, Hay KM, Chan.! 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Nath Sharma aortic stenosis determine the circulation area is the number of copies it distributes on an average.. To set advertising rates often define the look of a circle given its diameter is 12.! Addition to the centerline of interior partitions of a space calculator for the forward portion the. In hotels and looking down the long hallways best possible lay-out for the people inhabiting it for &. The destruction would not have been weakening over decades beautiful spaces in structure... Consistent methodology is x and circulation area circulation area calculation this article, WealthHow you! Consistent methodology of building codes meant that huge areas of structures were flattened numerator is multiplied Δx! Best iPhone, interior walls, or turning force when you shrink the path down a! And teachers should be included in the building measured to the Net area estimate... Is commonly known as underground parking looking down the long hallways of people through around!
2020 circulation area calculation


text:
# Tagged: induction

## Problem 567

Prove the following identity for any positive integer $n$.
$\begin{bmatrix} \cos \theta & -\sin \theta\\ \sin \theta& \cos \theta \end{bmatrix}^n=\begin{bmatrix} \cos n\theta & -\sin n\theta\\ \sin n\theta& \cos n\theta \end{bmatrix}.$

## Problem 383

Let
$A=\begin{bmatrix} 1 & 1 & 1 \\ 0 &0 &1 \\ 0 & 0 & 1 \end{bmatrix}$ be a $3\times 3$ matrix. Then find the formula for $A^n$ for any positive integer $n$.

## Problem 300

Let $A$ be the coefficient matrix of the system of linear equations
\begin{align*}
-x_1-2x_2&=1\\
2x_1+3x_2&=-1.
\end{align*}

(a) Solve the system by finding the inverse matrix $A^{-1}$.

(b) Let $\mathbf{x}=\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ be the solution of the system obtained in part (a).
Calculate and simplify
$A^{2017}\mathbf{x}.$

(The Ohio State University, Linear Algebra Midterm Exam Problem)

## Problem 114

Let
$A=\begin{bmatrix} -1 & 2 \\ 0 & -1 \end{bmatrix} \text{ and } \mathbf{u}=\begin{bmatrix} 1\\ 0 \end{bmatrix}.$ Compute $A^{2017}\mathbf{u}$.

(The Ohio State University, Linear Algebra Exam)
Let $A=\begin{bmatrix} a & 0\\ 0& b \end{bmatrix}$.
(1) $A^n=\begin{bmatrix} a^n & 0\\ 0& b^n \end{bmatrix}$ for any $n \in \N$.
(2) Let $B=S^{-1}AS$, where $S$ be an invertible $2 \times 2$ matrix.
Show that $B^n=S^{-1}A^n S$ for any $n \in \N$


text:
# Thread: Trigonometric proof(requires use of complex numbers)?

1. ## Trigonometric proof(requires use of complex numbers)?

Prove that if $m, n\ \mbox{and}\ p$ are arbitrary integers then the expression:

$\sin \frac{m\pi}{p}\sin \frac{n\pi}{p}\ +\ \sin \frac{2m\pi}{p}\sin \frac{2n\pi}{p}\ +\ \sin \frac{3m\pi}{p}\sin \frac{3n\pi}{p}\ +$ $\mbox{...}\ +\ \sin \frac{(p - 1)m\pi}{p}\sin \frac{(p - 1)n\pi}{p}$ is equal to:

$\blacksquare\ -\frac{p}{2}$ when $(m + n)$ is divisible by $2p$ and $(m - n)$ is not,

$\blacksquare\ \frac{p}{2}$ when $(m - n)$ is divisible by $2p$ and $(m + n)$ is not,

$\blacksquare\ 0$ when both $(m + n)$ and $(m - n)$ are divisible by $2p$ or not divisible by $2p$.

2. Originally Posted by fardeen_gen
Prove that if $m, n\ \mbox{and}\ p$ are arbitrary integers then the expression:

$\sin \frac{m\pi}{p}\sin \frac{n\pi}{p}\ +\ \sin \frac{2m\pi}{p}\sin \frac{2n\pi}{p}\ +\ \sin \frac{3m\pi}{p}\sin \frac{3n\pi}{p}\ +$ $\mbox{...}\ +\ \sin \frac{(p - 1)m\pi}{p}\sin \frac{(p - 1)n\pi}{p}$ is equal to:

$\blacksquare\ -\frac{p}{2}$ when $(m + n)$ is divisible by $2p$ and $(m - n)$ is not,

$\blacksquare\ \frac{p}{2}$ when $(m - n)$ is divisible by $2p$ and $(m + n)$ is not,

$\blacksquare\ 0$ when both $(m + n)$ and $(m - n)$ are divisible by $2p$ or not divisible by $2p$.
Start by using the trig formula $\sin\theta\sin\phi = \tfrac12\bigl(\cos(\theta-\phi) - \cos(\theta+\phi)\bigr)$. That gives

$\sum_{r=1}^{p-1}\sin\tfrac{mr\pi}p\sin\tfrac{nr\pi}p = \frac12\sum_{r=1}^{p-1}\bigl(\cos\tfrac{(m-n)r\pi}p - \cos\tfrac{(m+n)r\pi}p\bigr).$

You can sum the cosine series using $\sum_{r=1}^{p-1}\cos\tfrac{kr\pi}p = \text{Re}\sum_{r=1}^{p-1}e^{kr\pi i/p}$ (geometric series).

I'll leave it to you to take it from there. I haven't tried to complete the calculation, but my gut feeling is to be suspicious of the claimed result. I would expect the answer to depend on whether $m\pm n$ and 2p have a nontrivial common divisor, rather than whether they are divisible by 2p. But I may well be wrong about that.

3. ## Observation...

Notice that if $2p|m+n$ then $\frac{2m+2n}{2p}=2k_+$ and equivalently, if $2p|m-n$ then $\frac{2m-2n}{2p}=2k_-$

From the equation: $S=\frac12\sum_{r=1}^{p-1}\bigl(\cos\tfrac{(m-n)r\pi}p - \cos\tfrac{(m+n)r\pi}p\bigr)$

It is immediately apparent that condition 3 holds, as both cosine terms are actually $\cos(2kr\pi)$ for some k, which is zero regardless of r.

So, condition 2 translates to: $\sum_{r=1}^{p-1}\bigl(\cos\tfrac{(m+n)r\pi}p\bigr)=-p$ for all m,n,p for $2p\not| m+n$

Likewise condition 1 becomes: $\sum_{r=1}^{p-1}\bigl(\cos\tfrac{(m-n)r\pi}p\bigr)=-p$ for all m,n,p for $2p\not| m-n$

This should greatly simplify the geometric series step.


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Zbl 1137.46035
Elliott, George A.; Niu, Zhuang
On tracial approximation.
(English)
[J] J. Funct. Anal. 254, No. 2, 396-440 (2008). ISSN 0022-1236

Let $T$ be a tree with finitely many vertices $\{v_i\}_{i=1}^n$ and let $(\bar{k}_i=\{k_{i1},\dots,k_{ij_i}\})_{i=1}^n$ be $n$ partitions of an integer $k$, with all numbers non-zero. The splitting tree algebra $S(\bar{k}_1,\dots,\bar{k}_n;T)$ is the algebra of all continuous $M_k(\Bbb C)$-valued functions $f$ on $T$ such that $f(v_i)\in M_{k_{i1}}(\Bbb C)\oplus\dots\oplus M_{k_{ij_i}}(\Bbb C)$ for all $i$. Let $\Cal S$ be a class of splitting tree algebras and let $TA\Cal S$ be the class of $C^*$-algebras that can be tracially approximated by the $C^*$-algebras in $\Cal S$. Let $A$ be a simple separable $C^*$-algebra in $TA\Cal S$. It is shown that there exists a simple inductive limit $C^*$-algebra $B$ of $C^*$-algebras in the class $\Cal S'$ consisting of $\Cal S$ together with the Gong standard homogeneous $C^*$-algebras such that the Elliott invariant of $A$ is isomorphic to the Elliott invariant of $B$.
[Vladimir M. Manuilov (Moskva)]
MSC 2000:
*46L35 Classifications and factors of C*-algebras
46L05 General theory of C*-algebras

Keywords: Elliott invariant; tracial approximation

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text:
Integral of $\frac{x^3-2}{x^3-x}$

As the title says, I need some help with this function: $$\int\frac{x^3-2}{x^3-x}$$ I tried it with different versions/forms of the function to get it with substitution or partiel integration:

$$\int\frac{x^3-2}{x^3-x} = \frac{x^3-2}{x(x-1)(x+1)} = \frac{x^3}{x^3-x} - \frac{2}{x^3-x}$$

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$\int\big(\frac a x + \frac b {x+1} + \frac c {x-1} +d\big)$ – Jan Dvorak Jan 17 at 13:33
Try long division – anorton Jan 17 at 13:49

try this $$\frac{x^3-2}{x^3-x} = 1 + \frac{x-2}{x^3-x}$$ Now use partial fraction.

-
Why is it the same? – Giesi Jan 17 at 13:36
take LCM and see – experimentX Jan 17 at 13:37
@Giesi what is $(x^3-2)-(x^3-x)$? – Jan Dvorak Jan 17 at 13:38
Thx I get it ;) – Giesi Jan 17 at 13:43


text:
Solve For B In Terms Of A

1. May 24, 2008

LLS

Did I do this one correctly?

1. The problem statement, all variables and given/known data

If a = f - 2 and b = 4f + 5, what is an expression for b in terms of a?

It's multiple choice:

4a
4a - 4
4a + 2
4a + 7
4a + 13

3. The attempt at a solution

a = f - 2 and b = 4f + 5

add 2 to each side ----> a = f and b = 4f + 7

my answer is 4a + 7

2. May 24, 2008

Tedjn

You can add 2 to both sides of one equation, but you cannot just add two to one side of two equations. For example, what would happen if a = f, and b = 3f? When f = 1, we have a = 1, b = 3. Add two to both sides. a = f + 2 and b = 3f + 2. Now, to get the same answer of a = 1, we plug in -1 for f. This time, b = -1 instead of 3.

In this problem, what is the only thing that connects the two equations, i.e. what is the only thing to appear in both equations? How could you use that fact?

3. May 24, 2008

LLS

I am having trouble posting a response-getting an error message. I can't quote the above post.

"F" is common to both sides. I'm not sure what to do if that is the correct path.

4. May 24, 2008

Tedjn

Yes, you're right. What is f equal to? How can you get rid of the f in the second equation?

5. May 24, 2008

tiny-tim

Hi LLS!

It's a sort of chain-rule:

To write b in terms of a, you must write b in terms of f, and write f in terms of a.
So what is f in terms of a?

6. May 24, 2008

LLS

I need to divide by f on both sides.

a = -2 and b = 9

Would that make f = 0? And then b = 5?

Oy I'm confused.

I'll be out for the night at a family dinner in a bit. Please don't be upset if I don't get back to to the problem until tomorrow. I do appreciate the help.

7. May 24, 2008

Tedjn

No problem. Whenever you do anything to one side of an equation, you need to do it to the other side. In your case, dividing by f, what you end up with is

$$\frac{a}{f} = -2\hspace{0.35in}\text{ and }\hspace{0.35in}\frac{b}{f} = 4 + \frac{5}{f}\hspace{0.35in}\text{ if } f \neq 0$$

You cannot just divide one side of an equation by something. Always do it to both sides of the equal sign.

Read Tim's comment again. Why would finding f in terms of a help if you know b in terms of f?

8. May 24, 2008

HallsofIvy

Staff Emeritus
??? Why in the world would you "need to divide by f"?

You know that a = f - 2 . Okay, so solve for f "in terms of a": f= ?

Now just replace f by that in b= 4f + 5.

9. May 25, 2008

LLS

b = 4(f - 2) + 5

b = 4f - 8 + 5

b = 4f - 13

Thank you

10. May 25, 2008

tiny-tim

Nooo … b = 4f + 5, and you must write f in terms of a.

So f = … ?

11. May 25, 2008

HallsofIvy

Staff Emeritus
Since that is not any of the given answers, I'm going to hope that was a typo!


text:
# Math Help - Independent Random Variables

1. ## Independent Random Variables

3. Suppose we roll a dice n times. Define the random variable Xi to be equal to 1 if the
ith roll results in an even number, and 0 otherwise. Define the the random variable Yi to be
equal to 1 if the ith roll results in the number 6, and zero otherwise.
(a) Are X1 and Y1 independent random variables?
(b) Find the distribution of Z1 = X1 + Y1.
(c) Suppose n is 500. What is the approximate distribution of
1/500 (500)sigma(i=1)Xi +1/500 (500)sigma(i=1)Yi

2. Originally Posted by farshyfaroo
3. Suppose we roll a dice n times. Define the random variable Xi to be equal to 1 if the
ith roll results in an even number, and 0 otherwise. Define the the random variable Yi to be
equal to 1 if the ith roll results in the number 6, and zero otherwise.
(a) Are X1 and Y1 independent random variables?
$X_1$ and $Y_1$ are independednt if:

$P( (X_1=a) \wedge (Y_1=b))=P(X_1=a)P(Y_1=b)$

for all $a$ and $b$ in the appropriate sample spaces. Let $a=b=1$, does this hold?

(b) Find the distribution of Z1 = X1 + Y1.
The possible values of Z_1 are 0, 1, 2 calculate their probabilities.

(c) Suppose n is 500. What is the approximate distribution of
1/500 (500)sigma(i=1)Xi +1/500 (500)sigma(i=1)Yi
$U=\frac{1}{500}\sum_{i=1}^{500} X_i+\frac{1}{500}\sum_{i=1}^{500} Y_i=\frac{1}{500}\sum_{i=1}^{500} Z_i$

Now the $Z_i$'s are iid so the $U$ is the mean of a sample of size $500$ for a sample from the same distribution as $Z_1$ ... so is approximatly normally distributed with mean and variance ...

CB


text:
# Anova: Predicted R-Squared¶

Predicted R-squared is a measure of how well the model predicts a response value. It is computed as:

$Pred. R^2 = 1 - \begin{bmatrix}\frac{PRESS}{SS_{residual}\ +\ SS_{model}}\end{bmatrix} = 1\ - \begin{bmatrix}\frac{PRESS}{SS_{total}\ -\ SS_{curvature}\ -\ SS_{block}}\end{bmatrix}$

The Adjusted R-squared and Predicted R-squared should be within approximately 0.20 of each other to be in “reasonable agreement.” If they are not, there may be a problem with either the data or the model.

“Anova: PRESS” on page 1


text:
# C++ Rotating 2D shape in list

## Recommended Posts

Hay everyone

This is going to be a really basic concept but I am struggling with this immensely and am finding my self getting more and more confused as time goes on so any help with this would be a god send.

Basically I have a list which takes vertices for a shape, print out shape to the screen, such as a triangle and am trying to then implement a function which will rotate the triangle by a give input e.g 20 degrees. I have look over formulas and translating to the origin, perform the rotation and then move back to the centeroid but I an struggling to implement all this, I struggling to to see what the code is doing. Any help would be great thanks.

I have attached some .cpp and header files to the post for anyone to review, thanks. I have renamed the extensions as .txt to allow me to attach them.

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Fragment from shape.cpp:
    // rotates the vertices of a shape by a specified angle in degrees
//
// a simple method for rotating a point around 0,0 is
// x = x * cos(degrees in radians) - y * sin(degrees in radians)
// y = y * cos(degrees in radians) + x * sin(degrees in radians)
// first translate the object to the origin of the graph (0,0)
// then apply the rotation formula to each vertex of the shape
// finally, translate the object back to its original centroid
// remember that 0,0 is at the top left, not the bottom left, corner of the console
// this means that your shapes may appear upside down

Is this home work?

The three steps are literally listed above.

Do you know how to translate? I'd suggest to start with implementing a translation on a vertex, and then on the shape.

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Is this home work?

Seems like homework to me too.  But even if it isn't, this is how you learn to program.  Figure out what the code is doing.

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Yes it is homework, ok ill look into translating thanks

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Fragment from shape.cpp:

    // rotates the vertices of a shape by a specified angle in degrees
//
// a simple method for rotating a point around 0,0 is
// x = x * cos(degrees in radians) - y * sin(degrees in radians)
// y = y * cos(degrees in radians) + x * sin(degrees in radians)
// first translate the object to the origin of the graph (0,0)
// then apply the rotation formula to each vertex of the shape
// finally, translate the object back to its original centroid
// remember that 0,0 is at the top left, not the bottom left, corner of the console
// this means that your shapes may appear upside down

Is this home work?

The three steps are literally listed above.

Do you know how to translate? I'd suggest to start with implementing a translation on a vertex, and then on the shape.

Would I use a matrix to translate to point of origin?

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// x = x * cos(degrees in radians) - y * sin(degrees in radians)
// y = y * cos(degrees in radians) + x * sin(degrees in radians)

Instead, I would do this like so:

// x2 = x * cos(degrees in radians) - y * sin(degrees in radians)
// y = y * cos(degrees in radians) + x * sin(degrees in radians)

.. So x won't change in the first line, since it's used in the second one too (you're going to need the original x, not the rotated one for y rotation) :)

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// x = x * cos(degrees in radians) - y * sin(degrees in radians)
// y = y * cos(degrees in radians) + x * sin(degrees in radians)

Instead, I would do this like so:

// x2 = x * cos(degrees in radians) - y * sin(degrees in radians)
// y = y * cos(degrees in radians) + x * sin(degrees in radians)

.. So x won't change in the first line, since it's used in the second one too (you're going to need the original x, not the rotated one for y rotation) :)

There's a good habit I follow which is not to reuse variables. If the coordinates of the unrotated point are x and y, the coordinates of the rotated point should be something else, like x_rotated, y_rotated. Edited by Álvaro

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Thanks for the good advice guys, I have been told before not to reuse variables, can anyone point me in the direction of a tutorial of translation etc?

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You posted the steps yourself. You don't need another tutorial, you already have the steps documented in your files as a tutorial.  If you want you can put the search terms in your favorite search engine and find more, but you don't need them.

Your teacher has spoon-fed you exactly the information you need, right there in those files you posted.

If you need more help with it, contact your teacher.  It is their job to help you here, and if you don't talk to your teacher they won't know where you are struggling and where you need more instruction.  The teacher will assume you understand the material and move on to more complex material.

Edited by frob

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You posted the steps yourself. You don't need another tutorial, you already have the steps documented in your files as a tutorial.  If you want you can put the search terms in your favorite search engine and find more, but you don't need them.

Your teacher has spoon-fed you exactly the information you need, right there in those files you posted.

If you need more help with it, contact your teacher.  It is their job to help you here, and if you don't talk to your teacher they won't know where you are struggling and where you need more instruction.  The teacher will assume you understand the material and move on to more complex material.

Its not the math part that is the issue its applying it to each vertices

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Thanks to all the members for there advice. Its a shame the moderators a so pretentious and self righteous. Not a pleasant experience on this site, does not live up to its reputation.

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We have a strict rule of not doing people's homework on the site.

We will gladly help with a specific error or a specific problem you are having on homework when someone is clearly working on it and stumbling over something small like a syntax error, but anything more is directed back to your teacher.

This pattern is common to nearly all programming forums and discussion boards.

You are not asking about that type of problem, something where there is a subtle error in your code.  You have given us what appear to be the contents of your assignment and asked questions about topics your teacher should have covered.

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All I'm asking is how to apply rotation to a vertices, whats the point in knowledge when you don't share it? I'm not asking you to do an assignment Im basically asking to be pointed in the direction of an explanation of the logic, the files provided were just to show what the aim was.

Asking you guys to do the assignment would not benefit me in the slightest, I don't know whether you have miss interpreted my intentions with this thread but I want to learn how to do this, I am not a programming wiz kid or pick up a computer and automatically find myself in the matrix but I am committed to learning programming,

Sorry but it makes me angry that your implying I am here to get someone to do an assignment for me just to get a grade.

I'm not a little kid fresh in school, I'm 25 studying a degree and just need a helping hand which is difficult to get when 500 other people are trying to get that as well.

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So how far did you get with my suggestion to start with a translation of a single vertex?

We don't mind helping you, but "explaining the logic" is either the sentences you already have as comment in your assignment, or the actual code. To me, there is nothing in between, and the latter is exactly the solution you don't even want.

As such, I have no idea whatsoever what you don't understand, or where you are stuck. You haven't shown anything either, so I have nothing to point out where you make the wrong turn.

No doubt it's a simple communication problem, but that happens when you discuss something with a bunch of strangers across a text-only medium. A face-to-face discussion with someone like your teacher, is really a lot better in these cases.

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So how far did you get with my suggestion to start with a translation of a single vertex?

We don't mind helping you, but "explaining the logic" is either the sentences you already have as comment in your assignment, or the actual code. To me, there is nothing in between, and the latter is exactly the solution you don't even want.

As such, I have no idea whatsoever what you don't understand, or where you are stuck. You haven't shown anything either, so I have nothing to point out where you make the wrong turn.

No doubt it's a simple communication problem, but that happens when you discuss something with a bunch of strangers across a text-only medium. A face-to-face discussion with someone like your teacher, is really a lot better in these cases.

When you put it like that I see your position. I have gone away and had a look at working on applying the function to one vertices as such here is what I have so far, unfortunately it does not rotate the shape but moves it to the right instead.

Could you give a nod if I'm heading in the right direction or way off.

void Shape::rotate(double degrees)
{
int x, y, xx, yy;

x = vertices.back().getX() - centroid.getX();
y = vertices.back().getY() - centroid.getY();

int testX;
int testY;

list<Vertex>::iterator current = vertices.begin();

while (current != vertices.end())
{
testX = (*current).getX();
testY = (*current).getY();

radians = degrees * PI / 180;

testX = testX + xx +x;
testY = testY + yy +y;

(*current).setX(testX);
(*current).setY(testY);

current++;
}
}

The function is them called like so

list<Shape*>::iterator itr = shapes.begin();
while(itr!=shapes.end())
{
(*itr)->drawShape();
(*itr)->rotate(20);
Console::clear();
(*itr)->drawShape();
itr++;
}

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Your x and y variables are set in the wrong place: they use the position of ONE vertex instead of the loop vertex. So you're doing the same calculation for every vertex.

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I have revised the code again and can easily manipulate one of the vertices, the issue is when I iterate through the list it does do the other vertices

This allow to manipulate one of the vertices

int x, y, xx, yy;

x = centroid.getX();
y = centroid.getY();

x = vertices.back().getX() - centroid.getX();
y = vertices.back().getY() - centroid.getY();

radians = degrees * PI / 180;
xx = xx + centroid.getX();
yy = yy + centroid.getY();
vertices.push_back(Vertex(xx, yy));

	int x, y, xx, yy;
list<Vertex>::iterator itr = vertices.begin();
while (itr != vertices.end())
{
x = centroid.getX();
y = centroid.getY();

x = vertices.back().getX() - centroid.getX();
y = vertices.back().getY() - centroid.getY();

radians = degrees * PI / 180;
xx = xx + centroid.getX();
yy = yy + centroid.getY();
vertices.push_back(Vertex(xx, yy));

itr++;
}

}

I now get no output to the console?

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It still has problems. I don't think I should provide any more hints - you should be able to solve this, but the code you're writing indicates that you don't understand some pretty fundamental things about the programming and debugging process.

Read the code that you have written, line by line, think about what it does. Think about what is going into each variable, and how it's being used. ESPECIALLY consider how each line of code will affect the OTHER lines of code that are going to execute later.

Compare this with what you think it should be doing. If you don't know what it should be doing, that's what you need to figure out first. Just copy/pasting code in different arrangements and saying it doesn't work is not how programming works.

You should step through it line-by-line in a debugger, watch what it's doing, and see where it isn't doing what you expect it to do. Edited by Nypyren

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Here, free code commentary. Don't get used to it, though; Nypyren is absolutely right that you should learn to go through the code yourself, either executing the code by hand or with the help of a debugger.

	int x, y, xx, yy; // Why are these variables here? I don't know what they mean. They should be declared where you use them, not here.
double radians; // Same thing. At least this one has a better name. Although it's only the units...
list<Vertex>::iterator itr = vertices.begin(); // One would normally put this line and the two marked with (*) in a single for loop.
while (itr != vertices.end()) // (*)
{
x = centroid.getX(); // This would be a good place to declare the type of x. But the value you are assigning never gets used.
y = centroid.getY();

x = vertices.back().getX() - centroid.getX(); // Oh, so this is what you wanted x to be.
y = vertices.back().getY() - centroid.getY();

radians = degrees * PI / 180; // If degrees doesn't change inside the loop, why is this computation inside the loop?
xx = round(x * cos(radians) - y * sin(radians)); // And why are you using integer coordinates? Rotation wonn't work too well..
xx = xx + centroid.getX(); // Here you are modifying xx, which means the variable meant something up to this point, and something different from now on.
yy = yy + centroid.getY();
vertices.push_back(Vertex(xx, yy));

itr++; // (*) It's probably better to use ++itr, but you are not ready to understand why, so don't worry about it too much for now.
}

}

I now get no output to the console?

What output did you expect? I didn't see any line of code in what you posted that printed anything.

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Just for fun, here's how I would probably implement something like that:

#include <iostream>
#include <vector>
#include <complex>

typedef std::complex<float> C;

std::vector<C> rotate_points(std::vector<C> const &points, float angle, C center_of_rotation) {
std::vector<C> result;
C rotation = std::polar(1.0f, angle);
for (auto point : points)
result.emplace_back(center_of_rotation + (point - center_of_rotation) * rotation);
return result;
}

const float TAU = std::atan(1.0f) * 8.0f;

int main() {
std::vector<C> points = {C(1.0f, 1.0f), C(-2.0f, -1.0f)};
float angle = TAU * 0.25f; // A quarter of a turn, a.k.a. 90 degrees
std::vector<C> rotated_points = rotate_points(points, angle, C(1.0f, 1.0f));
for (auto point : rotated_points)
std::cout << point << '\n';
}

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Just for fun...

Dammit, why did I never learn cool properties of complex numbers when I was in school?!

I feel like there's all this awesome math I could be using, but the courses I took never introduced any of this. Edited by Nypyren

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Dammit, why did I never learn cool properties of complex numbers when I was in school?!

Yeah dude. Complex numbers, bro: http://acko.net/blog/how-to-fold-a-julia-fractal/

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That was something I asked my math teacher way back then when discovering the notation for complex numbers and graphing were the same, and conversions between polar coordinates and Cartesian coordinates gave the same results as converting complex numbers in the calculator. The teacher just said "Huh, I didn't realize that". It was the first time I had that type of mind-blowing experience with math. Various things clicked in my brain, these are all the same thing just applied in different types of math.

Then when I was introduced to quaternions in my programming book (still in high school before the Interwebz were a thing, the book was published in the 1970s) I saw the math and realized they were doing exactly the same thing when converting between Euclidean coordinates and quaternion coordinates.

Math is stupid, until you figure out what it means or where it helps you. Then math is amazing.

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# Calculus 1 : How to find integral expressions

## Example Questions

### Example Question #71 : Equations

Find the indefinite integral.

None of these

Explanation:

For this integral you need to know an integration rule for each part of the problem.

Using this rule we can get the integral of the first part

.

For the second part we must know that

so

.

and finally the last part follows the rule exactly

.

We must also include a C for the constant of integration as the integral is indefinite.

Thus the final answer is

.

### Example Question #71 : Equations

Give the integral expression for the velocity of an object falling from initial velocity  towards the earth with acceleration , between  and  seconds after it was released.

Explanation:

We know that the definite integral of acceleration between 0 and 4 seconds is the velocity. We know that the acceleration due to gravity is . We also know that the inititial velocity is . This means that when . Therefore, the velocity function can be written as

### Example Question #71 : Equations

Evaluate the following integral:

Explanation:

To integrate, we must make the following substitution:

We used the following rule to derivate:

Now, rewrite the integral and integrate:

We used the following rule for the integration:

Now, replace u with the original x-containing term to finish:

### Example Question #74 : Equations

Evaluate the following integral:

Explanation:

To integrate, we must make the following subsitution:

We used the following rule to derivate:

Now, after rearranging, we get the following integral:

and after integrating we get

We used the following rule to integrate:

To finish, we plug our x term back in place of u:

### Example Question #75 : Equations

Evaluate the following integral:

Explanation:

After integrating, we get

The integration was performed using the following rules:

### Example Question #76 : Equations

Evaluate the following integral:

Explanation:

In order to evaluate this integral, we must split the integrand into two seperate integrals:

For the first integral, we need to make the following substitution:

We found the derivative using the following rule:

Now, rewrite the integral in terms of u and solve:

We integrated using the following rule:

For the second integral, we just integrate:

and we use the rule

Now, just add together the results of the two integrals (the integral of the sum is equal to the sum of the integrals):

### Example Question #71 : Writing Equations

Evaluate the following integral:

Explanation:

To integrate, we must make the following substitution:

We found the derivative using the following rule:

Now, rewrite the integral in terms of u, and solve:

We used the following integration rule:

Now, replace u with the original term containing x:

### Example Question #78 : Equations

Evaluate the following integral:

Explanation:

When we integrate, we get

which was found using the rule

### Example Question #1122 : Functions

Evaluate the following indefinite integral:

Explanation:

To evaluate the integral the integral, use the inverse power rule:

Applying that rule to this problem gives us the following for the first term:

The following for the second term:

And the following for the third term:

We can combine these terms and add our "C" to get the final answer:

### Example Question #80 : Equations

Evaluate the indefinite integral: