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$D=dx÷206 265$

What does this number mean?

The number 206265 arcseconds/radian is often used in astronomy for angular conversions. It is simply derived from the product of 3600 arcseconds/degree and 57.2958 degrees/radian.

Edit based on symbols as defined in comment below:

With the distance to an object, $d$, and its lateral dimension, $D$, and using the small angle approximation where $D ll d$, the angular extent of the object in radians is given by $D/d$. Assuming that you have a measured angle, $x$, in arc seconds, you would first need to use the conversion factor above to find the angle in radians, resulting in the following equation.

$$frac{x}{206265} = frac{D}{d}$$

From that relation, knowing the measured angle and distance, you can caclulate the transverse linear dimension.

## Why do we use the sin x = x approximation for calculating waves on a rope?

If we’re going to do away with the small-angle approximation, by the same logic we might as well use general relativity to predict projectile motion it has the added benefit of working in extreme gravitational fields. Or why not use the Navier-Stokes equations to analyze the sound waves generated by a handheld tuning fork?

I do not intend to be mean I am only trying to point out the absurdity of your suggestion. I am no physicist, but it is my understanding that simplifying assumptions are often made in physics. Of course, the theories generated by these assumptions are only accurate in certain regimes, but for experimental purposes they are often good enough. Most importantly, these simplifications allow physicists to save time they would have spent wrestling with the (sometimes intractable) general theory, instead spending that time probing the implications of the simpler-but-still-accurate theory within its domain of validity. I imagine there are entire fields of physics based on making these assumptions in the context of a more general theory.

## The Time Value of Money

The money you have on hand today is worth more than the same amount of money in the future. The dollar you have in your wallet today has more buying power than a dollar a year from now. That’s because of different factors, like the effect of rising inflation.

The time value of money is the reason why you discount cash flows. In the example investment opportunity above, the buying power of the $6,000 generated in each year decreases as time goes on. To find out if the project is a good investment opportunity, you would discount the future cash flows to find the present value of the money. Simply put, you’re finding out how much $6,000 a year from now is worth in today’s time.

## Why Value Propositions Are Important (And How To Create Them)

What does your startup do? Who does it serve, and how are you different or unique?

These are questions startups need to answer clearly, effectively and quickly. If not, target audiences (consumers, partners, investors, media) move on because they’re time-strapped, and there are so many competitive options.

**The best way to address these questions is developing strong value propositions.**

There are many ways to define a value proposition but it comes down to telling people what you offer, the key benefits (aka what’s in it for users) and why you’re the best competitive option.

It sounds straightforward but it can be challenging to create effective value propositions because they need to be distinct, concise and interesting. They also need to be user-centric, which means focusing on what users need as opposed to what you offer.

**So how does a startup develop strong value propositions?**

To begin, you need to think about what your target audiences need to know. Of course, it helps to identify them and create detailed profiles, or personas, about their needs, motivations, goals, roles and buying behaviour.

The language that a startup uses to develop value propositions must reflect and connect with its target audiences. It’s that simple….and that complicated.

The upside is most startups can develop solid value propositions by investing the time and the willingness to embrace new ideas and approaches, even they feel uncomfortable (and chances are, they will feel strange initially).

A startup also has figure out how it’s different or unique, and how it can convince users it’s the best choice. We’re not talking about features, we’re talking benefits. When a customer is starting down the consideration funnel, they don’t care much about features they need to know at a high level if you and/or your product can solve their problems.

What makes a startup different or unique doesn’t have to be dramatic it just has to make a startup stand out from the crowd by highlighting a particular benefit(s) or angle(s).

An important part of this process is having in-depth knowledge of the competition. A startup has to know what its rivals offer, their key benefits and target audiences. By having a strong grasp on the competitive landscape, it is easier to effectively position yourself in different ways, including value propositions.

With my clients, I like to start the value proposition process by having a lengthy discovery session (aka brain dump), which involves a huge amount of territory. It means getting as much information about a startup, the competition and the market as possible.

From there, you look to identify the most interesting themes or ideas that can be built upon based on product, benefits, competitors and opportunities. Fast-forward, and what you get is a two or three sentence value proposition that tells the world what you do and why anyone should care.

Developing strong value propositions not only makes it easier to connect with target audiences but it establishes a foundation upon which a company’s marketing and sales activities can be built upon. It also provides employees with a consistent and cohesive way to talk about what they’re doing.

## 3. Sum of Squared Errors(SSE):

So let’s take the squares instead of the absolutes. The loss function will now become:

which is very much differentiable at all points and gives non-negative errors. But you could argue that why cannot we go for higher orders like 4th order or so. Then what if we consider to take 4th order loss function, which would look like:

Hence its *gradient* will vanish at 3 points. So it will have local minima as well — which are not our optimal solution. We need to find the point at global minima to find the optimal solution. So let’ s stick with the squares itself.

**4. Mean Squared Errors (MSE):**

Now consider we are using SSE as our loss function. So if we have a dataset of say 100 points, our SSE is, say, 200. If we increased data points to 500, our SSE would increase as the squared errors will add up for 500 data points now. So let’s say it becomes 800. If we increase the number of data points again, our SSE will further increase. Fair enough? Absolutely not!

The error should decrease as we increase our sample data as the distribution of our data becomes more and more narrower (referring to normal distribution). The more data we have, the less is the error. But in the case of SSE, the complete opposite is happening. Here, finally, comes in our warrior — Mean Squared Error. Its expression is:

We take the average or mean of SSE. So more the data, lesser will be the aggregated error, MSE.

Here as you can see, the error is decreasing as our algorithm is gaining more and more *experience*. The Mean Squared Error is used as a default metric for evaluation of the performance of most regression algorithms be it R, Python or even MATLAB.

**5. Root Mean Squared Error (RMSE):**

The only issue with MSE is that the order of loss is more than that of the data. As my data is of order 1 and the loss function, MSE has an order of 2. So we cannot directly correlate data with the error. Hence, we take the root of the MSE — which is the Root Mean Squared Error:

Here, we are not changing the loss function and the solution is still the same. All we have done is reduced the order of the loss function by taking the root.

## Example 2: Display the Number of Days Until an Opportunity Closes on a Report

You can also use formula fields in reports to increase the visibility of important information. Say, for example, you wanted a report column that displays the number of days until an opportunity is closed. First, create an Opportunity to test our formula.

If you’ve never created an Opportunity before, click and select **Opportunities**. go to the Opportunities tab and click **New**. Fill in any value for the Opportunity Name, select any Stage, and set a close date that’s at least 3 days in the future. Click **Save**.

Then take these steps to create a custom formula field called Days to Close on the Opportunities object with a Number return type.

- From Setup, open the Object Manager and click
**Opportunity**. - In the left sidebar click
**Fields & Relationships**. - Click
**New**. - Select
**Formula**and then click**Next**. - In the Field Label text area, type Days to Close .
- Select the
**Number**radio button. - Click
**Next**to open the formula editor. - To find the difference between the opportunity close date and today’s date subtract one from the other.
- Click
**Insert Field**and select**Opportunity | Close Date**and click**Insert**. - From the
**Insert Operator**menu, select**- Subtract**.

- Click
- But how do we tell our formula that we need today’s date? Luckily, there’s a function called TODAY() that updates to match the current date.
- In the
**Functions**menu on the right side of the editor, select**TODAY**. - Click
**Insert Selected Function**.

- In the
- Click
**Check Syntax**. If there are no syntax errors, click**Next**. - Click
**Next**to accept the field-level security settings, then click**Save.**

Now it's time to put your new formula field in a report.

- From , open the
**Reports**tab and click**New Report**. - Enter
**Opportunities**in the Search Report Types. field. Select**Opportunities**and click**Continue**. Your opportunity appears in the Report Preview panel. - Make sure
**Update Preview Automatically**is enabled. - In the Add column. field on the left side of the page, enter
**Days to Close**. This field is the formula field you just created. A column with the field containing the calculated value is automatically added to the report.

You won't need the report again for this unit. You can discard it and move on to the next example.

## As time period doesnot depend on amplitude so then why it is said that in measurement of time period amplitude must be kept small and also that amplitude must be zero ?

Because large amplitudes (beyond #10^@# ) invalidate the assumptions in the equation used.

#### Explanation:

If you look at the derivation it includes the line that the restoring force is proportional to # sin theta

theta# (the ‘small angle rule’)

This is less true as #theta# increases.

The Explanation describes why that is true for a simple pendulum.

#### Explanation:

You did not say what was oscillating. I will answer for the case where it is a simple pendulum that is oscillating. If it is a pendulum, amplitude must be small because the "time period does not depend on amplitude" rule applies to pendulums *only* if it is exhibiting simple harmonic motion.

Simple harmonic motion of a physical system requires that the force restoring the object (bob) to the equilibrium position must be proportional to the displacement from the equilibrium position.

Go to the site

http://hyperphysics.phy-astr.gsu.edu/hbase/pend.html

Scroll down to the section with a heading of "Period of Simple Pendulum". The first formula in that section is

#F_"net" = m*g*sintheta# ,

The angle theta (in radians) is the displacement. This equation comes from valid application of trigonometry.

In that formula, m and g are constants. Therefore this formula says that

are proportional. But the rule for simple harmonic motion says that

must be proportional. You know that sine is not a linear relation to the angle. It plots a sine wave. But, when the value of #theta# is small, it is a very good approximation to say that

When this approximation is valid that initial formula can be written

So in that form, we see that #F_"net" " and " theta# are proportional as required.

So, when amplitude is kept small (allowing use of the #sintheta = theta# approximation), time period is independent of amplitude.

## Opportunity Cost and Time Value of Money

Time value of money varies and involves an opportunity cost. That means that if you&aposre putting the $1000 in the CD, you may be foregoing an opportunity to use the money as a good faith deposit on a home. Calculating the time value of your money should tell you that instead of investing at all, you should have instead paid down expensive variable rate credit card debt that&aposs costing you hundreds a month.

An instrument or loan with a variable rate recalculates the interest paid or charged periodically.

For instance, a borrower may take out an adjustable rate mortgage (ARM) that has a low introductory rate of 2%, which will begin to adjust five years into the loan to a spread over a benchmark like the one-year Treasury bill. That means it adds the 2% to whatever the T-bill rate is and that&aposs the interest rate you pay on your loan for one year. The following year, it adjusts again depending on what T-bill rates are then, and so on until the loan matures.

Borrowers calculating the time value of money for these loans like that the interest cost is postponed for several years. The loan is designed to attract borrowers who may not otherwise be able to afford a fixed-rate mortgage and who then pin their hopes on interest rates remaining low once the interest rate begins to adjust to market value.

When that happens, the mortgage rate can jump higher suddenly depending on how much interest rates have increased since he first took out the loan. There&aposs no way to predict how much interest rates will riseਏive years from now, making it impossible to calculate the time value of money on the loan, so these can be risky.

Savvy investors who plan to sell their homes within the next several years make a TVM calculation to reduce their borrowing costs with an adjustable rate mortgages versus higher-interest fixed-rate loans. Taking advantage of the lower introductory rate, they aren&apost worried that the rate will reset much higher because they won&apost have the mortgage long enough to pay the higher rates.

## Why do we use the value 206265 in the small angle formula? - Astronomy

Plate scale relates the angular field of view of a frame, d(theta), (typically measured in arcseconds), to the size of the detector, ds, either in terms of a physical dimension (typically millimetres) or per pixel. Whilst a useful quantity in its own right - since knowledge of the plate scale tells you the field of view for a given detector - the plate scale also gives you the total length of the focal path, f, via:

It is necessary in several Astrolab projects to be able to determine the plate scale of the telescope-CCD combination, and the focal length of the telescope. Whilst the details of the various current telescope-CCD combinations are outlined on the Astrometrica tutorial page here, you may find that your project requires you, for example, to perform astrometry on an archived image that was taken when the telescope-CCD combination was different - perhaps because the CCD has been upgraded or a motorised focusser has been added to the telescope. We'll therefore describe the theory of how to extract these parameters from an archived image and go through a worked example.

To find the plate scale for a particular image you'll need at least two well defined reference targets within the frame if the frame you have doesn't allow you to accurately locate (in RA and Dec) two such targets, you may wish to calculate the plate scale using an observation from a similar date (for which you can be confident that the telescope-CCD combination is the same), and then apply your derived results in your astrometry of the original image. For our tutorial, we'll use an old image of the asteroid **(15) Eunomia**, taken from the second 10" telescope back in 2002 (E:10_220022_11_10EUNOMIA.001)

**It is essential that you do not use a moving target in your frame, such as an asteroid or comet, as a reference target!** The error on their RA and Dec will be too great to accurately determine their location in RA and Dec at the time of observation. Always use stars.

Armed with a suitable image containing at least two references, you'll need to identify each target in order to determine their RA and Dec. Open the image with CCDOPS and click on **Display > Parameters** to find the time and date of the observation, which is automatically stamped into the image's file header by the CCD. In our example here, the observation date is 11/10/2002 **(remember that the date format is the US standard MM/DD/YYYY, so this image is from the 10th November 2002)**, and the time of observation is 18:35. Also note down the image size (x pixels by y pixels) and the pixel size (x microns by y microns), as these will help to to identify the CCD model used for the imaging - we'll need this info later. Here we can see the image is 512x512 pixels in size.

Now open Earth Centred Universe and set the time and date to match those of the observation. Next, search for the primary target and centre the view on it - you may find the **Center** menu useful for this. Then left-click the primary target of the frame (i.e. what the image is of, not the first reference target) this will bring up the target info menu, and click the **place target** button, which should enclose the target with a box whose size reflects the field of view of the image, allowing you to determine which reference stars can be used. Find at least two stars in the field of view that you can see in the image, and left-click them to view their RA and Dec coordinates. Note these values down. If you can use more than 2 reference stars, do so - it will (in principle) reduce the error on your plate scale measurement, and offers a sanity check if the numbers don't all agree! If you have difficulty matching stars in the image with the Earth Centred Universe view, try rotating the image through 180 degrees in CCDOPS (**Utilities > Flip/Rotate Utilities**)- remember that the CCD might be upside down on the telescope! This is in fact the case for our example here.

Return to CCDOPS and find the x-y pixel coordinates of the reference centroids. This can be done accurately using the crosshair tool - you'll find a show/hide toggle for the crosshair tool under the **Display** menu. The position of the crosshair can be fine-tuned with the arrow keys, allowing you to place the crosshairs over the centre of the magnified image.

From our example, we'll identify the following reference stars:

**Star1:**

RA of 22:43:06.11 at x = 112

Dec of 09deg 25'08.7" at y = 96 **Star2:**

RA of 22:42:50.18 at x = 347

Dec of 09deg 28' 05.7" at y = 84 **Star3:**

RA of 22:43:11.46 at x = 144

Dec of 09deg 19' 44.2" at y = 459

With the RA and Dec coordinates for at least two targets, you can find the distance between the two in terms of angle - generally, you'll want this angle in arcseconds which means that we must convert the difference in RA from the usual units of "time" into arcseconds, using this formula:

RA difference [arcsecond] = RA difference [seconds of time] x 15 cos(Dec)

It may be confusing to think that the conversion is dependent upon the declination it may be useful to consider that at the poles, all 24 hours of RA are located at a single point on the sky, so the conversion is meaningless - effectively, RA tends to 0. At the equator, we recover the standard relationship of there being 24 hours of time in 360 degrees of rotation, and hence:

360/24 deg/hr = 15deg/hr => 15arcsec/sec

In our example here, we find the following differences (make the second to arcsec conversion using the mean declination of the two targets):

**Star1 - Star2:**

delta_RA(secs) = 24.05 at a mean dec

9.5 deg ==> delta_RA(arcsec) = 355.8

delta_Dec(arcsec) = 12.6 **Star1 - Star3:**

delta_RA(secs) = 2.77 at a mean dec

9.4 deg ==> delta_RA(arcsec) = 40.9

delta_Dec(arcsec) = 548.9 **Star2 - Star3:**

delta_RA(secs) = 21.28 at a mean dec

9.4 deg ==> delta_RA(arcsec) = 314.9

delta_Dec(arcsec) = 561.5

So with differences in x and y in terms of arcseconds, find the angular separation of the targets in the usual manner:

r (arcsecs) = sqrt[ ( RA_2(arcsec)-RA_1(arcsec) )**2 + ( Dec_2(arcsec)-Dec_1(arcsec) )**2 ]

**Star1 - Star2:**

R = 356 arcsecs **Star1 - Star3:**

R = 550 arcsecs **Star2 - Star3:**

R = 644 arcsecs

In the same fashion, find the number of pixels separating the targets:

r (pixels) = sqrt[ ( x_2(pixels)-x_1(pixels) )**2 + ( y_2(pixels)-y_1(pixels) )**2 ]

**Star1 - Star2:**

#pix = 235.3 pixels ==> d(theta)/ds = 1.51 arcsec / pixel **Star1 - Star3:**

#pix = 364.0 pixels ==> d(theta)/ds = 1.51 arcsec / pixel **Star2 - Star3:**

#pix = 426.4 pixels ==> d(theta)/ds = 1.51 arcsec / pixel

Obviously, the ratio r(arcsec) / r(pixels) therefore gives you the plate scale in units of arcsec/pixels. To convert this to a plate scale in terms of arcsec/mm, we'll need to know the size of the pixels. To determine this, first find out which CCD was used to capture the image - open the image in CCDOPS and check the **Display > Parameters** menu to find the image dimensions, and read off the pixel size from the table below:

CCD | Image Dimensions (pixels) | Pixel Size (microns) |

ST6 | 375 x 242 | 23 x 27 |

ST7 (double binning mode) | 382 x 255 | 18 x 18 |

ST7 (standard binning mode) | 765 x 510 | 9 x 9 |

ST9 | 512 x 512 | 20 x 20 |

Now convert from pixels to a physical scale by multiplying by the pixel size. There are two important things to remember: 1) the ST6 had rectangular pixels, so be sure to account for this, and 2) pixel sizes are quoted in microns, not millimetres!

Our example frame was taken with an ST9 CCD - recall earlier we noted that the image dimensions were 512x512 pixels. Hence we apply a pixel size of 20 microns to determine a final plate scale of 75.5 arcsec / mm.

Finally, to determine the focal length of the telescope-CCD combination (in millimetres), use the following relationship:

## Indirect

Use the **INDIRECT function** in **Excel** to convert a text string into a valid reference. You can use the & operator to create text strings.

### Cell Reference

Use the INDIRECT function in Excel to convert a text string into a valid cell reference.

1. For example, take a look at the INDIRECT function below.

Explanation: =INDIRECT(A1) reduces to =INDIRECT("D1"). The INDIRECT function converts the text string "D1" into a valid cell reference. In other words, =INDIRECT("D1") reduces to =D1.

2. The simple **INDIRECT** function below produces the exact same result.

3. Do we really need the INDIRECT function? Yes. Without using the INDIRECT function, this would be the result.

4. Use the & operator to join the string "D" with the value in cell A1.

Explanation: the formula above reduces to =INDIRECT("D1"). Again, =INDIRECT("D1") reduces to =D1.

### Range Reference

Use the INDIRECT function in Excel to convert a text string into a valid range reference. For example, use SUM and INDIRECT.

Explanation: the formula above reduces to =SUM(INDIRECT("D3:D6")). The INDIRECT function converts the text string "D3:D6" into a valid range reference. In other words, =SUM(INDIRECT("D3:D6")) reduces to =SUM(D3:D6).

### Named Range

Use the INDIRECT function in Excel to convert a text string into a valid named range.

1. For example, the AVERAGE function below uses the named range Scores.

Explanation: the named range Scores refers to the range D1:D3.

2. However, the AVERAGE function below returns an error.

Explanation: =AVERAGE("Scores") returns an error because Excel cannot calculate the average of a text string!

3. The INDIRECT function below does the trick.

Explanation: =AVERAGE(INDIRECT("Scores")) reduces to =AVERAGE(Scores).

### Worksheet Reference

Use the INDIRECT function in Excel to create a dynamic worksheet reference.

1. This is what a simple worksheet reference looks like.

Note: cell A1 on Sheet1 contains the value 10. Cell A1 on Sheet2 contains the value 20. Cell A1 on Sheet3 contains the value 30.

2. On the Summary sheet, enter the INDIRECT function shown below. Use the & operator to join the sheet name in cell A1 with "!A1".

Explanation: the formula above reduces to =INDIRECT("Sheet1!A1"). The INDIRECT function converts the text string "Sheet1!A1" into a valid worksheet reference. In other words, =INDIRECT("Sheet1!A1") reduces to =Sheet1!A1.

3. If your sheet names contain spaces or other special characters, enclose the sheet name in single quotation marks. Modify the INDIRECT function as shown below.