Don’t forget to grab a copy of my new book Motivation Beyond the Quotes by clicking this link now!

Don’t forget to grab a copy of my new book Motivation Beyond the Quotes by clicking this link now!

Comeback later this week when I go in deeper details on other functions inside the banner rotator at TEToolbox.

Maybe it goes without saying but whenever you decide to do something you always have to start somewhere. The problem is many of us wait for the perfect situation to get going. Take 90 seconds of your day to hear my thoughts about this very quote.

Don’t forget to grab a copy of my new book Motivation Beyond the Quotes by clicking this link now!

Head over to Patrick’s blog by clicking here and help us choose the cover to our new book.

Don’t forget to comeback here tomorrow to find out how you can get a copy!

When you are clear and sharp on your decision, others will not second guess you; instead, they’ll back you up. There is no magic in developing the ability to be decisive. It is simply a mindset; a mental attitude that comes with practice and persistence.

Don’t forget to checkout my blog post, by clicking here, which goes into deeper detail on this topic.

However Jim Rohn said it best “Motivation alone is not enough, if you have an idiot and you motivate him, now you have a motivated idiot.”

Today we’ll be going over how I did the main Treasure text on Melvin’s Treasure mini game card.

Fun Medieval Text Tutorial

• Software: Photoshop CS3 or higher
• Skill Level: Novice
• Time: 15-20 minutes

Before we get started you’re going to need Adobe Photoshop. Don’t worry if you don’t have a copy of it yet Adobe gives you a free 30 day trial. Here is a link to start your FREE no obligation trial so you can follow along on my super easy to do tutorial.

Step 1.

Setup a new file inside Photoshop. Name it “fun medieval text”.

Step 2.

Let’s get this image started off with a dark background color. I’m using #311909

I’m going to assume some of us have never used Photoshop before so I will show you how to do this. The rest of you PS pros skip to Step 3.

Step 3.

Find an image of some old, sun stained paper. There are a ton of great free resources for stock images, one of my favorites is http://www.sxc.hu/

Here is the link to download the same piece of paper I use in this tutorial.

We need to add this old paper image into our canvas, to do this click File > Place. A window will popup allowing you to search for the image you want to place in your canvas, once you find your image double left click on it. You will now see the old piece of paper centered perfectly inside our canvas. It should look like the image below if you didn’t skip ahead.

To get rid of the gray crossing lines just hit the ENTER key or right click on the old paper and left click Place.

Step 4.

The font I’m using for this tutorial is called Magic Medieval, you can download it here for free.

Change your Foreground Color to #1d98b9 if you don’t know how to do this please watch the video in Step 2. Select the Horizontal Type Tool and put in whatever text you want, I just typed in the word Treasure.

Duplicate this text layer so we now have 2 of them. You’ll see why we do this later.

Step 5.

Select your top layer and click Layer > Layer Style > Blending Options

Just copy the attributes I’ve given you for each category.

Drop Shadow

Inner Shadow

Outer Glow

Bevel and Emboss

Contour

Satin

Your image should now look like this.

Step 6.

Now that our top text layer is complete we can work on the bottom text layer to make this graphic look like a million bucks.

Select the bottom text layer and click Layer > Layer Styles > Blending Options. Once again just copy my attributes.

Drop Shadow

Your image should now look like this.

The hard work is done now you can lock these two text layers together so you can rotate and move it where ever your heart desires on the canvas.

I hope you found this tutorial useful and I look forward to making more of these for you in the near future. Don’t forget to post a link of your attempt at this tutorial in the comments section below. I’d love to see what you did!

What would you like me to show you on the next “How did you do that?” tutorial?

The term is commonly used in computer graphics to describe the three-dimensional region which is visible on the screen, the “viewing frustum”, which is formed by a clipped pyramid; in particular, frustum culling is a method of hidden surface determination.

In the aerospace industry, frustum is the common term for the fairing between two stages of a multistage rocket (such as the Saturn V), which is shaped like a truncated cone.

Each plane section is a floor or base of the frustum. Its axis if any, is that of the original cone or pyramid. A frustum is circular if it has circular bases; it is right if the axis is perpendicular to both bases, and oblique otherwise.

The height of a frustum is the perpendicular distance between the planes of the two bases.

Cones and pyramids can be viewed as degenerate cases of frusta, where one of the cutting planes passes through the apex (so that the corresponding base reduces to a point). The pyramidal frusta are a subclass of the prismatoids.

Two frusta joined at their bases make a bifrustum.

Let’s checkout the formula!

Volume

The volume of a conical or pyramidal frustum is the volume of the solid before slicing the apex off, minus the volume of the apex:

where B₁ is the area of one base, B₂ is the area of the other base, and h₁, h₂ are the perpendicular heights from the apex to the planes of the two bases.

Considering that

the volume can also be expressed as the product of the height h = h₂−h₁ of the frustum, and the Heronian mean of their areas:

Heron of Alexandria is noted for deriving this formula and with it encountering the imaginary number, the square root of negative one.

In particular, the volume of a circular cone frustum is

where π is 3.14159265…, and R₁, R₂ are the radii of the two bases.

The volume of a pyramidal frustum whose bases are n-sided regular polygons is

where a₁ and a₂ are the sides of the two bases.

Surface area

For a right circular conical frustum

and

where R₁ and R₂ are the base and top radii respectively, and s is the slant height of the frustum.

The surface area of a right frustum whose bases are similar regular n-sided polygons is

where a₁ and a₂ are the sides of the two bases.

Anyone else  excited as I am over how awesome sauce this Math Monday was? Comeback next week when we talk about Rhombicuboctahedron!

Al-Jayyani’s The book of unknown arcs of a sphere in the 11th century introduced the general law of sines.The plane law of sines was later described in the 13th century by Nasīr al-Dīn al-Tūsī. In his On the Sector Figure, he stated the law of sines for plane and spherical triangles, and provided proofs for this law.

According to Glen Van Brummelen, “The Law of Sines is really Regiomontanus’s foundation for his solutions of right-angled triangles in Book IV, and these solutions are in turn the bases for his solutions of general triangles.”

Regiomontanus was a 15th-century German mathematician.

The Law of Sines says that in any given triangle, the ratio of any side length to the sine of it’s opposite angle is the same for all three sides of the triangle. The Law of Sines is written formally as:

This is true for any triangle, not just right triangles!

a, b and c are sides.

A, B and C are angles.

(Side a faces angle A, side b faces angle B and side c faces angle C).

So if you divide side a by the sine of angle A it is equal to side b divided by the sine of angle B, and also equal to side c divided by the sine of angle C.

This is just beyond cool but what is it used for?

A triangle has three sides and three angles. The Law of Sines is one of the tools that allows us to solve the triangle. That is, given some of these six measures we can find the rest. Depending on what you are given to start, you may need to use this tool in combination with others to completely solve the triangle.

Now that we’ve got the basics lets check this bad boy out in action!

a/sin A = 8 / sin (62.2°) = 8 / 0.885… = 9.04…

b/sin B = 5 / sin (33.5°) = 5 / 0.552… = 9.06…

c/sin C = 9 / sin (84.3°) = 9 / 0.995… = 9.05…

The answers are almost the same!
(They would be exactly the same if I used perfect accuracy).

So to wrap this up, if we are given one side and its opposite angle we can find the “law of Sines” ratio for the triangle. Then, using that ratio and the other given elements, we can solve the triangle.

Triangles are AWESOME SAUCE!