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# Centering Ring Reinforcement Concept

### Help Support The Rocketry Forum:

#### eugenefl

##### Well-Known Member
TRF Supporter
I'm not sure if this is new or not, but I thought I'd share a "breakthrough" I had while building my MEGA-Gyroc MMT.

Purpose: Increase the "blockage" area ahead of the centering ring to better distribute the forces of liftoff thrust on the motor mount. To prevent the MMT from breaking loose inside the airframe and flying through the airframe.

Please see attached drawing. I'm curious if this has been done before or has ever been given any thought. Typically, "fin cans" or instances where the fins are incorporated into the MMT take the place of this design. In essence though, I kinda see it as extra insurance.

#### Bill

##### Well-Known Member
Originally posted by eugenefl
Purpose: Increase the "blockage" area ahead of the centering ring to better distribute the forces of liftoff thrust on the motor mount. To prevent the MMT from breaking loose inside the airframe and flying through the airframe.
One word: gussets. Cut three or four pieces of scrap balsa wood as wide as the distance from the outside of the motor-mount tube to the inside of the outer body tube and as long as the distance between your centering rings. Glue them onto the MMT as you would fins. Apply fillets where the gussets meet the MMT and centering rings. Adds lots of strength with little added weight unless you go crazy with the glue. Balsa sheeting is also cheaper and more available than a coupler.

If yours is a long motor-mount/stuffer tube, then cut triangular pieces instead of rectangles to form braces between the MMT and centering ring.

Bill

#### DynaSoar

##### Well-Known Member
Originally posted by eugenefl
I'm not sure if this is new or not, but I thought I'd share a "breakthrough" I had while building my MEGA-Gyroc MMT.

Purpose: Increase the "blockage" area ahead of the centering ring to better distribute the forces of liftoff thrust on the motor mount. To prevent the MMT from breaking loose inside the airframe and flying through the airframe.

Please see attached drawing. I'm curious if this has been done before or has ever been given any thought. Typically, "fin cans" or instances where the fins are incorporated into the MMT take the place of this design. In essence though, I kinda see it as extra insurance.

Some older Estes kits had motor mounts which used a piece of coupler tube the whole length between the centering rings. That much contact would distrubute far more power than engines had in those days.

To make that design, or yours, even stroonger, as Bill says, "gussets": pieces of balsa cut at an angle and glued between the centering ring and the tubes. All my 18mm to 24mm conversions get them. It may be overbuilding, but better that than the opposite.

#### Stewart32

##### Well-Known Member
can't get much more surface area than that?

#### Stewart32

##### Well-Known Member
can't get much more surface area than that?

#### Micromeister

##### Micro Craftman/ClusterNut
TRF Supporter
Good Ideas Eugene, but not at all new. The en-ga-neer has to agree with Bill: quick and easy reinforcements are to add gussets to your centering rings-MMt joint. That is by the way why TTW fins seem to work better then surface mounted fins. the second attachment point creates a gusset plate between the MMT and airframe.

#### eugenefl

##### Well-Known Member
TRF Supporter
Originally posted by Micromister
Good Ideas Eugene, but not at all new. The en-ga-neer has to agree with Bill: quick and easy reinforcements are to add gussets to your centering rings-MMt joint. That is by the way why TTW fins seem to work better then surface mounted fins. the second attachment point creates a gusset plate between the MMT and airframe.
Haha...I went to EMRR to look up "gussett".

Gusset - A roughly triangular shaped piece of material added to reinforce a joint, often in a corner.

This whole time I was picturing something to the effect of spokes on a CR or something like a wagon wheel.

#### Micromeister

##### Micro Craftman/ClusterNut
TRF Supporter
Traditionally a gusset is a series of triangular pieces welded or mounted perpendicular to a base plate and its 90 degree projecting member. Guessets do not have to be triangles, any profile can be used. an example would be a piece of 2" x 1/8" rolled scroll steel connecting a horizontal projecting arm member to a vertical support post acting as a reinforcing gusset for a cantilevered sign.
Yes your mental picture of a wagon wheel is accurate for this MMT/centering ring application. The "gussets" are the spokes, and would be connected to the MMT, centering ring (base plate) and the outer ring or Airframe. Minimum in this example would be 4 pieces, but any number could be used.

Honeycomb construction is another form of greatly increasing the strength of a spanning member (centering ring) without dramatically increasing the overall weight or thickness of the original part. ie two cardstock disc (centering rings) with the same cardstock thickness used for the gusset walls 3/16" to 1/4" high (combined thickness of the completed honeycomb part) will be stronger than say a 1/8" plywood ring alone and maybe 1/3 the weight.

See Yes-at-day I couldnt speel En-GA-neer...now I are one

#### limd21

##### Well-Known Member
For those of us who cut our own centering rings, the nice thing about using gussets is that the rings themselves can be built out of relatively light and, most importantly, easy to cut material (e.g. card stock). It's just plain easier to manually cut accurate rings with light material than it is with heavy, thick material. (You guys using fly cutters are cheating ;-)

The gussets take the structural load, the centering rings do the, well, centering.

#### Micromeister

##### Micro Craftman/ClusterNut
TRF Supporter
Wow! Sorry WAY off topic but I almost moved to boulder about 28 years ago...was going to open a sign shop there but got cold feet and stayed in this #4#$#$ forsaken part of the country
Is boulder still as clean and clear skied as I remember....memories....

#### DJ Delorie

##### Well-Known Member
In the diagram, you have both thrust rings on the same side of the centering ring. What I would have done (and almost did on my last model, but will on the next) is insert the coupler into the airframe first, put the MMT braces behind the centering ring, and then glue it all together. That way, the engine thrust is always pushing parts together rather than pulling them apart and relying only on glue strength.

My last model had balsa spars (i.e. rectangular sticks) glued ahead of the front CR just to give more glue area between the MMT and the body tube, and to add shear strength. The next one will have the spars between the two CRs, with a coupler ahead of it.

The other thing I do is put extra glue inside the BT. After inserting the MMT, I stand the rocket up and let the glue puddle in front of the CR. That adds a little extra strength and burn-proofs the CR.

#### Micromeister

##### Micro Craftman/ClusterNut
TRF Supporter
Just so we keep this stuff in perspective. All these wonderful strength reinforcements we've been discussing here help make the motor mount centering rings stiffer, providing less flex in these bulkheads that would otherwise be transfered to the area actually keeping the motor mount in place. What is that you ask? those butt mating, slide-in glue joints at the motor tube and the airframe tube. That's to say all that holds our models together is the glues and a bit of friction. I think we sometimes really get beyond the most important issure, fitting of parts and overall weight to thrust ratios.
It's more the shock absorbing elasticity of the adhesives we use then all the strength we "think" we're building into our models that keeps them flying. All these super reinforced motor mounts will fly straight through the model if those tiny little glue joints fail. Try to keep in mindl elasticity and shear strength! Or come up with a mechanical fastening system for cardboard to wood.
Just to keep us humble.

A

#### Austin

Micro makes a good point and it is valid for small to medium rockets...especially on those relying on the bodytube to take the brunt of stress that is applied during flight. Some may be necessary to act as a buffer.

However, when it comes to the big guns, the best structural designs usually include an internal airframe that takes the total lifting stress of the model, while the body tube is only along for the ride. Most of the power is distributed through the airframe and the adhesives are selected and used for the strength they will add rather than the "give" they can provide. I don't see West Systems epoxy giving much under stress, but it's holding and streghtening qualities are right where they need to be...

Moreover, most of the thrust on large airframes is applied at the rear, and sometimes at the top if secured there. this thrust is distributed through the airframe rather than the body tube. The lip you use for a bottom CR/thrustplate will play much into distributing power at launch. I personally use aluminum bottom plates, which prevent the "fly through" effect and distribute the power evenly across the bottom with little flex ; it has worked well.

As mentioned, joint types, materials and adhesives play a major part in modeling and use much of the bodytube for power distribution, while larger airframes have the major components attached to the frame itself, while the bodytube goes along for the ride.

Carl

#### Elapid

##### Well-Known Member
Butt Joints

a butt joint will take huge amounts of force and does not depend on adhesives for strength, PLUS adhesives are strongest with tensile loads rather than shear forces (rem the super glue commercials with the guy hanging from a steel beam? twist with the same force and the glue would fail).

for the forward centering ring, make two pieces, one that slides over the end of the tube, the other butts against the end of the tube, having the same I.D. as it does. these rings are then glued together. for the rear CR, for ultimate strength, butt it against the body tube AND the MMT tube, and like the forward CR, do it with two rings, one that fits around the MMT tubing and inside the body tube, the other would butt against both the MMT and BT.

epoxies get soft with heat. faster-curing epoxies are worse than slow-cure. I prefer JB Weld atm for its hig heat tolerance and strength.

hope this helps

#### powderburner

##### Well-Known Member
I put in another vote for gussets . . . sort of.

I also agree with DJ Delorie's comments about modifying the stack of the components, and MicroMister's observation that a tight fit is more important than a gob of glue.

Back to CR's and reinforcing webs: The deal is, structural loads follow the stiffest path. If you have a CR that is infinitely strong and does not deflect under load (ridiculous example, I know, but I'm trying to make a point) then that CR will transmit ALL THE LOAD between the motormount and the BT, regardless of what else, how many, what orientation, what attachment, etc, is used for reinforcement.

Now your mother-of-all-centering rings mega-plywood creation is going to be pretty stiff, even when compared to typical peak-thrust levels of mid-power motors. Addition of plywood gussets would help distribute the load and would be structurally effective, but would also be quite heavy. Addition of cardboard (or other soft material like balsa) gussets will probably not add much in the way of structural reinforcement.

If you were to use a thick cardboard CR (a relatively softer material that would deflect under thrust loads) then the gussets would just about 'take over' as the primary load path. Mind you, the solid CR would still be needed to help react the internal structural loads, and to help hold things together. Using thick cardboard for all components, you could build significantly lighter structure this way---this is only a half-step away from the 'honeycomb' structure that someone else mentioned.


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# Conditional Formatting to Create 100% Stacked Bar Chart in Excel

A colleague asked me if there was a way to create a dynamic in-cell 100% stacked bar chart in Excel for three product sales.

As usually is the case, there were thousands of rows of data. There is no way she could have used Excel in-built charts as that would have taken her ages to create charts for each set of data points.

This got me thinking, and fortunately, conditional formatting came to the rescue. I was able to quickly create something neat that fits the bill.

##### Creating a 100% Stacked Bar Chart in Excel

Suppose you have sales data for 12 months for three products (P1, P2, and P3). Now you want to create a 100% stacked bar chart in Excel for each month, with each product highlighted in a different color.

Something as shown below:

##### How to create this:

First, you need to calculate the percentage breakup for each product for each month (I was trying to make a 100% stacked chart remember!!).

To do this, first create three helper columns (each for P1, P2, and P3) for all the 12 months. Now simply calculate the % value for each product. I have used the following formula:

=(C4/SUM($C4:$E4))*100)

Once you have this data in place, let’s dive in right away to make the stacked chart

1. Select 100 columns and set their column width to 0.1.
2. Select these 100 cells in the first data row (K4:DF4) in this case.
3. Go to Home –> Conditional Formatting –> New Rule.
4. In New Formatting Rule Dialogue box, click on ‘Use a formula to determine which cells to format’ option.
5. In the ‘Edit the Rule Description’ put the following formula and set the formatting to Blue (in ‘Fill’ tab)
=COLUMNS($K$4:K4)<=$G4  1. Now again select the same set of cells and go to Home – Conditional Formatting – Manage Rules. Click on New Rule tab and again go to ‘Use a formula to determine which cells to format’ option. Now put the formula mentioned below and set the formatting to green color. =AND(COLUMNS($K$4:K4)>$G4,COLUMNS($K$4:K4)<=($G4+$H4))
1. And finally again repeat the same process and add a third condition with the following formula and set the formatting to orange color.
=AND(COLUMNS($K$4:K4)>($G4+$H4),COLUMNS($K$4:K4)<=100)
1. Now click Ok and you would get something as shown below:

Hide the helper columns, and you have your dynamic 100% stacked bar chart ready at your service.

Now it’s time to bask in the glory and take out some time to brag about it

format: plain


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# Automorphism groups of virtually cyclic groups

Let $V$ be a virtually cyclic group. Then is $Aut(V)$ also a virtually cyclic group?

This is true when $V$ is a finite group (zero-ended) and when $V = C_\infty, D_\infty$ (both two-ended).

-

In Finitely generated groups with virtually free automorphism groups, by M.R. Pettet, it is proved in theorem 3.4 that the automorphism group $Aut(G)$ of a finitely generated group is virtually cyclic if and only if both $Z(G)$ and $G/Z(G)$ are virtually cyclic.


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# How do you convert (7pi)/5 from radians to degree?

Jan 31, 2017

$\pi \text{ radian="180^@ rArr 7/5pi" radian=} {\left\{\left(\frac{7 \pi}{5}\right) \left(\frac{180}{\pi}\right)\right\}}^{\circ} = {252}^{\circ}$


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### Home > MC1 > Chapter 8 > Lesson 8.2.1 > Problem8-29

8-29.

Using this ratio, each side should become three times as long.

If you were to enlarge the above square so that each side is three times as long, the labeled sides would each be six units.
Try drawing this enlarged square, if you are having trouble.

Make each side half the length of the original shape.


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# Thermal Physics “Thermal Equilibrium” Confusion on statement

I am reading a book regarding the work of Anders Celcius that states the following -

By definition, any object at 100ºC must be in thermal equilibrium with boiling water.

I am confused by this statement.

Surely in thermal equilibrium, it must be between two different objects with no net flow of thermal energy.

However, an object at 100ºC does not have to have boiling water to be at that temperature. I may just be confused.

Can someone explain on this statement please?

Thanks.

• Pretty sure it just means that any object with a temperature of $100^{\circ}$ C, should it come into contact with boiling water, will be in thermal equilibrium with the water -- not that boiling water is in any way necessary to heat an object to $100^{\circ}$ C. – quanticbolt Mar 30 '17 at 17:07
• @quanticbolt ahhh got it! Add that as an answer please so I can give it a rating and a tick! Thanks. – vik1245 Mar 30 '17 at 17:48

The book does not imply that boiling water is necessary to heat an object to a temperature of $100^{\circ}$ -- there are several other methods by which an object can be heated to a desired temperature.

Rather, the book merely states that should we place an object with a uniform temperature of $100^{\circ}$ C in a body of boiling water, that the object would be in thermal equilibrium with that water, as it should be.

## Equilibrium

System in equilibrium

It is a fact of experience that when two systems come into thermal contact (energy is transferred with a mechanism different from work, Heat) generally changes happen to both of them.

However if we isolate them we will see that eventually they reach a state in which no further change is perceptible, no matter how long one waits.

For simple systems meaning that they have no magnetic or electrical properties their equilibrium state can be completely described by their volume V and pressure P.

Systems in equilibrium with each other

If we have two systems $S_1$ and $S_2$ in thermal contact and we consider the two as one system $S_{12}$, then if the system $S_{12}$ is in equilibrium then they two systems $S_1$ and $S_2$ are said to be in equilibrium with each other.

## Zeroth Law

The zeroth law states that if two systems A and C are in thermal equilibrium with a third one B(which we later will see that can be defined as the thermometer) then they are in equilibrium with one another as well.

## Criterion of Equilibrium and Empirical Temperature

Suppose we have three systems A,B and C that are in equilibrium states.

$$A(P_A,V_A),B(P_B,V_B),C(P_C,V_C)$$

The condition under which A and B are in equilibrium may be expressed bu the equation: $$F_1(P_A,V_A,P_B,V_B)=0 \Rightarrow P_B = f_1(P_A,V_A,V_B)$$

The same applies for the equilibrium of systems C and B.

$$P_B = f_2(P_C,V_C,V_B)$$

From the two above equations turns out that:

$$f_1(P_A,V_A,V_B)=f_2(P_C,V_C,V_B) \Rightarrow G(P_A,V_A,V_B,P_C,V_C)=0$$

From the zeroth law turns out that if A and C are separately in equilibrium with B the they are with each other as well. Therefore:

$$F_3(V_A,P_A,P_C,V_C)$$

Since the last two equations are to be equivalent, then the implicit function G is independent from $V_B$. Therefore going backwards:

$$φ_1(P_A,V_A)=φ_2(P_C,V_C)$$

Therefore we can define a function $$φ(P,V)=θ$$ Which we call empirical temperature and it has the property that systems that take the same value are in equilibrium with each other.

We have defined that water boils at $100^oC$. Therefore according to the property of the emperical temperature function every system at $100^oC$ is in equilibrium with boiling water.


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# Math Help - Surface area and Volume

1. ## Surface area and Volume

A Circle C is given by the equation x^2 +(y-1)^2 = 1. It is rotated about the x-axis to give a surface of revolution called a torus.

a) Find the total surface area of the torus.

(b) Find the volume of the torus and verify that it is 2pi times the area of C.

2. ## re: Surface area and Volume

Have you covered the theorems of Guldin?

3. ## re: Surface area and Volume

No, what we have covered are that :

the surface area is the integral of 2pi y (1+(dy/dx)^2)^1/2 dx, and the volume is the integral of pi y^2 dx. I have used these equations, integrated from -1 to 1 and got 2pi(2+pi) for the area, and 10pi/3 +pi^2. But I think I'm doing something wrong. Can you help me?

4. ## re: Surface area and Volume

Better work in parametric equations $C\equiv x=\cos t,\;y=1+\sin t\;(t\in[0,2\pi])$ . Then, the area is $A=2\pi\int_0^{2\pi} y\sqrt{(dx/dt)^2+(dy/dt)^2}\;dt=\ldots$ . Let's see what do you obtain.

5. ## re: Surface area and Volume

Originally Posted by FernandoRevilla
Better work in parametric equations $C\equiv x=\cos t,\;y=1+\sin t\;(t\in[0,2\pi])$ . Then, the area is $A=2\pi\int_0^{2\pi} y\sqrt{(dx/dt)^2+(dy/dt)^2}\;dt=\ldots$ . Let's see what do you obtain.

When I integrated and used the limits 0 and 2pi, I got an answer of 4pi^2, but when I changed to 0 and pi, I got 2pi(pi+2) which is the same answer as when I worked it out using the cartesian equation rather than the parametric. Which one is correct?

And what about the volume? I worked it out using parametric as well but got an answer of 2pi^2. The question says that it should be 2pi times the answer of the area...

Thanks

6. ## re: Surface area and Volume

Originally Posted by perla
When I integrated and used the limits 0 and 2pi, I got an answer of 4pi^2
Right.

And what about the volume? I worked it out using parametric as well but got an answer of 2pi^2. The question says that it should be 2pi times the answer of the area...
Right, the volume is $V=2\pi^2=2\pi\cdot \pi=2\pi \cdot \textrm{Area}(C)$


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# Knowledge base: Warsaw University of Technology

Back

## Proposal of the methodology for determining the stiffness matrix of the crankshaft

### Mateusz Paweł Wojciechowski

#### Abstract

The aim of the study was to propose a method for determining the stiﬀness matrix for the linear-elastic model of the crankshaft. In the ﬁrst part of the work described tools, that were used in calculations. The Maxwell-Mohr’s method was ﬁrst described one the example of a beam on one ﬁxed supported at the end and loaded with concentrated moment on the other. The energy and matrix method were used. Additionally, the principle of operation of the ﬁnite element method were presented, its advantages and disadvantages, were given, also the aplication were describet In the next chapter, using the SolidWorks program, a 3D model of the existing crankshaft was created, its material parameters were determined and the accuracy of the model relative to the real object was veriﬁed. After checking the accuracy of the model, a fragment of the shaft was separated, which had two main spigots and one crank pin. Thanks to this choice, it will be possible to use this method for each shaft length by multiplying the given model. Next, the discretized object was discretized, the boundary conditions were determined, the load was given and subjected to static analysis. The results were read and a susceptibility matrix for the CAD model was created. In the next stage, a beam model was created. The obtained frame was loaded identically as 3d model and displacements were determined using the Maxwell-Mohr method. Next, the susceptibility matrix of the analyzed frame was determined. Young’s modulus and the moment of inertia were selected so that matrices determined analytically from the beam model and simulated from the spatial model were as close as possible to each other. In the last part of the work, the error of simplifying the beam model relative to the matrix of susceptibility of the CAD model was determined. The conclusions summarize the conclusions drawn from the research.
Diploma type
Engineer's / Bachelor of Science
Diploma type
Engineer's thesis
Author
Mateusz Paweł Wojciechowski (FACME) Mateusz Paweł Wojciechowski,, Faculty of Automotive and Construction Machinery Engineering (FACME)
Title in Polish
Propozycja metodyki wyznaczania macierzy sztywności wykorbienia
Supervisor
Robert Zalewski (FACME/IMDF) Robert Zalewski,, Institute of Machine Design Fundamentals (FACME/IMDF)Faculty of Automotive and Construction Machinery Engineering (FACME)
Bogumił Chiliński (FACME/IMDF) Bogumił Chiliński,, Institute of Machine Design Fundamentals (FACME/IMDF)Faculty of Automotive and Construction Machinery Engineering (FACME)
Certifying unit
Faculty of Automotive and Construction Machinery Engineering (FACME)
Affiliation unit
Institute of Machine Design Fundamentals (FACME/IMDF)
Language
(pl) Polish
Status
Finished
Defense Date
16-04-2019
Issue date (year)
2019
Internal identifier
SIMR; D-2174
Reviewers
Bogumił Chiliński (FACME/IMDF) Bogumił Chiliński,, Institute of Machine Design Fundamentals (FACME/IMDF)Faculty of Automotive and Construction Machinery Engineering (FACME) Jakub Lasocki (FACME/IV) Jakub Lasocki,, Institute of Vehicles (FACME/IV)Faculty of Automotive and Construction Machinery Engineering (FACME)
Keywords in Polish
Macierz sztywnościModel belkowy Model liniowo-sprężysty Metoda Elementów Skończonych
Keywords in English
Stiﬀnes matrix Beam model Linear elastic model Finite element method
Abstract in Polish
File
• File: 1
Mateusz_Wojciechowski_Praca_inżynierska_-_Propozycja_metodyki_wyznaczania_macierzy_sztywności_wykorbienia.pdf
Request a WCAG compliant version
Local fields
Identyfikator pracy APD: 28870

Uniform Resource Identifier
https://repo.pw.edu.pl/info/bachelor/WUTef9037d1c5f94ab2a943332dc4ea05fa/
URN
urn:pw-repo:WUTef9037d1c5f94ab2a943332dc4ea05fa

Confirmation
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Probability Formula

# Probability Formula

Probability is based on observations of certain events. Probability of an event is the ratio of the number of observations of the event to the total numbers of the observations. An experiment is a situation involving chance or probability that leads to results called outcomes. An outcome is the result of a single trial of an experiment. The probability of an event is the measure of the chance that the event will occur as a result of an experiment.

Probability of an event A is symbolized by P(A). Probability of an event A is lies between  0 ≤ P(A) ≤ 1.

## Probability Formula

Probability is the measure of how likely an event is. And an event is one or more outcomes of an experiment. Probability formula is the ratio of number of favorable outcomes to the total number of possible outcomes.

Probability Formula

Probability of an Event =  $\frac{Number\ of Favorable\ Outcomes}{Total\ Number\ of\ Possible\ Outcomes}$

Measures the likelihood of an event in the following way:

- If P(A) > P(B) then event A is more likely to occur than event B.
- If P(A) = P(B) then events A and B are equally likely to occur.

## What is the Formula for Probability

The probability of an event tells that how likely the event will happen. Situations in which each outcome is equally likely, then we can find the probability using probability formula. Probability is a chance of prediction. If the probability that an event will occur is "x", then the probability that the event will not occur is "1 - x". If the probability that one event will occur is "a" and the independent probability that another event will occur is "b", then the probability that both events will occur is "ab".
Probability of an event  A can be written as:

P(A) = $\frac{Number\ of favorable\ outcomes}{Total\ number\ of\ possible\ outcomes}$

## How to Solve Probability Problems

A probability is determined from an experiment, which is any activity that has an observable outcome like tossing a coin and observing whether it lands heads up or tails up. The possible outcomes of an experiment are called sample space of the experiment.

Steps to find the probability:

Step 1: List the outcomes of the experiment.

Step 2: Count the number of possible outcomes of the experiment.

Step 3: Count the number of favorable outcomes.

Step 4: Use the probability formula.

## Solved Examples

Question 1: Two dice are rolled once. Calculate the probability that the sum of the numbers on the two dice is 5.
Solution:
Possible outcomes (Sample Space) = {(1, 1), (1, 2),...............,(1, 6), (2, 1), (2, 2),................,(2, 6), (3, 1), (3, 2),...........,(3, 6), .............,(4, 1), (4, 2),..........,(4, 6), (5, 1), (5,2),...............,(5, 6), (6, 1), (6, 2),......................,(6, 6)}

Total possible outcomes = 36

Number of outcomes of the experiment that are favorable to the event that a sum of two events is 6

=> Favorable outcomes are: (1, 5), (2, 4), (3, 3), (4, 2) and (5, 1)

Number of favorable outcomes = 5

Use, probability formula = $\frac{Number\ of favorable\ outcomes}{Total\ number\ of\ possible\ outcomes}$

= $\frac{5}{36}$

The probability of a sum of 6 is $\frac{5}{36}$.

Question 2: What is the probability of getting head when tossing a coin.

Solution:
Sample Space = {H, T}

Number of possible outcomes = 2

Number of favorable outcomes = 1
(because of only one head)

=> Probability of getting head = $\frac{Number\ of favorable\ outcomes}{Total\ number\ of\ possible\ outcomes}$

= $\frac{1}{2}$

=> Probability of getting head is $\frac{1}{2}$.