Thursday 30 June 2016

Lesson Note On Definitions of Trigonometric Functions and Its Equations

Definitions of Trigonometric Functions

Draw a unit circle with center O. Let a central angle with initial side OP and terminal side OQ contain x radians (that is, the arc PQ has length x). Drop a perpendicular from Q to OP meeting it at R. Then OR = cos(x) and RQ = sin(x). If those directed line segments are up or to the right, the lengths are positive. If they are down or to the left, the lengths are negative.
Values at special angles:
  x      sin(x)     cos(x)      tan(x)      cot(x)       sec(x)       csc(x)
   
  0        0          1           0          ---           1           ---
 /6      1/2      sqrt(3)/2   sqrt(3)/3   sqrt(3)     2 sqrt(3)/3      2
 /4   sqrt(2)/2   sqrt(2)/2      1           1         sqrt(2)      sqrt(2)
 /3   sqrt(3)/2     1/2       sqrt(3)     sqrt(3)/3       2       2 sqrt(3)/3
 /2       1          0          ---          0           ---           1
2/3   sqrt(3)/2    -1/2      -sqrt(3)    -sqrt(3)/3      -2       2 sqrt(3)/3
3/4   sqrt(2)/2  -sqrt(2)/2     -1          -1        -sqrt(2)      sqrt(2)
5/6      1/2     -sqrt(3)/2  -sqrt(3)/3  -sqrt(3)    -2 sqrt(3)/3      2
          0         -1           0          ---          -1           ---

More values at special angles:
  x              /10                  /5
sin(x)      (-1+sqrt[5])/4      sqrt(10-2 sqrt[5])/4
cos(x)   sqrt(10+2 sqrt[5])/4      (1+sqrt[5])/4
tan(x)    sqrt(1-2/sqrt[5])      sqrt(5-2 sqrt[5])
cot(x)    sqrt(5+2 sqrt[5])      sqrt(1+2/sqrt[5])
sec(x)    sqrt(2-2/sqrt[5])         -1+sqrt[5]
csc(x)        1+sqrt[5]          sqrt(2+2/sqrt[5])

Use the above values and the identities below to obtain values of trigonometric functions of the following multiples of /10:
    3/10 = /2 - /5,
    2/5  = /2 - /10,
    3/5  = /2 + /10,
    7/10 = /2 + /5,
    4/5  =    - /5,
    9/10 =    - /10.
    



Bounds

   |sin(x)| <= 1,
   |cos(x)| <= 1,
   |sec(x)| >= 1,
   |csc(x)| >= 1.



Identities

   sec(x) = 1/cos(x),
   csc(x) = 1/sin(x),
   cot(x) = 1/tan(x),
   tan(x) = sin(x)/cos(x),
   cot(x) = cos(x)/sin(x).

   sin(-x) = -sin(x),
   cos(-x) = cos(x),
   tan(-x) = -tan(x),
   cot(-x) = -cot(x),
   sec(-x) = sec(x),
   csc(-x) = -csc(x).

   sin(/2-x) = cos(x),
   cos(/2-x) = sin(x),
   tan(/2-x) = cot(x),
   cot(/2-x) = tan(x),
   sec(/2-x) = csc(x),
   csc(/2-x) = sec(x).

   sin(/2+x) = cos(x),
   cos(/2+x) = -sin(x),
   tan(/2+x) = -cot(x),
   cot(/2+x) = -tan(x),
   sec(/2+x) = -csc(x),
   csc(/2+x) = sec(x).

   sin(-x) = sin(x),
   cos(-x) = -cos(x),
   tan(-x) = -tan(x),
   cot(-x) = -cot(x),
   sec(-x) = -sec(x),
   csc(-x) = csc(x).

   sin(+x) = -sin(x),
   cos(+x) = -cos(x),
   tan(+x) = tan(x),
   cot(+x) = cot(x),
   sec(+x) = -sec(x),
   csc(+x) = -csc(x).

   sin(2+x) = sin(x),
   cos(2+x) = cos(x),
   tan(2+x) = tan(x),
   cot(2+x) = cot(x),
   sec(2+x) = sec(x),
   csc(2+x) = csc(x).

   sin2(x) + cos2(x) = 1,
   tan2(x) + 1 = sec2(x),
   1 + cot2(x) = csc2(x).

   sin(x+y) = sin(x)cos(y) + cos(x)sin(y),
   cos(x+y) = cos(x)cos(y) - sin(x)sin(y),
   tan(x+y) = [tan(x)+tan(y)]/[1-tan(x)tan(y)],
   cot(x+y) = [cot(x)cot(y)-1]/[cot(x)+cot(y)].

   sin(x-y) = sin(x)cos(y) - cos(x)sin(y),
   cos(x-y) = cos(x)cos(y) + sin(x)sin(y),
   tan(x-y) = [tan(x)-tan(y)]/[1+tan(x)tan(y)],
   cot(x-y) = [cot(x)cot(y)+1]/[cot(y)-cot(x)].

   sin(2x) = 2 sin(x)cos(x),
   cos(2x) = cos2(x) - sin2(x),
            = 2 cos2(x) - 1,
            = 1 - 2 sin2(x),
   tan(2x) = [2 tan(x)]/[1-tan2(x)],
   cot(2x) = [cot2(x)-1]/[2 cot(x)].


   |sin(x/2)| = sqrt([1-cos(x)]/2),
   
   |cos(x/2)| = sqrt([1+cos(x)]/2),
   
   |tan(x/2)| = sqrt([1-cos(x)]/[1+cos(x)]),
   
   tan(x/2) = [1-cos(x)]/sin(x),
            = sin(x)/[1+cos(x)].
            

   sin(3x) = 3 sin(x) - 4 sin3(x),
   cos(3x) = 4 cos3(x) - 3 cos(x),
   tan(3x) = [3 tan(x)-tan3(x)]/[1-3 tan2(x)].

   sin(4x) = 4 sin(x)cos(x)[2 cos2(x)-1],
   cos(4x) = 8 cos4(x) - 8 cos2(x) + 1.

   sin(5x) = 5 sin(x) - 20 sin3(x) + 16 sin5(x),
   cos(5x) = 16 cos5(x) - 20 cos3(x) + 5 cos(x).

   sin(6x) = 2 sin(x)cos(x)[16 cos4(x) - 16 cos2(x) + 3],
   cos(6x) = 32 cos6(x) - 48 cos4(x) + 18 cos2(x) - 1.

   sin(nx) = 2 sin([n-1]x)cos(x) - sin([n-2]x),
   cos(nx) = 2 cos([n-1]x)cos(x) - cos([n-2]x),
   tan(nx) = (tan[(n-1)x]+tan[x])/(1-tan[(n-1)x]tan[x]).

   sin(x)cos(y) = [sin(x+y) + sin(x-y)]/2,
   cos(x)sin(y) = [sin(x+y) - sin(x-y)]/2,
   cos(x)cos(y) = [cos(x-y) + cos(x+y)]/2,
   sin(x)sin(y) = [cos(x-y) - cos(x+y)]/2.

   sin(x) + sin(y) = 2 sin[(x+y)/2]cos[(x-y)/2],
   sin(x) - sin(y) = 2 cos[(x+y)/2]sin[(x-y)/2],
   cos(x) + cos(y) = 2 cos[(x+y)/2]cos[(x-y)/2],
   cos(x) - cos(y) = -2 sin[(x+y)/2]sin[(x-y)/2],
   tan(x) + tan(y) = sin(x+y)/[cos(x)cos(y)],
   tan(x) - tan(y) = sin(x-y)/[cos(x)cos(y)],
   cot(x) + cot(y) = sin(x+y)/[sin(x)sin(y)],
   cot(x) - cot(y) = -sin(x-y)/[sin(x)sin(y)].

   [sin(x)+sin(y)]/[cos(x)+cos(y)] = tan[(x+y)/2],
   [sin(x)-sin(y)]/[cos(x)+cos(y)] = tan[(x-y)/2],
   [sin(x)+sin(y)]/[cos(x)-cos(y)] = -cot[(x-y)/2],
   [sin(x)-sin(y)]/[cos(x)-cos(y)] = -cot[(x+y)/2],
   [sin(x)+sin(y)]/[sin(x)-sin(y)] = tan[(x+y)/2]/tan[(x-y)/2].

   sin2(x) - sin2(y) = sin(x+y)sin(x-y),
   cos2(x) - cos2(y) = -sin(x+y)sin(x-y),
   cos2(x) - sin2(y) = cos(x+y)cos(x-y).
   
   sin2(x) = (1 - cos[2x])/2,
   cos2(x) = (1 + cos[2x])/2,
   tan2(x) = (1 - cos[2x])/(1 + cos[2x]),

   sin3(x) = (3 sin[x] - sin[3x])/4,
   cos3(x) = (3 cos[x] + cos[3x])/4,

   sin4(x) = (3 - 4 cos[2x] + cos[4x])/8,
   cos4(x) = (3 + 4 cos[2x] + cos[4x])/8,

   sin5(x) = (10 sin[x] - 5 sin[3x] + sin[5x])/16,
   cos5(x) = (10 cos[x] + 5 cos[3x] + cos[5x])/16,

   sin6(x) = (10 - 15 cos[2x] + 6 cos[4x] - cos[6x])/32,
   cos6(x) = (10 + 15 cos[2x] + 6 cos[4x] + cos[6x])/32,



Relations in Right Triangles

In the right triangle ABC with right angle C = /2,
   A + B = /2,
   c2 = a2 + b2,
   sin(A) = cos(B) = a/c,
   cos(A) = sin(B) = b/c,
   tan(A) = cot(B) = a/b,
   cot(A) = tan(B) = b/a,
   sec(A) = csc(B) = c/b,
   csc(A) = sec(B) = c/a,
   ha = b,
   hb = a,
   hc = ab/c.

Solving Right Triangles

Case I: You are given a and A.
B = /2 - A, c = a csc(A), b = a cot(A).
Case II: You are given a and B.
A = /2 - B, c = a sec(B), b = a tan(B).
Case III: You are given c and A.
B = /2 - A, a = c sin(A), b = c cos(A).
Case IV: You are given a and b.
tan(A) = a/b, B = /2 - A, c = a csc(A).
Case V: You are given a and c.
sin(A) = a/c, B = /2 - A, b = a cot(A).



Relations in Oblique Triangles

   A + B + C = ,
   s = (a+b+c)/2, half the perimeter,
   r = radius of inscribed circle,
   R = radius of circumscribed circle,
   K = area.
The Law of Sines:
   a/sin(A) = b/sin(B) = c/sin(C) = 2R.
This implies that a <= b <= c if and only if A <= B <= C.
The Law of Cosines:
   a2 = b2 + c2 - 2bc cos(A),
   b2 = c2 + a2 - 2ca cos(B),
   c2 = a2 + b2 - 2ab cos(C).
The Law of Tangents:
   (a+b)/(a-b) = tan[(A+B)/2]/tan[(A-B)/2],
   (b+c)/(b-c) = tan[(B+C)/2]/tan[(B-C)/2],
   (c+a)/(c-a) = tan[(C+A)/2]/tan[(C-A)/2].
Newton's Formulae:
   (a+b)/c = cos[(A-B)/2]/sin(C/2),
   (b+c)/a = cos[(B-C)/2]/sin(A/2),
   (c+a)/b = cos[(C-A)/2]/sin(B/2).
Mollweide's Equations:
   (a-b)/c = sin[(A-B)/2]/cos(C/2),
   (b-c)/a = sin[(B-C)/2]/cos(A/2),
   (c-a)/b = sin[(C-A)/2]/cos(B/2).
Other relations:
   a = b cos(C) + c cos(B),
   b = c cos(A) + a cos(C),
   c = a cos(B) + b cos(A).

   tan[(A-B)/2] = [(a-b)/(a+b)]cot(C/2),
   tan[(B-C)/2] = [(b-c)/(b+c)]cot(A/2),
   tan[(C-A)/2] = [(c-a)/(c+a)]cot(B/2).

   sin(A) = 2K/(bc),
   sin(B) = 2K/(ca),
   sin(C) = 2K/(ab).

   K = sr = sqrt[s(s-a)(s-b)(s-c)],
   K = aha/2 = bhb/2 = chc/2,
   K = ab sin(C)/2 = bc sin(A)/2 = ca sin(B)/2,
   K = a2 sin(B)sin(C)/[2 sin(A)],
     = b2 sin(C)sin(A)/[2 sin(B)],
     = c2 sin(A)sin(B)/[2 sin(C)].

   R = abc/(4K) = a/[2 sin(A)] = b/[2 sin(B)] = c/[2 sin(C)],
   r = K/s,
     = sqrt[(s-a)(s-b)(s-c)/s],
     = c sin(A/2)sin(B/2)/cos(C/2),
     = ab sin(C)/(2s),
     = (s-c)tan(C/2).

   sin(A/2) = sqrt[(s-b)(s-c)/(bc)],
   sin(B/2) = sqrt[(s-c)(s-a)/(ca)],
   sin(C/2) = sqrt[(s-a)(s-b)/(ab)].

   cos(A/2) = sqrt[s(s-a)/(bc)],
   cos(B/2) = sqrt[s(s-b)/(ca)],
   cos(C/2) = sqrt[s(s-c)/(ab)].

   tan(A/2) = sqrt[(s-b)(s-c)/{s(s-a)}] = r/(s-a),
   tan(B/2) = sqrt[(s-c)(s-a)/{s(s-b)}] = r/(s-b),
   tan(C/2) = sqrt[(s-a)(s-b)/{s(s-c)}] = r/(s-c).

   (a+b)/(a-b) = [sin(A)+sin(B)]/[sin(A)-sin(B)] = cot(C/2)/tan[(A-B)/2],
   (b+c)/(b-c) = [sin(B)+sin(C)]/[sin(B)-sin(C)] = cot(A/2)/tan[(B-C)/2],
   (c+a)/(c-a) = [sin(C)+sin(A)]/[sin(C)-sin(A)] = cot(B/2)/tan[(C-A)/2].

   ha = a sin(B)sin(C)/sin(B+C) = a/[cot(B)+cot(C)] = b sin(C) = c sin(B),
   hb = b sin(C)sin(A)/sin(C+A) = b/[cot(C)+cot(A)] = c sin(A) = a sin(C),
   hc = c sin(A)sin(B)/sin(A+B) = c/[cot(A)+cot(B)] = a sin(B) = b sin(A).
   
   cos(A) + cos(B) + cos(C) = 1 + r/R.



Solving Oblique Triangles

Case I: You are given any two angles and one side c.
The third angle is determined from A + B + C = . Now the Law of Sines can be used to find b = c sin(B)/sin(C) and a = c sin(A)/sin(C).
Case II: You are given two sides and the angle opposite one of them, say a, c, and A.
Subcase A: a < c sin(A). There is no solution.
Subcase B: a = c sin(A). There is one solution:
C = /2, B = /2 - A, b = c cos(A).
Subcase C: c > a > c sin(A). There are two solutions. Use the Law of Sines to find sin(C) = c sin(A)/a < 1. There are two angles C and C' =  - C having that sine, one acute and one obtuse. Then compute B =  - A - C and B' =  - A - C'. Now use the Law of Sines again to find b = a sin(B)/sin(A) and b' = a sin(B')/sin(A). The solutions are (a,b,c,A,B,C) and (a,b',c,A,B',C').
Subcase D: a >= c. There is one solution. Use the Law of Sines to find sin(C) = c sin(A)/a <= 1. Then angle C must be acute, so it can be found uniquely from sin(C). Then compute B =  - A - C. Now use the Law of Sines again to find b = a sin(B)/sin(A).
Case III: You are given two sides and the included angle, say a, b, and C.
You can compute the third side c by using the Law of Cosines. Then the Law of Sines can be used to find the sines of the other two angles sin(A) = a sin(C)/c andsin(B) = b sin(C)/c. The angles opposite the two shortest sides are then acute, and uniquely determined from their sines, and the third, largest angle is found fromA + B + C = .
Alternatively, you can use the Law of Tangents. You know that (A+B)/2 = (-C)/2,which is easily computable. Then by the Law of Tangents,tan[(A-B)/2] = cot(C/2) (a-b)/(a+b), so you can find (A-B)/2 uniquely. ThenA = (-C)/2 + (A-B)/2, and B = (-C)/2 - (A-B)/2. Then c = a sin(C)/sin(A).
Case IV: You are given all three sides.
You can use the Law of Cosines to find A, then use the Law of Sines to computesin(B) = b sin(A)/a and sin(C) = c sin(A)/a.
Alternatively, you can find r = sqrt[(s-a)(s-b)(s-c)/s], and use tan(A/2) = r/(s-a) to find A/2, and hence A, and similarly for B and C.
Alternatively, you can use sin(A/2) = sqrt[(s-b)(s-c)/(bc)] to find A/2 (since A/2 < /2), and hence A, and similarly for B and C.
In any case, the results can be checked by using Mollweide's Equations.


Inverse Trigonometric Functions

   x = Arcsin(y)  ==>  y = sin(x),  -/2 <= x <= /2,
   x = Arccos(y)  ==>  y = cos(x),     0 <= x <= ,
   x = Arctan(y)  ==>  y = tan(x),  -/2 <  x <  /2,
   x = Arccot(y)  ==>  y = cot(x),     0 <  x <  ,
   x = Arcsec(y)  ==>  y = sec(x),     0 <  x <  ,
   x = Arccsc(y)  ==>  y = csc(x),  -/2 <  x <  /2.

   y = sin(x)  ==>  x = Arcsin(y) + 2n  or   - Arcsin(y) + 2n,
   y = cos(x)  ==>  x = Arccos(y) + 2n  or  -Arccos(y)    + 2n,
   y = tan(x)  ==>  x = Arctan(y) +  n,
   y = cot(x)  ==>  x = Arccot(y) +  n,
   y = sec(x)  ==>  x = Arcsec(y) + 2n  or  -Arcsec(y)    + 2n,
   y = csc(x)  ==>  x = Arccsc(y) + 2n  or   - Arccsc(y) + 2n,
where n is an arbitrary integer.
   Arcsin(y) + Arccos(y) = /2,
   Arctan(y) + Arccot(y) = /2,
   Arcsec(y) + Arccsc(y) = /2.

   Arcsin(y) = Arccsc(1/y),
   Arccos(y) = Arcsec(1/y),
   Arctan(y) = Arccot(1/y),  y > 0
   Arccot(y) = Arctan(1/y),  y > 0
   Arcsec(y) = Arccos(1/y),
   Arccsc(y) = Arcsin(1/y).

   Arcsin[sin(x)] = (-1)Floor[1/2 - x/] (x +  Floor[1/2 - x/]),
   Arccos[cos(x)] = /2 - (-1)Floor[-x/] (x + /2 +  Floor[-x/])
   Arctan[tan(x)] = x - Floor[x/ + 1/2] .

   sin[Arcsin(y)] = cos[Arccos(y)] = y,
   cos[Arcsin(y)] = sin[Arccos(y)] = sqrt[1-y2],
   tan[Arcsin(y)] = cot[Arccos(y)] = y/sqrt[1-y2],
   cot[Arcsin(y)] = tan[Arccos(y)] = sqrt[1-y2]/y,
   sec[Arcsin(y)] = csc[Arccos(y)] = 1/sqrt[1-y2],
   csc[Arcsin(y)] = sec[Arccos(y)] = 1/y,
   sin[Arctan(y)] = cos[Arccot(y)] = y/sqrt[1+y2],
   cos[Arctan(y)] = sin[Arccot(y)] = 1/sqrt[1+y2],
   tan[Arctan(y)] = cot[Arccot(y)] = y,
   cot[Arctan(y)] = tan[Arccot(y)] = 1/y,
   sec[Arctan(y)] = csc[Arccot(y)] = sqrt[1+y2],
   csc[Arctan(y)] = sec[Arccot(y)] = sqrt[1+y2]/y,
   sin[Arcsec(y)] = cos[Arccsc(y)] = sqrt[y2-1]/y,
   cos[Arcsec(y)] = sin[Arccsc(y)] = 1/y,
   tan[Arcsec(y)] = cot[Arccsc(y)] = sqrt[y2-1],
   cot[Arcsec(y)] = tan[Arccsc(y)] = 1/sqrt[y2-1],
   sec[Arcsec(y)] = csc[Arccsc(y)] = y,
   csc[Arcsec(y)] = sec[Arccsc(y)] = y/sqrt[y2-1].

   sin[2 Arcsin(y)] = 2y sqrt[1-y2],
   cos[2 Arccos(y)] = 2y2 - 1,
   tan[2 Arctan(y)] = 2y/(1-y2).

Lesson Note On Cylindrical Coordinates

Cylindrical Coordinates
Cylindrical coordinates are obtained by using polar coordinates in a plane, and then adding a z-axis perpendicular to the plane passing through the pole. This gives three coordinates (r,theta,z) for any point.To transform from rectangular Cartesian coordinates (x,y,z) to cylindrical ones and back, use the following formulas:

x = r cos(theta),
y = r sin(theta),
z = z,

r = ±sqrt(x2+y2),
theta = arctan(y/x),
z = z.
The sign of r is determined by which of the values of the arctangent function is chosen:

  Sign of x    Sign of y    Quadrant of theta    Sign of r  
++I+
++III
+II+
+IV
I
III+
+II
+IV+
The quadrant of theta can always be chosen to make r positive, if it is so desired.



Common Uses

The most common use of cylindrical coordinates is to give the equation of a surface of revolution. If the z-axis is taken as the axis of revolution, then the equation will not involve theta at all.
Examples:

  • A paraboloid of revolution might have equation
    z = r2.
    This is the surface you would get by rotating the parabola z = x2 in the xz-plane about the z-axis. The cartesian coordinate equation of the paraboloid of revolution would be z = x2 + y2.
  • A right circular cylinder of radius a whose axis is the z-axis has equation
    r = R.
  • A a sphere with center at the origin and radius R will have equation
    r2 + z2 = R2.
  • A right circular cone with vertex at the origin and axis the z-axis has equation
    z = m r.
As another kind of example, a helix has the following equations:
r = R,
z = a theta.
Sometimes a change from rectangular to cylindrical coordinates makes computing difficult multiple integrals simpler.


Compiled by Robert L. Ward.

Monday 27 June 2016

How To Write A Business Proposal Template

Step : Write The Proposal
Once you have identified the information necessary to answer your client’s questions, you need to organize this information for maximum impact and clarity. Here is a standard business plan template which you can adjust to fit your specific needs.
How To Write A Business Proposal Template 
Section 1. Introduction: Start by introducing your company and mission in a way that relates your company to your prospective client’s needs. You can include a brief story that gives your client a feel for you brand’s character and helps build trust. Highlight what distinguishes your company, your accomplishments, credentials, and any awards.  This should be no more than 1 page.
Section 2. Executive Summary: The Executive Summary is one of the most important sections in your proposal. This is where you should present the case for why you are the right company for the job, and give the reader the takeaway message of the proposal. You should not try to summarize every aspect of the proposal, but rather focus on the conclusions you want the reader to reach after reading it. Use direct, factual language that is objective and persuasive, and make sure that it feels like it is about your potential client, not just your company. Show that you are familiar with their company, and can meet their specific needs.  This section should be also be kept to 1 page.
Section 3. Table of Contents: The Table of Contents is an optional section that is helpful for longer proposals with lots of details.
Section 4. Body: Once you have presented your overall case in the Executive Summary, you can outline the specifics of your proposal. This is where you can answer the “who, what, when, where, how, and why” questions that you identified in step 2. Include information on scheduling, logistics, and pricing. You can also include testimonials from past clients and a link to your website.
Section 5. Conclusion: Once you have outlined the details of your proposal re-emphasize the exceptional results your company can provide. You should conclude with a call to action that encourages the reader to contact you, or visit your website for more information. Ideally, you want your client to make an immediate connection with your company, even if it is something small.
Section 6: Appendix: The Appendix is an optional section that you can use to include information that might not fit well in the body of your proposal. You can include resumes, or additional graphs, projections, and customer testimonials.
You should strive to strike a balance that provides enough information to answer potential questions, but keeps the proposal lean and free of superfluous information that will only confuse your client.

The Tone And Language You Should Use

Make sure you use clear concise and simple language that avoids lots of industry jargon and technical terms. Also, avoid using hyperbole that exaggerates your company or service, as this may undermine the trust you are trying to establish with your potential client. Focus on accurately and confidently presenting the results your company can offer. You can include some of your methodology, but this should be a supporting factor, not the focus of the proposal. Remember, your client cares most about the end result and the financial costs.

How Long Should The Proposal Be?

If possible, keep the total length of the proposal within an 8-minute read in order to capitalize on the short attention span of many readers. Remember, your client may be reading many proposals and will not necessarily appreciate a proposal stuffed with superfluous information. If you want to add more information that is not directly relevant to the job, put it in the appendix.
While there is no set format, the formatting should be clear and easy to understand. Break the proposal into clear sections and subsections that use effective headings and subheadings.

Be Sure To Proofread and Edit

Once you are done writing, make sure to carefully proofread and edit your proposal.  After you’re done, wait a day or two and the double check your proposal to verify that you’ve included all the information your client might need. If your potential client sent a Request for Proposal, go back and make sure that your proposal provides all of the information requested. It should be easy for your client to find information within your proposal, so if you included a table of contents make sure that all of the headings and subheadings are included with accurate page numbers.

Always Keep The Client In Mind

The most important thing is to try and think like your client. If you can put yourself in their shoes, you will be better able to explain why your company is best for the job, and anticipate all the questions they may have. If you follow this guide, you will be well on your way to a great job-winning proposal.


DCP pavement design and charts


DCP pavement design and charts


Introduction
This is a very simplified introduction to the use of the DCP. The DCP can be used at all levels from simple investigation to very sophisticated design. It can be integrated with FWD testing, linear elastic layer design (CIRCLY, ELSYM, LED). It’s a great tool for catalogue design (TRH 4, AustRoads). I know the South Africa and Australian design systems well, the British and American design systems reasonably well, and the DCP adds value to all of them. Great value. The DCP can be used on roads (of all types includes cemented layers - just drill through the cement), and runways (up to and included PCN 100 pavements for 747s). As well as carparks, container terminals, building foundations. About the only thing that it doesn't do well is open coconuts. I was on Christmas Island, and tried to open a freshly fallen coconut with the DCP - coconut stayed intact and I broke the DCP.

The following words are intentionally simple so they can be read by people who have English as second language. The DCP can be described in very difficult and complex language just as easily, so do not be fooled by the simple words.

The DCP results from the analysis described below can be used to divide the runway into uniform sections, get design subgrade strengths and thus feed into FAA design charts, or to estimate elastic moduli and feed into the various (APSDS, LED) computer elastic layer design programmes for runway thickness. They can also be used to help understand FWD and HWD test results.

Above all, I find the DCP is absolutely marvellous for sorting out the confusion that arises when all the various test results are put together. Usually nothing makes sense, and one test shows a section is about to fail, while another test shows this is the strongest section. The DCP is the "honest" tool to help sort out the mess.


DCP and pavement investigation
The DCP is a most useful tool for pavement investigation because it can 'look inside' a road, a runway or any other pavement. It shows field bearing strength which is the most important material property for roads.

Gradings, PI, and Atterberg limits are other tests used, but these are indirect or indicator tests and are less relevant than bearing strength. Because it is a quick, inexpensive pavement test, and because it relates very directly to the performance of a surfacing, DCP testing is often the main one suggested for use.



Effect of moisture on DCP and CBR results
Most road/runway materials become weaker if they become soaking wet, particularly clayey materials. In the field, the materials in the road/runway are usually drier than soaking wet. The DCP measures field CBR, and this is probably higher than a laboratory soaked CBR test. So a dry clayey gravel may have a good DCP-CBR (over 80), while its soaked CBR could be 45. The effect is most pronounced in the basecourse, because the lower layers are generally wetter than the basecourse.

Most of the time, this moisture effect does not matter much because the bitumen surfacing keeps the road/runway dry inside, even when it is raining.  But if water does get into the road/runway from underneath (if it is a wet area), or from on top (if the surfacing is cracked or potholed), it could matter.

Climate has an effect as well - in a wet climate (where the grass is green all year round, there are lots of lovely trees, rivers, ponds, etc), then the materials tend to be wetter. The DCP tests should be done in the wet season when the materials are at their weakest. The DCP tests should be compared to laboratory soaked CBR test results to see what the effect of soaking is. In a dry country, this is less important.

There are two three methods to estimate what the material will be like when it is soaked:
take a sample, and do a laboratory soaked CBR test, or
check the Plasticity Index (PI; from the Atterberg Limit tests). Low PI materials (PI < 6) will not weaken too much when wet, so field DCP is a good measure of soaked CBR. Higher PI materials will weaken when wet, so field DCP could be deceptively high.
See the appendix to his document


DCP test frequency
DCP tests should be done at the following frequency:
Upgrading a road/runway 2-4 DCP tests per kilometre, with the tests staggered as outer wheeltrack/inner wheeltrack one side, outer wheeltrack/inner wheeltrack other side, centreline, etc; at least 8 DCP tests per likely uniform section are needed to provide adequate data for the analysis.
Failure investigation  2-4 DCP tests in the failed section, and the same in a nearby unfailed section.

It is useful to take at least 2 samples per kilometre for new roads, or 2 samples per section for failure investigations, to check laboratory soaked CBR, Atterbergs and insitu moisture content of each layer.  Also test the insitu basecourse density (optional for new roads, needed for failure investigations). Local soils laboratories can do all these soil tests.

In any event, never come back from the field without a heap of DCP tests been done. They are quick to do, and even half a day will give you 16 tests.


DCP test method
Assemble the DCP.  Place the DCP point on the surface, hold the DCP upright and start. Lift the hammer to the top of its travel, and release; do not throw it down, but let it drop. Record the depth of penetration every 5 blows. Continue to below 800mm below the surfacing.  Hitting a stone/rock will show as a horizontal line for 5-25 blows. When this happens during testing, just keep hammering with the DCP and usually it will break through.

Plot the depth versus blows directly on the sheet. Direct plotting guards against minor errors in reading.  Since the sheet only reads 50 blows across, transfer the depth at 50 blows across to the left hand side of the sheet and keep recording down


Drawing the DCP lines
Essentially, join up the dots on the DCP sheet to show continuous lines. It is helpful to draw this as only a few straight lines of "best-fit". The CBR at each depth can be read off the chart directly. Since most pavement layers are 150mm thick, it is usually possible to interpret the DCP line into actual pavement layers.


Use of DCP computer programmes
The use of DCP computer programmes is not recommended except by extra highly qualified and experienced pavement engineers. All current programmes necessarily have built-in assumptions, and these can be changed by experienced users. However if left unchanged, the programme will still operate and can make gross mistakes. The hand-drawn DCP charts are much safer to use. I only use hand drawn charts today, even though I have a number of the computer programmes available.


Pavement strength analysis

The DCP results can be used for determining uniform pavement sections, design subgrade strength, design strength of unbound (unstabilised) sub-bases and basecourses. It can even be used for quality control during construction of non-standard materials. The following steps give some illustration of how some designers use them.


STEP 1 DO DCP TESTING ALONG THE ROAD/RUNWAY
DCP testing is performed along the length of road/runway.  The frequency of tests should generally be in accordance with the standards here, but the visual inspection may indicate adjustments to the frequency.  If the road/runway is very uniform the frequency can be reduced, and if it is variable then it should be increased.  The basic frequency should be:

• test at the rate of 5 DCP tests per kilometre, with the tests staggered as centreline, outer wheeltrack one side, outer wheeltrack other side, centreline, etc;
• perform an additional test at every significant location picked up in the visual survey, such as particular failure areas;
• ensure that at least 8 DCP tests are performed per likely uniform section to provide adequate data for the statistical analysis.


STEP 2 DIVIDE ROAD/RUNWAY INTO UNIFORM SECTIONS
The results of the investigation, including the DCP testing and visual assessment, enable the length of road/runway to be divided in relatively uniform sections for the purposes of rehabilitation.

The minimum length of section should be 100 metres, and desirably 1000 metres. On long lengths of road with uniform conditions, it can be 10 000 metres.  Note that construction of sections shorter than 500 metres is awkward.  It may be that a low DCP result occurs in a spot which was identified in the visual survey as an isolated problem area; these are typical of an isolated drainage problem and consideration should be given to repair of these individually rather than taking them as representative of the section.

For runways with heavy aircraft, the design pavement thickness can be quite a lot, especially if the subgrade is weak. Very often on existing runways, the subgrade strength (and depth of existing pavement layers) is amazingly variable.  It is even more variable if you test outside the central 15 metres!  It is not unknown to have outer edges of runways with only half the strength of the centre of the runway - even though they have been built to the same thickness. The trafficking and compaction in the centre make it so much stronger there. So the DCP can be used to delineate the weakest areas. Then selective reconstruction can be used to fix those areas.

As an example, at Broome International Airport the 10/28 runway PCN was 28. Parts of the runway were even weaker than that and had failed. However by reconstructing 300m of the 2000m, the weakest areas and failures were replaced. The whole runway could be re-rated to PCN 35. Then an thin asphalt overlay could be used to bring the PCN up to 45.  The alternative was a thick asphalt overlay over the whole lot - the weak, failed and good areas of the runway. The cost savings amounted to 70% of the cost of the thick overlay.


STEP 3 CALCULATE THE DESIGN CBR VALUE FOR EACH LAYER IN EACH SECTION
The design CBR is found for each layer (such as subgrade, subbase or basecourse) in each section and calculated statistically to provide a safety margin against the variability of material within the section.  A normal distribution of data is assumed and the Student's T distribution at the 80% level is used:

First the design DCP is found:

Design DCP = average DCP  - .9 * (standard deviation)

Example   The DCP/CBR results of a basecourse in a section were as follows:
DCP/CBR:  125,  143,  120,  100,  145,  115,  140,  135
Mean (average) = 127,9       Standard deviation = 15,7
DCP = average DCP  - .9 * (standard deviation)
= 127,9 - .9 * 15,7   = 114
Note that the equation is using a one-tailed T-distribution for 8 samples and is reasonably robust for sample sizes from 5 to 30.

Then the design CBR is found by considering the relationship between CBR and DCP.  This is rather too complex to cover here. Instead, let me say what some designers do. In Australia, for drier areas, the DCP is taken as equal to the CBR for the subgrade and sometimes for the upper layers.  Easy, for dry areas.

In South Africa, they'll do a CBR test in the laboratory at soaked and at unsoaked conditions, and see how the strength varies with moisture. Then they'll consider how wet the inside of the pavement will get, before they pick the relationship between DCP and CBR. Often, a simple rule of thumb for areas in moderate climates is to reduce the DCP by 20% - so a design DCP of 80 becomes equivalent to CBR 60. The answer also depends on the assumptions of the design method being used, whether they are based on soaked or insitu strength, and how that has been handled in determining elastic moduli, etc. No easy answer there, sorry. But at least it keeps me in consulting work !



APPENDIX

Extract from:
THE PREDICTION OF MOISTURE CONTENT IN UNTREATED PAVEMENT LAYERS AND AN APPLICATION TO DESIGN IN SOUTHERN AFRICA
Division of Roads and Transport Technology, Bulletin 20, CSIR Research Report 644, Pretoria, 1988
Author: S. J. Emery

Table 22  Variation of CBR with moisture content

Material class    Soaked                  mc/OMCm ratio
(TRH 14)          CBR (%)         1,0           0,75        0,5
                                  Ratio of unsoaked to soaked CBR
G4                  80            1,34          1,88        2,5
G5                  45            1,77          2,57        3,6
G6                  25            2,35          3,56        5,4
G7                  15            3,00          4,71        7,6
G8                  10            3,65          5,88        9,9
G9                  7             4,33          7,16        12
G10                 3             6,50          11,41       22

Notes
1. Material class is South African material class - other countries can use the Soaked CBR column to identify their materials.
2. mc/OMCm ratio is the ratio of field moisture content (mc) to optimum moisture content (OMC at Mod AASHTO compaction). Thus a material, which has been prepared in the laboratory for an unsoaked CBR test, is at OMC because it has been compacted at OMC, and so the mc/OMCm ratio is 1.00.  A material in the field with a field m.c of 6% and an OMC of 8% has an mc/OMCm ratio of 0,75.
3. So a material with a soaked CBR of 45 can be expected to have an unsoaked CBR (at OMC) strength of 45 * 1,77 = 80. The same material, in the field, dried back to say 6% with an OMC of 8% which means a ratio mc/OMC of 0.75, would have a field CBR of 45 * 2,57 = 115.
4. These relationships are approximate and variable, and they also depend on particle size distribution, soil suction and the amount of clay in the material, none of which are explicitly addressed in the modelling. However they give an indication of the strength gain with drying out.

Great Tips for Success in Both Life and Business

Great Tips for Success in Both Life and Business

Sometimes what helps us to be successful in our professional lives is not such a great idea in our personal lives — competition is a quality that comes to mind. At the same time, we all have a limited amount of time each day to do the things that we want to do.
So for the sake of saving time and energy, I’m sharing a list of tips that will help you be successful in both life and in business.

Learn how to balance life:“There is an immutable conflict at work in life and in business, a constant battle between peace and chaos. Neither can be mastered, but both can be influenced. How you go about that is the key to success.”

Do not be afraid of failure. “Failure is simply the opportunity to begin again, this time more intelligently.”

Have an unwavering resolution to succeed.
“I made a resolve then that I was going to amount to something if I could. And no hours, nor amount of labor, nor amount of money would deter me from giving the best that there was in me. And I have done that ever since, and I win by it. I know.”

Add Value

No matter what you do and where you go, you can’t go wrong with adding value. Simply put value is anything that people are willing to pay for. In your professional life, the more value you can offer the more money you can make. In your personal life, more value translates to closer relationships and strong personal growth. The best way to add value is to find the intersection between what people are willing to pay for and what service or product you can offer that is aligned with your values, strengths and goals.
How are you adding value to your employers and loved ones today? What can you do to increase your ability to add value?

Follow Your Passion

Reading numerous biographies on great people and from my own personal observations and encounters, I’ve realized that those who achieve greatness professional and personally follow their passion. The reason why great people are few and far in-between is because most people don’t even know what their passion is. For those that do figure out their passion, most of them don’t follow their passion consistently. This is one of the main reasons why people don’t reach their goals.
Do you know what your passion is? If not, what are you going to do to find out? If  you do know what you passion is, are you following it?

Be a man of action. “It had long since come to my attention that people of accomplishment rarely sat back and let things happen to them. They went out and happened to things.”

Avoid conflicts.
“The most important single ingredient in the formula of success is knowing how to get along with people.”

Don’t be afraid of introducing new ideas.

 “A person with a new idea is a crank until the idea succeeds.”  

Believe in your capacity to succeed.

“If you can dream it, you can do it.”

Always maintain a positive mental attitude

“Nothing can stop the man with the right mental attitude from achieving his goal; nothing on earth can help the man with the wrong mental attitude.”

Don’t let discouragement stop you from pressing on.

“Let no feeling of discouragement prey upon you, and in the end you are sure to succeed.”

Be willing to work hard.

“Unless you are willing to drench yourself in your work beyond the capacity of the average man, you are just not cut out for positions at the top.”

Be brave enough to follow your intuition.

“Have the courage to follow your heart and intuition. They somehow already know what you truly want to become. Everything else is secondary.”
Set a goal for yourself
"Failure is unavoidable in our lives. But if I can learn something from it every time, the chance of success will only increase."

Be Extraordinary

If you do the same thing as everyone else, it’s hard to be successful. It is important to find the edge and then push past it. That is how you become noticed and get what you want. Whether it is money, meaningful relationships and/or a sense of personal accomplishment, the extraordinary person attracts them all.
How are you extraordinary?  If you feel just ordinary, what are you going to do to become extraordinary? For those who don’t know, you may want to check out articles on my blog and also How to go from Ordinary to Extraordinary.


Volumes

Volumes
Volume calculations for rectangular prism and pyramid are shown below:

A truncated pyramid is a pyramid which top has been cut off.
If the A1+A2 is almost equal in size then the following formula can be used instead:
V = h × (A1 + A2) / 2


A prismoid is as a solid whose end faces lie in parallel planes and consist of any two polygons, not necessarily of the same number of sides as shown opposite, the longitudinal faces may take the form of triangles, parallelograms, or trapeziums.


Lesson Notes On Areas

Areas
Area calculations refer usually to rectangular and triangular shapes. If you need the trigonometric function for calculations click here.
There are different ways to calculate the area of the opposite figure. Try to minimise the amount of calculation. The figure could be divided in three distinct areas
a=10.31x5.63+
b=6.25x5.76+
c=10.39x4.79

or the whole rectangle minus the hole (d)
A =16.67x10.31-6.25x4.55.
As you can see the 2nd method is easier. Look at the shape and try to shorten the calculations.
If you know only the sides of a triangle then use the formula given in the figure below.

An area can usually be divided it in triangles (rectangles, parallelograms, trapeziums etc).
Parallelograms has opposite sides parallel and equal. Diagonals bisect the figure and opposite angles are equal..

The trapezium has one pair of opposite sides parallel.
(A regular trapezium is symmetrical about the perpendicular bisector of the parallel sides.)

An arc is a part of the circumference of a circle; a part proportional to the central angle.
If 360° corresponds to the full circumference. i.e. 2   r then for a central angle of   (see opposite figure) the corresponding arc length will be b =  /180 x r  .