Friday, 25 May 2018

Optimal Car Separation at Traffic Lights - Current Habits Require a Rethink

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In fifty two years of driving, I have had five accidents.  Four of these were somebody running into the back of my car at a traffic light or at a roundabout.  My experience is supported by data that nearly 25% accidents are rear-end collisions and short separation between vehicles is the cause.  

Why do we pack closely at traffic lights? - waiting to accelerate away when the lights turn green?  Everybody does it and it is considered more time-efficient.  But is it so?

A recent study at Virginia Tech. by Ahmadi et al. conclusively shows that short separation distances at traffic lights do not save time and may serve only to increase the number of rear-end collisions.  They find that the separation that cars maintain during free flow - the 2 seconds rule - is a good guide for  separation distances (S meters) at traffic lights.  Separations up to about 8 m do not impact on travel efficiency in a 30 mph zone.  

The experimental study is described in the slide:
(Click on the slide to see full page image, press Escape to return to text)


In figure B, we notice that at smaller separations, cars in the queue take longer before start to accelerate - for example, for S = 0.38 m, the third car in the queue is not moving even after 6 seconds from when the first car began to accelerate (light going green).  This is because a car would start to move (accelerate) only when the car in front has moved to a safe separation appropriate to a free flow driving (remember the 2-second rule).
In contrast, when S = 7.6 m, even the fifth car is able to move within the initial 6 seconds.

Notice; independent of the bumper-to-bumper distances from 0.38 m to 7.6 m, the time for all the ten cars to cross the intersection was constant to within 1 second at 23 seconds.  Only for S = 15 m, where the separation is comparable to the minimum distance for comfortable driving, the time increased to 27 +- 3 second.  

It would seem that keeping a larger separation in stop-go driving conditions does not impact on travel time and is much safer.  Drivers should be made aware of this observation and encouraged to follow the conclusions.  I might have saved my time spent in pursuing four insurance claims which thankfully settled in my favour.

The study explains these experimental observations using a theoretical model and I encourage you to visit their publication for details.

Post Script:  The 2-second rule gives the following safe spacing for driving at different speeds

70 mph  S = 62 m;  50 mph  S = 44 m;  30 mph  S = 26 m and  20 mph  S = 17 m.

Considering that the reaction time of the driver might be of the order of 1 second (she may be tired too), the 2 seconds rule over-estimates the recommended separations by a factor of may be two for 20 or 30 mph zones.  

Monday, 21 May 2018

Surface to Volume Ratio for a Spheroid, Cylinder, Cone and Rectangular Box - A Quantitative Analysis

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Among all 3-D solids, for a given volume V, a sphere has the smallest surface area S.  

This statement is universally made. But for me, it has been almost impossible to find a proof of this statement. Another question I have often wondered is, how different the ratio S/V is for other 3-D shapes compared to a sphere.  In the following, I provide a full analysis for a cylinder, a cone, a rectangular box and a spheroid (of which the sphere is a special case). Let us first look at the sphere to set the baseline.
(Click on a slide to view its full page image, Press Escape to return to text)

Sphere:  For a sphere, the situation is straightforward.  The volume and the surface area of a sphere are completely determined by its radius R.  This is shown in slide 1.  

Cylinder:   The volume and surface area of a cylinder are expressed as in slide 2.



Cone:

 Rectangular Box:


In the above discussion, we have obtained general expressions for a cylinder, a cone and a rectangular box.  These are very useful to obtain insight into the situation: for example, we found that minimum surface area of a cylinder happens when its height is twice its radius, or for a cone the condition is that the slant height is three times the radius of the circular base, etc.
We can, of course, use a spreadsheet (I have used EXCEL) to check the general results also and obtain a good understanding how the surface area increases with the parameter of interest.  
This is shown in the next 3 slides for a given volume of 1000 cm^3.




The minimum surface areas are significantly greater than for a sphere of volume 1000 cm^3 which has a surface area of 483.6 cm^2.

Spheroid:  I now return back to the case of a sphere which is a special member of the ellipsoid family.  An ellipsoid is a surface that may be obtained from a sphere by deforming it by means of directional scalings.  The three principal axes of an ellipsoid are of unequal lengths.      
In order to keep the discussion simple, we shall consider a simplified ellipsoid - a spheroid that has two of its axes equal in length. This is shown in the next three slides:




I have shown here that, for the solids considered, a sphere indeed has the smallest surface area for a given volume.  
It might be interesting for you to find examples where this fact is manifested in nature and design.

Final Word:  You might have noticed that for cylinders, cones and boxes, when the length parameter goes to zero or to very large values, the surface area tends to infinity.  This is simply because the 3-D surface tends to approximate a 2-D plane of large dimensions.  
  

Sunday, 13 May 2018

Climbing a Conical Mountain -- Almost Impossible Shortest Path, and a Realistic Path You Can Use

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Suppose you wish to visit a point half way up a conical mountain.  
The question is -- what path would you choose?  
One could use the shortest path from your base at ground level expecting to save time and energy and  build the shortest road/railroads for cars/trains to travel on.  But is this feasible? - Apparently not, as the slopes one would face are just too high for all forms of transport.

Ever wondered why mountain roads generally loop round?  This is to manage the slope of the road that you can bicycle on, or drive a car.  In this blog, I shall analyse the design considerations of a mountain road with a reasonable slope and calculate the length of the road required to visit a point half way up the mountain.
We shall learn some interesting properties of cones using geometry and school level maths.  

As shown in the figure, we can 'develop' a 3-D cone into a 2-D surface -- this makes life much easier. For a right circular cone of slope length L, the 2-D shape is a segment of a circle of radius L.  For ease of discussion, let us consider that the segment subtends an angle of 120 degrees at the centre of  the circle.  
We shall express angles in radians  - 
1 radian is 180/Pi degrees where Pi = 3.142.
(click on the figure to see full page image)

1.  Shortest Path:  This has been discussed at several places and I refer you to these(1, 2, 34). The situation is really quite straight forward and is explained in the following figure:


Notice, how the path to point P (shown by the red curve) starts at a steep angle of 60 degrees from the horizontal plane (plane for the base of the cone).  The path also goes uphill beyond point P and then reaches P through the downhill part.  
Even if we were to take the shortest path from A to return back to A (path AB in the previous figure), it will follow the green curve with a slope of 48 degrees. 
These slopes are absolutely too steep for any transport and impractical for that reason.
Let us try to understand why paths that look straight on a 2-D cone template actually are curved on the 3-D cone.  This is explained in the next figure
As discussed previously, on folding the 2-D template, the circular segment BWA, becomes a closed circle and sits on a horizontal plane while radii VB, VS, VQ etc. lie on the sloping sides of the cone.  Paths from A (AB and AP) make different angles with these radii as one moves away from point A.  
Path AB:  The path starts pointing upwards with a positive slope relative to tangent at A. The tangent lies in a horizontal plane.  This angle decreases until point F where the path is parallel to the tangent at W - here the path has zero slope.  Beyond point F, the path has a negative slope as shown in the figure. F is the midpoint of path AB.  
Path AP:  The path follows same pattern as path AB but now the path has zero slope at position G, and the portion GP has a negative slope (goes downward).  

2.  A Realistic Solution: To construct a path that most modes of transport may be able to use, let us first look at how slopes in roads are defined and what the maximum slopes on existing roads are.

 
2. A constant Gradient Path:  We shall now consider a realistic path that cars can travel and shall choose a gradient of 1 in 6 (1:6) or approximately 10 degree slope of the road.  The path makes an angle of ~10 degrees (1/6 radian) with the horizontal (base of the cone) throughout the ascent. In the figure below, this is represented by the Greek letter theta (Θ) i.e. Θ = 1/6
First we do a general analysis:

Notice that after travelling a short distance AP, the car has moved up the slope by a distance ZP = dL and the line VA has turned through an angle ε.  Therefore, we can write

           AP = S1= L ε   and  ZP = dL = L ε Θ    eq. 1

The remaining length of the slope after one step ___    L1 = L - dL = L (1 - εΘ)


In the next step of angle ε
the car will travel a shorter distance equal to (L - dL)ε and the total distance travelled in the first two steps is equal to (use eq.1)

            S1 + S2 =  Lε + (L - dL)ε = Lε + (L - εΘ)ε = Lε (1 + (1 - εΘ)).

The remaining length of the slope after two steps

             L2 = L1 - L1 εΘ = L1 x (1 - εΘ) = L x (1 - εΘ)^2

In the third step, the distance travelled is 

           S1 + S2 + S3 = Lε {1 + (1 - εΘ) + (1 - εΘ)^2}

The remaining length of the slope after three steps

           L3 = L2 - L2 εΘ = L2 x (1 - εΘ) = L x (1 - εΘ)^3

and so on till a full circle of the cone is completed - 
this requires N = 120/ε steps (angles expressed in degrees).  We can write that the total distance travelled to complete one full circle of the cone is

          S  = ε {1 + (1 - εΘ) + (1 - εΘ)^2 + ...  N terms}     eq.2

and the length of the slope remaining is

         L(remaining) = L x (1 - εΘ)^N            eq.3

The car would have climbed a distance H equal to 

            H = L - L(remaining) = L (1 - r^N)  where r = 1 - εΘ       eq.4

In eq.2, the right hand side is a geometric series with a multiplier r =  (1 - εΘ)

The sum may then be written as 

         S = ε (1 - r^N)/ (1-r)  =  H ε/(εΘ)  =  H /Θ                eq.5

Now we are ready to calculate some numbers.  

Let us say that each step is equal to a turn of 1 degree or 1 x Pi/180 = 1/57.3 radian.  Therefore,

              ε =1/57.3 radian;   Θ = 1/6 radian and N = 120.

Equations 4 and 5 then give:

For One Loop:            H = L (1 - 0.705) = 0.295 L ~ 0.3 L
                                  S = H/(1/6) = 0.295 x 6 L  =  1.77 L

The circumference of the base of the cone is L times 2Pi/3 = 2.095 L.  
After one round, the car has moved up 29.5% of the slope towards the vertex of the cone and travelled a distance of 1.77 times the length of the slope.  

If we continue to a second loop around the cone then an exactly the same analysis can be carried out but with the starting value of the slope length appropriate for the remaining cone.  Alternately, we can use equations 4 and 5 with N = 240 to represent two loops.

For two loops (N = 240) the result will be 

                H = L (1 - 0.497) = 0.503 L
                S = H/(1/6) = 0.503 x 6 L  =  3.02 L

These results are shown in the following figure:

Some points to note here:  The choice of ε equal to one degree angle is quite good.  I checked by using ε = 0.1 degree and the result does not change suggesting that the calculation is secure.  Also, I have chosen a slope angle of 1/6 radian (about 10 degrees) that I think is  realistic.  The method described above should be accurate up to a slope of about 20 degrees.  For more steep slopes, it will be better to use actual trigonometric relations but that will make the analysis more complicated - and also then road gradient less practical.  

Interestingly, our choice of a 1:6 gradient takes the car to the half way point in two loops. Of course, one can use equation 4 and calculate the gradient of the road required to reach a certain point on the mountain after a given number of loops. 
For example, if we wish to reach the half way point in one loop, the we can use eq.4 for H = 0.5 L and find Θ (the slope required).  Simple calculator work  gives Θ = 0.33 radians or 18.9 degrees.  
Eq.5 then tells us that the length of the path S = 1.515 L.

We can summarize the results as follows:

To reach the half way point

Shortest Path:           Θ = 60 degrees,     S = 1.32 L
Road with one loop:   Θ = 18.9 degrees,  S = 1.52 L
Road with two loops:  Θ = 10 degrees,    S = 3.02 L

Hope you have enjoyed reading this blog and feel more comfortable about analyzing conical structures.

Sunday, 29 April 2018

A Couple of Challenging Geometry puzzles - Created by Martin Gardner - Recreational Mathmatics

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These Puzzles are mathematically challenging

I was inspired to look into Martin Gardner's (1914 - 2010) work by an article in the recent Physics World issue.  Gardner ran a column in Scientific American for 25 years and published books on recreational mathematics.  Some of his puzzles can take a life of their own and result in a few sleepless nights.

I have chosen two examples from the famous collection of his puzzles based on geometry.  I use these puzzles to show how trigonometry and geometry are equally efficient in solving problems.  Personally, I find trigonometry easier to use - it provides ready-made relations between the sides and angles of a triangle which might be more difficult to visualize in a geometrical context - particularly if you are using hand-drawn diagrams that may not be  constructed accurately.

The first problem that I have chosen is best illustrated by the following diagram (click on the figure to see a full page image)


I shall give the geometry based solution but let us first try the easier trigonometric solution of the above problem:




Geometrical Solution: If we look at the angles at the corner marked T in the diagram:

angle WTR = angle TRS = angle C
angle WTP = angle TPS = angle A

and angle WTR = angle WTP + angle PTR
or          C = A + angle PTR 

Essentially, to solve the puzzle, we need to prove that angle PTR = angle TQS 

One way to do this is to complete a right angled triangle that will include angle PTR.  This may be done by drawing the perpendicular PV from point P on an extended line TR - this gives us the following figure:




Line PV has been drawn parallel to the diagonal YR of the middle square to ensure that angle PVT is 90 degrees.  In the figure above, we can clearly see that right angled triangles QST and TVP are similar as all their sides are in the same ratio (1 :2), viz. 


QS  =  2,       ST  =  1     and   TQ  =  √5
TV  = 2√2,    VP = √2     and   PT = √10

This means that corresponding internal angles are equal.
 Therefore, angle facing side ST (angle B)  is equal to angle facing side PV  or  


         angle PTV  (or PTR)  =  angle B

Thus we can be sure that angle PTR = angle B, and 

                        angle C = angle A + angle B

Notice, we have used the properties of similar triangles to solve the above puzzle.  This method is used frequently in solving geometry problems and sometimes it can provide a quicker and more concise solution. To demonstrate this point,  I discuss another geometrical puzzle in the following:  

PUZZLE No. 2:


There are three different ways that one can solve this problem.  
Geometry, Trigonometry and Pythagoras Theorem. 
We shall try all three methods:

Solution using Geometry:  Here we have three similar right-angled triangles - their internal angles are the same.  It is straightforward to show that

Triangle ABC:  angle BAC = A, angle BCA = C and angle ABC = 90
Triangle BDA:  angle BAD = A, angle ABD = C and angle ABC = 90
Triangle BDC:  angle CBD = A, angle BCD = C and angle BDC = 90

In similar triangles, the ratios of corresponding sides are equal. This implies that

For angles 90 degrees and A ___  AC:BC = BC:DC = AB:BD
   or   25:a = a:16 = c:H
   or     a^2 = 25 x 16 = 400   or   a = 20 cm

For angles 90 degrees and C ___  AC:AB = BC:BD = AB:AD
  or   25:c = a:H = c:9 
  or     c^2 = 25 x 9 = 225    or   c = 15 cm

We also have a:H = c:9   or  20 x 9 = 15 x H   or  H = 12 cm

Therefore   H = 12 cm,  a = 20 cm and c = 15 cm

Solution using Trigonometry: 

In right angled triangle ABD,   tan A = BD/AD = H/9
In right angled triangle CDB,   tan C = BD/DC = H/16

But angle C = 90 - angle A;   therefore  tan C = tan (90-A) = 1/tan A

This gives us ---   tan A = H/9 and 1/tan A = H/16
On multiplying the two equations:  1 = H^2/(9x16)

            or   H^2 = 144   or H = 12 cm

Solution using Pythagorus Theorem:  

In the three triangles:

     a^2 + c^2 = 25^2     or   a^2 + c^2 = 625    eq.1 
     H^2 + 16^2 = a^2    or   a^2 - H^2 = 256     eq.2  
     H^2 +  9^2  = c^2    or   c^2 - H^2 =  81      eq. 3

Subtract the sum of equations 2 and 3 from equation 1 to obtain

2 x H^2 = 625 - (256 + 81) = 625 - 337 = 288

or H^2 = 144   or   H = 12 cm

Final Word:  Geometrical puzzles are really addictive.  I remember during my school days, my father had asked my younger brother and I to see if we can divide an angle in three equal parts just using a compass and ruler.  Of course, this is not possible but we spent weeks of our play time trying to work it out and used up a large amount of paper.  But it was lots of fun and we learnt many useful tricks and theorems in mathematics during this time.  It was one of the most productive period in term of return from our unsuccessful attempts. Struggle makes you wiser and more resourceful. 

Keep puzzling...

Sunday, 22 April 2018

What is the Number(s) of the Torn Page/Pages? - Two Seemingly Impossible Puzzles

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I have two seemingly impossible puzzles below:

PUZZLE 1: One page of a book is torn off and the sum of the remaining page numbers is 14001.  Can you find the torn page number?

PUZZLE 2: Some pages were torn off from a  magazine with page numbers from 1 to 32. If the sum of the  remaining page numbers is 495, how many pages were torn off and what are their numbers? 

Puzzle two is more tractable but puzzle one has created a lot of confusion and many wrong and misleading statements and solutions have been posted online.  
I write this blog to provide proper solutions to these two very interesting mathematical puzzles.

Both puzzles appear to provide too little information for solution.  
However, the details are in the way books and magazines are printed and their pages numbered.  Following standard practice, I shall call the leaf of a book a page which has two page-faces (two page numbers); I also note that all page-faces are numbered and start from number 1 on page one of the book.   

The statements of the puzzles may be supplemented by the following observations:

1.  Pages in a book are created by printing four page-faces (2 in the front and 2 in the back) on one sheet of paper and then folding the sheet in half.  Books are stitched and bound in a way that the page number of the end page-face is a multiple of 4
2.  The sum of all page numbers is an even number. (refer to point 1) 
3.  Each page starts with an odd number on the front and has an even number on its reverse side.  
4.  The sum of page numbers on the two page-faces of a page is an odd number (ref. point 3).
5.  All page numbers are integers.

In puzzle 1, one page is torn off the book; Therefore, the sum of remaining numbers must be odd (ref. points 2 and 4)
This has been a problem with several statements of puzzle 1 where the sum of remaining page numbers is given as 10,000 or 15,000 etc - an even number.  This is incorrect and makes the puzzle unsolvable.

SOLUTION (PUZZLE 1):  One page of a book is torn off and the sum of the remaining page numbers is 14001.  Can you find the torn page number?

Let the page numbers in the book go from 1 to N.
Also suppose that the page with page numbers P and P+1 has been torn.
Let us work out the sum S of  page numbers 1 to N in the book.  It is

        S = 1 +    2    +   3 + 4 +   ........+ (N-1) + N
OR    S = N + (N-1) + ....................    +    2    + 1

Add the two equations

       2S = (N+1) + (N+1) + (N+1) +........ N terms
            = N (N+1)

OR    S =  N (N+1)/2          eq.1

The sum of numbers on the torn page = P + (P+1) = 2P + 1

Therefore, the sum of remaining page numbers, 14001, may be written as

          N (N+1)/2 - (2P + 1)  = 14001                   eq.2

or      N^2 + N = 4P + 2 + 28002 = 4P + 28004      eq.3

We note that the minimum value of P is 1 (first page is torn) and the maximum value of P is N-1 (last page is torn).  
These limits of P, when used in eq.3, will define the range that N may have.

For P = 1 _____  N^2 + N = 28008
                         N^2 + N + 0.25 =  28008.25
                         (N + 0.5)^2 = (167.36)^2
Therefore           N = 167.36 - 0.5 = 166.86

For P = N-1_____  N^2 + N = 4N - 4 + 28004
                            N^2 -3N = 28000
                            N^2 -3N + 2.25 = 28000 + 2.25 = 28002.25
                            (N - 1.5)^2 = (167.34)^2
                            N = 167.34 + 1.5 = 168.84

Therefore, The maximum page number in the book must be in the range 167 to 169.  But N also has to be an even number.  This gives a unique value of N as 168. We can use this value of N in eq.2 to calculate P - the torn page number.

                      168 x 169/2 - (2P + 1) = 14001
or                     2P = 14196 - 14001 -1 = 194
or                         P = 97

The answer of puzzle 1 is that the page with page numbers 97 and 98 was torn off.  This answer is unique and no other number satisfies the equations.

SOLUTION (PUZZLE 2): Some pages were torn off from a magazine with page numbers from 1 to 32. If the sum of the  remaining page numbers is 495, how many pages were torn off and what are their numbers?

Sum of the page numbers (all pages present) of the magazine = 32 x (32+1)/2 = 528
After tearing off some pages, the sum of the remaining page numbers = 495

Sum of the page numbers torn off = 528 - 495 = 33

The sum of the front and back page numbers for a page with front face page number P is P + (P+1) = 2P+1

If an even number of pages are torn then the sum of their page numbers will also be even - this is inconsistent with the odd number 33.
Therefore, an odd number of pages were torn off. 

The answer cannot be 1 page torn - then the page numbers will have to be 16 and 17 to add to 33.  However on a page, the number on its front page face is odd - not even as in 16.

We can now find actual torn page numbers by trying 3, 5, 7... pages. 

For 3 pages, I calculate the following combinations which add to 33.

Front page face numbers 1, 3 and 11 give (1+2) + (3+4) + (11+12) = 3 + 7 + 23 = 33
Front face numbers 1, 5 and 9 give (1+2) + (5+6) + (9+10) = 3 + 11 + 19 = 33
Front face numbers 3, 5 and 7 give (3+4) + (5+6) + (7+8) = 7 + 11 + 15 = 33

For 5 pages, the lowest sum will be if the first five pages were torn.  This will give us 
(1+2) + (3+4) + (5+6) + (7+8) + (9+10) = 55 which is greater than 33.  Therefore 5 or more pages are not possible answers.

The three torn pages have page numbers:
1,3,11   or  1,5,9   or  3,5,7

Note that page numbers 3,5,7 represent consecutive pages.  If the puzzle is worded as consecutive pages were torn off the magazine then 3,5,7 becomes a unique answer.  
This may be found by the following method.  
The page numbers of the three torn pages are
            P, P+1, P+2, P+3, P+4 and P+5.  
Their sum is 6P+15. 
If this is equal to 33 then P = 3 as before. 

Hope you have enjoyed the puzzles.  Let me know by writing to ektalks@yahoo.co.uk or by leaving your comment on the blog site.    

Wednesday, 18 April 2018

Division of a Huge Cake: An Interesting Mathematical Puzzle




Mathematical puzzles are addictive and lately, my addiction to them has become quite strong.  The ones I like most are those that have some unexpected outcome and, of course, they also should require some struggle to find the solution.  

Division of a Huge Cake is a puzzle which was published in The Guardian a few years ago.  I thought the published solution of the puzzle did not quite do the justice to the beautiful and somewhat counter-intuitive outcome.  This encouraged me to write this blog - hope you enjoy the discussion of the result.  

The puzzle is (reformulated by me): An eccentric duke throws a big party for 100 of his good friends.  He wishes to give bigger portions of the cake to some of his selected important guests who are served first.  The duke concocts a strange (or shall we say eccentric) formula for dividing the cake  
The first guest gets 1% of the cake.  
The second guest gets 2% of the remaining part.  
The third guest gets 3% of the part remaining and so on...  .  
The 50th guest gets 50% of the part left, and 
the 100th guest gets 100% of whatever is left.
Who gets the biggest piece and how much does he get?  Did the duke achieve what he wanted?
Is this a good way to divide the cake?

Solution:  As the guests are served their share of the cake, the amount remaining gets progressively smaller and even though they are taking a bigger percentage of the remaining cake, after a certain stage the size of the serving starts to get smaller.  One of the guests will get the biggest piece and we need to work it out by using some algebra:

Let us say that for guest number the amount of cake left is a fraction P of the original size.  
Guest N will get N% of this; that is - Guest N receives a portion = p x N/100 
After guest N has had his portion, 
the cake left is equal to   P - P x N/100  = P (1 - N/100)
The next guest (number N+1) will take (N+1)% of this remaining size i.e., 
his share will be  (N+1)/100 x P (1 - N/100)

The question is - for what value of N,  the share that guest number (N+1) receives is smaller than the share taken by guest number N?  
Put it another way, we want to find out for what N, the following relation is true:

(N+1)/100 x P (1 - N/100) - P x N/100 < 0

Multiply by 10,000 and divide by P to get:

(N + 1) x (100 - N) - 100 x N < 0 

OR   N x 100 + 100 - N^2 - N - 100 x N < 0

OR   N^2 + N > 100

Equation in red tells us that when N is large enough for the inequality to hold, subsequent guests will get a smaller portion than the previous guest.

Obviously for N less than or equal to 9, the left hand side is smaller than 100; but for N = 10, the left hand side is 110 and is greater than 100.  
One can use a spreadsheet to work out the portion of the cake that each guest receives.  This is shown in the slide below (click on slide to see full page image)

Guest number 10 will receive the biggest slice of the cake and he will receive 6.28% of the initial cake.


Notice that under this plan, guest 30 onwards receive very little cake - guest 32 only receives 0.1% of the cake. Interestingly, guests number 85 onwards receive less than a molecule of the cake - there is no cake left for them to share.

Is there a more equitable way of dividing the cake?  The problem with the duke's method is that the portions served in the beginning are too large.  Consider a 10 kg cake with an average serving of 100 g per guest.  In duke's method, guest number 10 will receive 628 g portion!  
This situation may be moderated by modifying the formula by which the guests receive their portions as the fourth root of their number - Nth guest receives N^0.25 percents of the remaining cake and not N% as suggested by the duke.
If we follow this recipe then the distribution of portions is as shown: 
This method gives a reasonable distribution of cake portions and meets the duke's wishes of serving larger portions to his important guests.  The portion sizes vary from 0.263% for guest 99 to a maximum of 1.576% for guest 13.  The cake remaining is 8.1% after all guests have been served.