3 Round Trip Time Calculations Revision 10 111010

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3 Elevator Traffic Analysis andDesign

Selected Topics in Mechatronics 0908589

©Copyright held by the author 2011, Lutfi Al-Sharif of 25

Chapter 3Elevator Traffic Analysis and Design

(Revision 10.0, 10/10/2011)

1. Summary of the Traffic Analysis and Design Steps The following steps have to be followed in the elevator traffic analysis and designprocedure.

1) Inputs: The input parameters needed in order to carry out the traffic analysis for abuilding are as follows:

a) Type of building (office, residential, hotel, hospital, shopping centre.....etc).b) Number of floors above ground.c) Floor heights.d) Net areas for each floor.

e) In certain cases the client might specify the required interval and theexpected arrival rate.

2) Analysis: The designer will then carry out the following:

a. Find the recommended speed based on dividing the total travel by 20 andfinding the nearest preferred speed value.

b. Set the suitable values of jerk and acceleration depending on the type of traffic in the type of building given (e.g., for a hospital, select low values of acceleration and jerk).

c. If not set by the client, deciding on the suitable arrival rate for the type of

traffic in the type of building given.d. If not set by the client, deciding on the suitable interval for the type of traffic

in the type of building given.e. Decide on a starting number of lifts.f. Decide on a starting capacity for the lifts.g. Carry out the traffic analysis that provides the required interval at the

expected arrival rate.h. Change the number of lifts up or down until the required interval is

achieved.i. Once the interval is achieved, fix the number of lifts.

j. If it is not possible to find a suitable solution using a maximum number of 10 lifts then investigate zoning the building into two or three zones if needed.

k. Depending on the car loading percentage, reduce the capacity of the liftsused until the loading percentage is near 80%.

l. Select the suitable number, speed and capacity of the goods lift required.m. Select the suitable number of fie fighting lifts, using the 60 second rule to

decide on the speed of the lifts.

3) Outputs: The aim of the traffic is to find the following parameters:

a) Speed of the lifts.b) Number of lifts.






c) Capacity of the lifts.d) Zoning if required.e) Goods lift requirements.f) Fire fighting lifts requirements.

2. Legendτ is the round trip time in sint is the interval at the main terminal in sL is the number of the elevators in the groupU is the total building populationUi is the building population on the ith floorHC% is the handling capacity expressed as a percentage of the building population

in five minutes AR% is the passenger arrivals expressed as a percentage of the building population

in the busiest five minutes

N is the number of floors above the main terminalH is the highest reversal floor (where floors are numbered 0, 1, 2….N S is the probable number of stopsdf is the typical height of one floor in mv is the top rated speed in m/sa is the top acceleration in m/s2 j is the top rated speed in m/s3 tf is the time taken to complete a one floor journey in s assuming that the lift attains

the top speed v CC is the car carrying capacity in persons P is the number of passengers in the car when it leaves the ground (does not need to

be an integer)dG is the height of the ground in m where more than the typical floor heighttdo is the door opening time in s tdc is the door closing time in s tsd is the motor start delay in s tao is the door advance opening time in s (where the door starts opening before the

car comes to a complete standstill)tpi is the passenger boarding time in s tpo is the passenger alighting time in s

3.0 Introduction

The round trip time (τ ): The round trip time is defined as the time taken for theelevator to pick up the passengers from the main lobby, travel to the upper floors anddeliver the passengers to their destinations and then express back to the mainterminal again to pick up more passengers.

τ must be measured between one event and exactly the following event. Forexample it could be measured between the doors starting to open at the mainterminal until they start re-opening again at the main terminal.

The highest floor that the elevator reaches in one round trip is called the highestreversal floor, H. It can be less than or equal to N, the number of floor above themain terminal. It does not need to be an integer.

The number of stops that the elevator makes in one round trip is called theprobable number of stops, S. It can be less than or equal to the number of passengers in the car, P. It does not need to be an integer.






When more than one lift is present in the same group, then ideally they have to bespread evenly. The interval is the time between the arrivals of the lift in the mainterminal. The interval is obtained by dividing the round trip time by the number of elevators in the same group. Thus the interval can be improved either by increasingthe number of elevators or reducing the round trip time. The round trip time can bereduced by increasing the speed or reducing the number of stops or reducing thehighest reversal floor.

The various parameters used in the elevator traffic analysis are shown in the tablebelow:






Parameter Definition Typical valuesor comments

τ

The round trip time

The average time taken by the elevator to do acomplete cycle in the building during up peak

conditions (pick up passengers from the mainterminal, travel to the upper floors and thenexpress back to the main terminal again)

int The interval The average time between successive arrivals of

elevators in the main terminal30 s

LNumber of elevators in

the same groupNumber of elevators in the same group 2 to 8

U Population Total building population

HC% The handling capacity The percentage of the building population

transported in five minutes

AR% The arrival rate The passenger arrivals expressed as a percentageof the building population in the busiest five minutes

N Number of floors The number of floors above the main terminal

Note that the

total number of floors is N+1

(when the mainterminal isincluded)

H The highest reversal

floor

Highest reversal floor (where floors are numbered0, 1, 2….N) NH ≤

CC Car carrying capacity The car carrying capacity in persons. This is

theoretical capacity of the car.

P Passengers in the car The number of passengers in the car when it leavesthe main terminal (does not need to be an integer).

This is the effective capacity of the carCCP ⋅= 8.0

S The probable number

of stops

Probable number of stops the that elevator willmake in one round trip PS ≤

df Floor to floor height The typical height of one floor in m (finished floor

level to finished floor level, FFL to FFL)

No need toenter this value

for the topmost floor

4 mv Speed The rated speed in m/sa Acceleration The top acceleration in m/s

21

j J erk The top rated speed in m/s 0.5 to 1.5

tf One floor cycle time The time taken to complete a one floor journey in s

assuming that the lift attains the top speed v

dG Main terminal heightthe height of the main terminal in m where more

than the typical floor height5 m

tdo Door opening time The door opening time in s

2 s

tdc Door closing time The door closing time in s

3 s

tsd Motor start delay The motor start delay in s

0.5 s

tao Advance door opening

time

The door advance opening time in s (where thedoor starts opening before the car comes to a

complete standstill)

Do not use onsafety grounds

tpi Passenger boarding

time

The passenger boarding time in s 1 to 1.2 s

tpo Passenger alighting

time The passenger alighting time in s 1 to 1.2 s






4. Number of passengersIt is accepted in practice that the car does not fill up completely. In general it only fillsup to 80% of the car carrying capacity.

CCP ⋅= %80

5. The Probable Number of Stops The probable number of stops S is the average number of stops that the elevatormakes in one round trip journey

−−=

P

N

111NS

for equal floor populations

∑=

−−=

N

1i

P

i

U

U1NS

for unequal floor populations where Ui is the

population of floor i and U is the total buildingpopulation.

6. The Highest Reversal Floor The highest reversal floor H is the highest floor that the lift reaches in a round trip journey.

P1N

1i N

iNH ∑

−

=

−=

for equal floor populations

∑ ∑−

= =

−=

1N

1 j

P j

1i

i

UUNH for unequal floor populations where Ui is the floorpopulation of floor i and U is the total building

population.

7. The Round Trip Time EquationIn this section, the equation for the round trip time equation is derived from firstprinciples. It makes the following assumption:

1. The traffic is pure incoming traffic (up peak only). All passengers arrive at theentrance and board the elevator to go to his/her destination on one of the

upper floors.

2. In one trip time, the elevator makes S stops and reaches H highest reversalfloor. These two variables have been derived elsewhere. They depend on thenumber of floors above the main entrance (the lobby), N, and on the numberof passengers in the car, P. They can be derived for the general case wherethe floor populations are unequal or for the special case where the floorpopulations are equal.

3. The elevator collects P passengers from the main terminal (lobby) anddelivers them to their selected destinations. It then expresses back to the

main terminal to pick another set of P passengers.






4. Every time the elevators stops at a floor, it takes time for the doors to open,the passengers to transfer (in at the main terminal and out at the destinationfloors) and then for the doors to close. Further delay at each stop is causedby the start delay of the motor. The delay will be reduced in the case of advanced door opening.

5. The assumption will be made that the elevator will attain its top speed in onefloor journey.

6. The assumption will also be made that the floor heights are equal.

7. The assumption will also be made that there is only one entrance floor (arrivalfloor). All passengers arrive through this single entrance.

The round trip time is made up of main three parts: the time spent at the ground floor

collecting passengers, the time spent travelling to the upper floors and deliveringpassengers to their destination, and the third part is the express travel back from thehighest reversal back to the main terminal.

tacc is the time taken to accelerate up to the top speed from standstill in s tdec is the time taken to decelerate down from the top speed down to standstill in s

HSMT τ τ τ τ ++=

Where:

MTτ is the time spent at the main terminal in s

Sτ is the time spent travelling to the upper destination floors and delivering the

passengers in s

Hτ is the time spent expressing back to the main terminal from the highest reversal

floor in s

The time spent at the main terminal involves the opening of the door, the transfer of P passengers into the car and the closing of the door, in addition to the start delayminus the advanced door opening.

( ) piaosddcdoMT tPtttt +−++=τ

( ) ( ) ( )

( ) ( )

( )poaosddcdodecacc

f f f

poaosddcdodecacc

f

poaosddcdo

f

decaccS

tPttttttv

d

v

dS

v

dH

tPttttttSv

dH

tPttttSv

dHttS

+

−+++++−⋅+

⋅=

+−++++⋅+

⋅=

+−++⋅+

⋅++⋅=τ

But






j

a

a

v

v

dttttt

v

dt

f

decaccvdecacc

f

f ++=++=++=

( )poaosddcdo

f

f

f

S tPttttv

d

tSv

d

H +

−+++−⋅+

⋅=τ

The third part of the round trip time equation is the time taken to travel back from thehighest reversal floor express back to the main terminal:

−+

⋅=

++

⋅=

++

⋅=

v

dt

v

dH

j

a

a

v

v

dH

j

a

a

v

v

dH

f

f

f

f

f

Hτ

Adding all the three elements together gives:

( ) ( ) ( )

( ) ( )popiaosddcdo

f

f

f

f f

f

poaosddcdo

f

f

f

piaosddcdo

HSMT

ttPttttv

dtS

v

dH

vdt

vdH

tPttttv

dtS

v

dHtPtttt

++

−+++−⋅++

⋅⋅=

−+

⋅+

+

−+++−⋅+

⋅++−++=

++=

12

τ τ τ τ

The equation for the round trip time (RTT) can be written as follows:

( ) ( )

−⋅+++

−+++−⋅++

⋅⋅=

v

ddttPtttt

v

dtS

v

dH

f G

popiaosddcdo

f

f

f 212τ

Note that the last term has been added to the case where the main terminal (thelobby) has a floor height that is more than the typical floor height. As this distance iscovered in both the up and down directions, it has been multiplied by 2. dG is theheight of the main terminal floor.

In general for most buildings (especially office buildings) the main terminal orthe lobby has a height greater than the typical floor height, usually for aestheticreasons but sometimes for functional reasons as well.

8. The Handling Capacity The handling capacity is the percentage of the building population that can be movedin five minutes.






U

PLHC

⋅

⋅⋅=

τ

300%

9. The Interval

The interval is the average time between the arrivals of consecutive elevators in themain terminal.

Lint

τ

=

11. Kinematic Equations The time t taken to complete a journey of distance d, with a top speed of v, topacceleration a, and top jerk j can be calculated as follows under the three differentconditions:

If

+≥ aj

jvvad

22

then j

a

a

v

v

d

t ++=

If

+<≤

aj

jvvad

j

a2 22

2

3

then

2

j

a

a

d4

j

at

++=

If 2

3

j

a2d < then

3

1

j

d32t

=

12. Mathematical Proof the Probable Number of Stops for Equal floor PopulationStarting from first principles, derive an expression for the probable number of stops(S) and the probable highest reversal floor (H) that a lift will make in a round trip

journey during up peak (incoming traffic).Under up peak (incoming traffic) conditions, the lift will fill up with passengers at

the ground floor, and then deliver the passengers to their destinations in the upperfloors. It then expresses back to the ground floor to collect more passengers and soon.

The car has a capacity CC, but only fills up to P passengers. The number of floors above ground is denoted by N.

Assume that the floors have equal populations. State all the assumptions that you

make in your derivation.

Solution

We shall assume equal floor populations and that passenger destinations areindependent (i.e., one passenger’s choice of destination will not influence anotherpassenger’s choice of destination).

Derivation of the probable number of stops The probably that passenger j will stop at a floor i:

=

N

1)iflooratstopwill jpass(P






where N is the number of floors above the main lobby. Thus the probability thatpassenger j will not stop at a floor i is:

−=

N

11)iflooratstopNOTwill jpass(P

But the car contains P passengers. So the probability that none of them will stop atfloor i is the product of all of their respective probabilities:

P

N

11)iflooratstopNOTwillpassall(P

−=

where P is the number of passengers in the car when it leave the ground floor. Thecomplement of this quantity is the probability that at least one passenger will stop at

a floor:

P

N

111)iflooraatstopwillpassoneleastat(P

−−=

But this is the same as the probability that a stop will take place on floor i. So theprobability of stopping on a floor i is:

P

N

111)iflooratstopa(P

−−=

The expected value of the number of stops can be obtained by adding theprobabilities of stopping on all N.

( )

−−⋅=

−−⋅= ∑

=

PN

1i

P

N

111N

N

1111stopsof numberE

Thus the probable number of stops S is equal to:

( )

−−⋅==

P

N111Nstopsof numberES

13. Mathematical Proof the Highest Reversal Floor for Equal Floor PopulationsStarting from first principles, derive an expression for the probable highest reversalfloor (H) that a lift will make in a round trip journey during up peak (incoming traffic).

Under up peak (incoming traffic) conditions, the lift will fill up with passengers atthe ground floor, and then deliver the passengers to their destinations in the upperfloors. It then expresses back to the ground floor to collect more passengers and soon.

The car has a capacity CC, but only fills up to P passengers. The number of floors above ground is denoted by N.






Assume that the floors have equal populations. State all the assumptions that youmake in your derivation.

Solution

The probably that passenger j will stop at a floor i:

=

N



−=N



P

N


−=

The probability that the lift will not travel any higher than a floor i is the probability thatit will not stop on floor i+1 or i+2 or i+3 all the way to floor N. This is expressed asthe product of these individual probabilities:

PPPPP

1i

11

2i

11.......

2N

11

1N

11

N

11

)ifloorabovetravelnotwilllift(P

+−⋅

+−

−−⋅

−−⋅

−

=

This can be re-written as:

PPPPP

1i

i

2i

1i

.......2N

3N

1N

2N

N

1N


+⋅

+

+

−

−

⋅

−

−

⋅

−

=

Putting all terms inside the same bracket gives:

P

1i

i

2i

1i.......

2N

3N

1N

2N

N

1N


+⋅

+

+

−

−⋅

−

−⋅

−

=

This simplifies to:






P

N

i)ifloorabovetravelnotwilllift(P

=

Note that this is also equal to the probability that the highest reversal floor will be any

of the floors 1 to i. In other words, it is the probability that the higher reversal floorwill be floor 1 or 2 or 3…..or i, or the sum of these probabilities.

( ))().........3()2()1()( iHPHPHPHPifloorabovetravelnotwillliftP =+=+=+=

Applying the same argument to floor i-1, gives the following:

P

N

iifloorabovetravelnotwillliftP

−=−

1)1(

Note that this is also equal to the probability that the highest reversal floor will be anyof the floors 1 to i-1. In other words, it is the probability that the higher reversal floorwill be floor 1 or 2 or 3…..or i-1, or the sum of these probabilities.

( ))1().........3()2()1()( −=+=+=+= iHPHPHPHPifloorabovetravelnotwillliftP

If we subtract the two expressions from each other only one term remains (which isthe probability that the highest reversal floor is i. So the probability that the i

th floor isthe highest reversal floor is the probability that the lift does not travel above the i

th floor minus the probability that the lift does not travel above the (i-1)th floor. Thus:

PP

N

i

N

ifloorreversalhighesttheisifloorP

−−

=

1)(

But any of the floors from 1 to N could be the highest reversal floor. The expectedvalue of the highest reversal is the weighted sum of all the possible highest reversalfloors (i.e., 1 to N) each multiplied by their respective probabilities:

∑=

−

−

⋅=

=

N

1i

PP

N

1i

N

i

i)H(E

)floorreversalhighest(valueectedexp

Re-arranging the two terms will make it easier to simplify later on:

∑=

+

−−⋅=

N

1i

PP

N

i

N

1ii)H(E

Expanding gives:






( ) ( )

( ) ( )

⋅+

−⋅−+

−−+

−−−

⋅+

⋅−+

⋅+

⋅−+

+

−

=

+

−−⋅+

−+

−−⋅−

+

−−⋅+

+

−−⋅+

+

−−⋅=

PPPP

PPPPPP

PPPP

PPPPPP

N

NN

N

1NN

N

1N1N

N

2N1N

...N

33

N

23

N

22

N

12

N

1

N

0

N

N

N

1NN

N

1N

N

2N1N

...N

3

N

133

N

2

N

122

N

1

N

111)H(E

Simplifying gives:

NN

1N

.....N

3

N

2

N

1

0)H(E

PPPP

+

−−

−

−

−=

This can be rearranged as:

∑−

=

−=

1N

1i

P

N

iN)H(E

Note that the summation extends only to N-1 (i.e., not N). Also note that themaximum possible value of H is N, which is to be expected at high values of P.

14. Mathematical Proof the Lowest Floor Express for Equal Floor PopulationsStarting from first principles, derive an expression for the probable lowest floor towhich the lift will travel without stopping in a round trip journey during up peak(incoming traffic).

Under up peak (incoming traffic) conditions, the lift will fill up with passengersat the ground floor, and then deliver the passengers to their destinations in the upperfloors. It then expresses back to the ground floor to collect more passengers and soon.

The car has a capacity CC, but only fills up to P passengers. The number of floors above ground is denoted by N. Assume that the floors have equal populations.

State all the assumptions that you make in your derivation.

Solution

The probably that passenger j will stop at a floor i:

=

N








−=N


But the car contains P passengers. So the probability that none of them will stop at

floor i is the product of all of their respective probabilities:

P

N


−=

The probability that the lift will not stop at any floor lower than i is the probability thatit will not stop on floor 1, 2, 3… all the way to floor i-1. This is expressed as theproduct of these individual probabilities:

PPPP

iNNNN

ibelowflooranyatstopnotwillelevatorP

+−−

−−⋅

−−⋅

−

=

2

11.......

2

11

1

11

11

)(


PPPPP

iN

iN

iN

iN

N

N

N

N

N

N


+−

+−⋅

+−

−

−

−⋅

−

−⋅

−

=

2

1

1.......

2

3

1

21

)(


P

iN

iN

iN

iN

N

N

N

N

N

N


+−

+−⋅

+−

+−

−

−⋅

−

−⋅

−

=

2

1

3

2.......

2

3

1

21

)(

This simplifies to:

P

N

iNibelowflooranyatstopnotwillelevatorP

+−=

1)(

The term above is the probability that the elevator will not stop at any floor lower thanthe ith floor. In effect it is also that probability that any of the floors above the ith floorcould be the lowest call express.

In a similar manner, the probability that the lift will not stop at a floor below thei+1th floor can be calculated as shown below:

P

N

iNibelowflooranyatstopnotwillelevatorP

−=+ )1(






The term above is the probability that the elevator will not stop at any floor lower thanthe i+1th floor. In effect it is also that probability that any of the floors above the i+1th floor could be the lowest call express.

Subtracting the two expressions from each other, gives the probability that theith floor is the lowest floor express (i.e., is on average the floor at which the elevatormakes the first stop when travelling in the up direction on a round trip journey). Theprobability that the i

th floor is the lowest floor express is the probability that the liftdoes not stop below the ith floor minus the probability that the lift does not stop belowthe (i+1)th floor. Thus:

PP

N

iN

N

iNP

−−

+−=

1)expresscalllowesttheisifloor(

But any of the floors from 1 to N could be the lowest call express. The expectedvalue of the lowest call express is the weighted sum of all the possible lowest call

express floors (i.e., 1 to N) each multiplied by their respective probabilities:

∑=

−−

+−⋅=

=

N

i

PP

N

iN

N

iNiLCEE

valueected

1

1)(

)expresscalllowest(exp

Re-arranging the two terms will make it easier to simplify later on:

∑=

−

+

+−

−⋅=

N

i

PP

N

iN

N

iN

iLCEE1

1

)(

Expanding gives:

( ) ( )

∑−

=

+=

+

−+

−+

−+

=

+

⋅+

−

⋅−

−−

−⋅+

+

−−

−⋅+

−−

⋅=

1

1

1

01.......

3211

01121

...32

321

21

1)(

N

i

P

PPPPP

PPPP

PPPPPP

N

iLCE

NNN

N

N

N

N

N

NNN

NNN

N

N

N

N

N

N

N

N

N

N

N

NLCEE

Simplifying gives:

PPPPP

NNN

N

N

N

N

NLCE

+

−+

−+

−+=

01.......

3211






This can be rearranged as:

( ) ∑−

=

+=

1

1

1N

i

P

N

iLCEE

Note that the summation extends only to N-1 (i.e., not N). Also note that theminimum possible value of LCE is 1, which is to be expected at high values of P.

Comparing the equation for H and the equation for LCE we can see that thesame summation term exists in both. The difference between N and H is the samedifference between LCE and 1 (symmetrical values between 1 and N). The additionof H and LCE always results in the value of N+1.

1+=+ NLCEH

15. Assumptions in deriving the basis round trip time equationWhen deriving the basic equation for the round trip time, the following assumptionshave been made:

1. Equal floor populations in the derivation of H and S.2. Passenger choices of floors are independent of each other (this affects the

derivation of H and S).3. Constant passenger arrival rate. It is assumed that the arrival process of

passengers is not random and that passengers arrive in a uniform mannerwith equal time spacing between them. In reality passengers arrive randomlyin a process that is best represented by a Poisson arrival process.

4. An important assumption made in the derivation f the round trip time equationis that the top speed is attained in one floor journey. This is not correct inmany cases where the speed is above 2.5 m·s-1.

5. It has been assumed that only one passenger is boarding or alighting at thesame time.

6. The only type of traffic present is the incoming traffic (up peak traffic).7. Equal floor heights have been assumed.8. It has been assumed that the doors starts closing immediately after the last

passenger has boarded or alighted. In reality there will be delay depending onthe timer controlling the door operation, and depending on whether otherpassengers use the door close button.

9. It has been assumed that passengers enter the building from one singleentrance. In reality many buildings have underground car parks or differentlevel street entrances.

10. No door re-openings have been assumed.11. It has been assumed that all lifts in the same group serve all floors and that

any passenger regardless of his/her destination can board any available lift.

16. Solved Design Exercise The parameters for an office building are shown below.

Assume the following parameters:

a. Passenger arrival rate is 15%.b. Number of floors above ground is 14 floors.






c. Car capacity is 21 person (1600 kg).d. Floor height is 4.5 m (finished floor level to finished floor level).e. Three basement floors, B1, B2, B3 used as car parks.f. Basement floor height is 4.5 m.g. Top speed, v, is 1.6 m·s-1.h. Top acceleration, a, is 1 m·s-2.i. Top jerk, j, is 1 m·s-3.

j. Passenger transfer time out of the car is 1.2 s.k. Passenger transfer time into the car is 1.2 s.l. Door opening time is 2 s.m. Door closing time is 3 s.n. Advanced door opening is 0.5 s.o. Start delay is 1 s.p. Total building population of 1100 persons.q. Equal floor populations.

Find the number of lifts required assuming no arrivals from the basement.Find the actual interval and the actual handling capacity.

Solution The equation for the round trip time (τ ) can be written as follows:

( ) ( )

−⋅+++

−+++−⋅++

⋅⋅=

v

ddttPtttt

v

dtS

v

dH

f G

popiaosddcdo

f

f

f 212τ

Where:τ is the round trip time in sH is the highest reversal floor (where floors are numbered 0, 1, 2….N S is the probably number of stopsdf is the typical height of one floor in mv is the top rated speed in m/stf is the time taken to complete a one floor journey in s assuming that the lift attains

the top speed v P is the number of passengers in the car when it leaves the ground floor (does not

have to be an integer for this calculation, but is used as an integer for thepurposes of calculating H and S)

dG is the height of the ground in m where more than the typical floor heighttdo is the door opening time in stdc is the door closing time in stsd is the motor start delay in stao is the door advance opening time in s (where the door starts opening before the

car comes to a complete standstill)tpi is the passenger boarding time in stpo is the passenger alighting time in s

We need to check that the lift will attain top speed in a one floor journey.We use the following equation as a check:






16.41

16.16.11

aj

jvva5.4d

222

=

⋅+⋅=

+≥=

So the top speed of 1.6 m/s will be attained in a one floor jump. In this case, the time

taken to complete a one floor journey, tf will be:

s4.51

1

1

6.1

6.1

5.4

j

a

a

v

v

dtf =++=++=

We first start by finding the value of P based on 80% car loading. P works out as16.8 passengers.

Moving on to calculate H and S as follows:

1014

1

1114N

1

11NS

8.16P

=

−−⋅=

−−⋅=

62.13......0174.0075.0288.01414

i14

N

iNH

13

1i

8.161N

1i

P

=−−−=

−=

−= ∑∑

=

−

=

Substituting in the RTT equation gives:

( ) ( )

( ) ( )

s

v

ddttPtttt

v

dtS

v

dH

f G

popiaosddcdo

f

f

f

82.2053.40895.76

6.1

5.45.422.12.18.165.0132

6.1

5.44.5110

6.1

5.46.132

212

=++=

−⋅++⋅+

−+++−⋅++

⋅⋅=

−⋅+++

−+++−⋅++

⋅⋅=τ

In order to meet the expected arrival rate of 15%, the number of lifts required can becalculated as follows:

738.68.16300

82.205110015.0

300

%

300%

=⋅

⋅⋅=

⋅

⋅⋅=

⋅

⋅⋅=

P

UHCL

U

PLHC

τ

τ

So the nearest larger integer is 7 lifts. So seven lifts are required to achieve thehandling capacity that meets the expected arrival rate.

In practice the handling capacity achieved will be:

%6.15110082.205

8.167300300% =

⋅

⋅⋅=

⋅

⋅⋅=

U

PLHC

τ

which is more than the expected arrival rate of 15%.






The achieved interval will be:

sL

int 4.297

82.205===

τ

which is less than the 30 s limit specified for offices.

ProblemsFor each of the following cases, design a suitable VT system, by finding the suitablenumber of elevators and their speed, as well as the actual car loading.

Try to find the most economical solution by looking for the minimum number of elevators that meets the requirements.Assume equal floor population and equal floor heights.In using the round trip time equation ignore the fact that the top speed has not been

attained.Select the speed v by dividing the total travel distance and then rounding down to thenearest preferred speed (1, 1.6, 2, 2.5, 3.15, 4, 5, 6.3, 8, 10 m·s -1).

Find:L, v, Pact, intact and HC% Also calculate the actual car loading. Actual car loading = (Pact/CC)%

Assume that the door opening time is 2 sAssume that the door closing time is 3 sAssume that the start delay is 1 s

Assume that the advance door opening is 0.5 s.Assume that the passenger transfer time is 1.2 sAssume that the acceleration, a, = 1 m·s-2 Assume that the jerk, j, = 1.5 m·s-3

1. Office Building: An office building has the following parameters.N= 20 floors above the main terminal. AR%= 12% Target interval, inttar of 30 sCar capacity, CC, of 26 personsFloor height, df , = 4.5 m

Total population, U is 1600 persons

Assume one single main entrance.



Solve for the following two cases:






a) Single entrance without any basements.b) One main entrance and three car park basements, each basement with an arrivalpercentage of 6% for each.



Assume one single main entrance.

4. Residential Building: An office building has the following parameters.

N= 30 floors above the main terminal. AR%= 7% Target interval, inttar of 50 sCar capacity, CC, of 16 personsFloor height, df , = 3.2 m


Assume one single main entrance.■

References1. Basset J ones The probable number of stops made by an elevator GE Review

26(8) 583-587 (1923)






Appendix AThe Round Trip Time Equation for the Case of Unequal Floor Heights

In the case where the floor heights are unequal, this will have an effect on thecalculation of the round trip time equation. The equation for the round trip time canbe amended as follows in order to account for this case as follows.

The effect of the unequal floor heights can be taken into consideration byassuming an effective floor height df eff that can be inserted into the original round triptime equation.

The effective floor height df eff is the expected value fo the floor height. Theeffective floor height is the weighted average of all the floor heights multiplied by theprobability of the elevator passing through that floor. In order for the elevator to passthrough a floor it should travel to any of the floors above that floor. Thus it isnecessary to find the probability of the elevator travelling above a certain floor, i.

The probability of the elevator not stopping at a certain floor, assuming equal

floor populations is the probability that passenger j will stop at a floor i:

=

N

1)iflooratstopwill jpass(P (1)


−=N

11)iflooratstopNOTwill jpass(P (2)


P

N


−= (3)

The probability that the lift will not travel any higher than a floor i is the probability thatit will not stop on floor i+1 or i+2 or i+3 all the way to floor N. This is expressed asthe product of these individual probabilities:

PPPPP

1i

11

2i

11.......

2N

11

1N

11

N

11


+−⋅

+−

−−⋅

−−⋅

−

=

(4)


PPPPP

1i

i

2i

1i.......

2N

3N

1N

2N

N

1N


+⋅

+

+

−

−⋅

−

−⋅

−

=

(5)







P

1i

i

2i

1i.......

2N

3N

1N

2N

N

1N


+⋅

+

+

−

−⋅

−

−⋅

−

=

(6)

This simplifies to:

P

N

i)ifloorabovetravelnotwilllift(P

= (7)

Thus the probability that the lift will travel above the floor i is:

P

NiifloorabovetravelwillliftP −=1)( (8)

Thus the expected value of the travel distance can be calculated as the weightedaverage of the various floor heights as follows:

( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( )

( )

−−⋅−+

+

−⋅+

−⋅+=

−⋅+

−−⋅−+

+

−⋅+

−⋅+

−⋅=

P

f

P

f

P

f f total

P

f

P

f

P

f

P

f

P

f total

N

NNd

Nd

NdddE

N

NNd

N

NNd

Nd

Nd

NddE

111

...2

121

110

11

11

...2

121

110

10

(9)

Where:E(dtotal) is the expected value of the distance travelled in the up direction.

The last term above reduces to zero (as it is impossible for the elevator to passthrough floor N). The expected floor height is obtained by dividing the expected totatravel distance by the highest reversal floor, H. So the equation for the effective floorheight can be expressed as shown below (assuming equal floor populations):

( )( )

H

N

iid

dE

N

i

P

f

f

∑−

=

−⋅

=

1

0

1

(10)

Where:






( )idf is the floor height for floor i

f dE is the expected value of the floor heights (effective floor height)

H is the highest reversal floorN is the number of floors above the main terminal

P is the number of passengers boarding the car from the main terminal






Appendix BProof of the time for a one floor journey assuming that the top speed has been

attained

Assume the journey is split into 6 parts: d1, d2, d3, d4, d5, d6 and d7. Also assume thetime duration for each part is t1, t2, t3, t4, t5, t6 and t7 respectively. The total distanceof the journey is d,the top speed is v, the top acceleration is a and the jerk is constantat a value of j.

Note from symmetry the following equations apply:

7531 tttt ===

And:

62tt =

The distances are also equal as follows:

71 dd =

53 dd =

62dd =

At the end of the time t1, the acceleration would have attained its top value. Hence:

j

at =1

The speed at the time t1 is:






( ) j

at jtv

⋅=⋅⋅=22

1 2211

The speed at the time t1+t2 is:

( ) j

avttv

⋅−=+2

2

21

The distance covered at time t1 is:

2

3

0

21

62

11

j

adtt jd

t

⋅=

⋅

⋅⋅= ∫

The distance d3 is equal to the area of a rectangle less the area d1 (from symmetry).

( )

⋅−

⋅=

⋅−⋅=

2

3

2

3

1366 j

a

j

av

j

atvd

The time t2 can be calculated as follows:

j

a

a

v

a

j

a

j

av

t−=

⋅−

⋅−

=

22

22

2

The area d2 is the area of a trapezoid, calculated as follows:

⋅

⋅−

⋅=

−⋅

=⋅

⋅+

⋅−

= j

av

a

v

j

a

a

vvt

j

a

j

av

d2222

22 2

2

22

2

The total distance d1+d2+d3+d5+d6+d7 can be calculated as follows:

a

v

j

av

j

av

a

v

j

avdddddd

22

76532122

2 +⋅

=

⋅

⋅−

⋅+

⋅⋅=+++++

( )a

v

j

avddddddddd

2

7653214 −⋅

−=+++++−=





j

a

a

v

v

d

v

a

v

j

avd

t −−=

−

⋅−

=

2

4

++=

−−+

−⋅+

⋅=+⋅+⋅=

j

a

a

v

v

d

j

a

a

v

v

d

j

a

a

v

j

atttt 2424 421

3 Round Trip Time Calculations Revision 10 111010

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