ARCHITECTURAL | Design & Site Planning | 10 Questions (CHALLENGING)
TRICKY AND UNCOMMON DESIGN PROBLEMS
(10 Questions, Difficulty Level:CHALLENGING)
by Raison John J. Bassig
Q#1: You designed a supermarket having a total of five (5) aisles. The center aisle measures 3.20m in width, while the rest of the aisles measure 1.15m in width each. If you laid out a cross aisle where the said aisles will terminate, what is the smallest size of that cross aisle you can specify?
a. 3.20-meter wide cross aisle
b. 4.35-meter wide cross aisle
c. 5.50-meter wide cross aisle
d. 6.08-meter wide cross aisle
e. 7.80-meter wide cross aisle
According to the 2004 Implementing Rules and Regulations (IRR) of the National Building Code of the Philippines (PD 1096),
"Aisles shall terminate in a cross aisle, foyer, or exit. The WIDTH OF THE CROSS AISLE shall NOT BE LESS than the SUM OF THE REQUIRED WIDTH OF THE WIDEST AISLE PLUS 50% OF THE TOTAL REQUIRED WIDTH OF THE REMAINING AISLE leading thereto."
(Rule XII, Sec.1207.12.c. of the 2004 IRR of PD 1096)
In my given problem, we have a total of 5 aisles (including the center aisle). The center aisle is [3.2m] wide (which is the widest of all aisles). The remaining 4 aisles are [1.15m] wide each.
Based on the provisions of the Code, the minimum width of the cross aisle where all 5 aisles originate/terminate would be,
Minimum Width of Cross Aisle = [Widest Aisle] + [50% of the Sum of All Other Aisles]
Minimum Width of Cross Aisle = [3.2m] + 0.5 (4 other aisles x [1.15m per aisle])
Minimum Width of Cross Aisle = [3.2m] + 0.5 (4.6m)
Minimum Width of Cross Aisle = [3.2m] + 2.3m
Minimum Width of Cross Aisle = 5.5m
Therefore, the correct answer is c. 5.50-meter wide cross aisle.
Q#2: As their architect, a newly-married couple consulted you about the ceiling height of their existing circular bedroom, which is artificially-ventilated, with ample window openings, and located at the 3rd floor of their house. The room's circumference measures exactly 10.25m, while the existing ceiling height is 2400mm. Telling you that they want their room to conform with all applicable provisions of Light and Ventilation in the National Building Code, which of the following should you recommend to them?
We should increase the ceiling height by 950mm.
It is recommended to lower the ceiling by 300mm.
Your ceiling must be exactly 2.7m in height.
No problem, sir/ma'am, your ceiling height already conforms with the Code.
According to the 2004 Implementing Rules and Regulations (IRR) of the National Building Code of the Philippines (PD 1096),
"Habitable rooms provided with artificial ventilation shall have ceiling heights NOT less than 2.40 meters measured from the floor to the ceiling; provided that for buildings of more than one (1) storey, the minimum ceiling height of the first storey shall be 2.70 meters and that for the second storey 2.40 meters and the succeeding stories shall have an unobstructed typical head-room clearance of NOT less than 2.10 meters above the finished floor. Above-stated rooms with natural ventilation shall have ceiling heights of NOT less than 2.70 meters."
(Rule VIII, Sec. 805.1 of the 2004 IRR of PD 1096)
If ONLY the height of the ceililng is taken into account (using the above Sec. 805 of PD 1096), then, the ceiling of the couple's room is already above the minimum requirement for a habitable room with artificial ventilation located above the second story, and, thus, conforms with such code provision.
BUT!!! If you read the question carefully, the couple's DESIRE is to conform with ALL APPLICABLE provisions of the entire Rule VIII (Light and Ventilation) of the Code, which means that we must NOT ONLY consider Sec. 805 alone, but ALSO consider OTHER provisions based on the data given in the problem (note that I specifically mentioned the room size in the problem -- which is already a hint to the type of question you were looking at).
Since a bedroom is a type of HABITABLE ROOM, provisions in the code include its minimum room size (area and volume) and ceiling heights as seen in Sections 805, 806, and 807 of Rule VIII.
So, even though, requirements for Sec. 805 (Ceiling Heights) has already been met, we still need to check the other two (2) provisions (Sec. 806 and Sec. 807) if it does indeed meet the minimum code requirements.
In the code, Sec. 806 provides the minimum dimension of "Rooms for Human Habitations" as:
"Rooms for Human Habitations - 6.00 sq. meters with a least dimension of 2.00 meters;"
(Rule VIII, Sec. 806.1.a of the 2004 IRR of PD 1096)
Now, let's check if their existing room conforms with Sec. 806. Based on the given data in the question, the Circumference (C) of their circular bedroom is [10.25m]. Since [C = π x diameter], we can easily solve for it's area (A), where [A = π x (diameter)2 / 4],
A (of circle) = π x (diameter)2 / 4
A (of circle) = 3.14... x (3.26...m)2 / 4
A (of circle) = approx. 8.35 sq.m.
The calculated area of the couple's circular bedroom, at ~8.35 sq.m., conforms with the minimum provision of Sec. 806 (which is 6.00 sq.m. minimum).
Now we check Sec. 807 (Air Space Requirements). In the code, it states that,
"Minimum air space shall be provided as follows:...xxx...c. Habitable Rooms - 14.00 cu. meters of air space PER PERSON."
(Rule VIII, Sec. 807.1.c of the 2004 IRR of PD 1096)
Note that in the given problem, your client is a NEWLY-MARRIED COUPLE (i.e., 2 persons -- which is another hint). So, the minimum air space requirement of the bedroom can be calculated as,
Required Air Space (Volume) = [14.00 cu.m./person] x 2 persons = 28 cu.m.
Since we have previously solved the area of the bedroom [~8.35 sq.m.], let's check if their existing ceiling height, given in the problem as [2400mm] (or 2.4m), conforms with Sec. 807,
Existing Air Space (Volume) = [Area of Room] x [Existing Ceiling Height]
Existing Air Space (Volume) = [8.35 sq.m.] x [2.4m]
Existing Air Space (Volume) = approx. 20 cu.m. ==> does NOT conform with the Code!
Ooops! Based on our above computations, we have just found out that their existing room volume of approx. 20 cu.m. DOES NOT conform with the provisions stated in Sec. 807! As we have calculated, it should have been 28 cu.m. minimum.
This means that, to conform with the Code, we must ADJUST the ceiling height HIGHER to create MORE air space or room volume:
[Minimum Required Air Space or Volume] = [Existing Room Area] x Minimum Required Ceiling Height
Minimum Required Ceiling Height = [Minimum Required Air Space or Volume] / [Existing Room Area]
Minimum Required Ceiling Height = [28 cu.m.] / [8.35 sq.m.]
Minimum Required Ceiling Height = approx. 3.35m
From their existing ceiling height of 2.4m, the minimum required ceiling height to conform with the provisions of Rule VIII should be made HIGHER (to ~3.35m as computed above). So, the increase in height would be,
3.35m - 2.4m = 0.95m or 950mm
So, your recommendation to your newly-wed clients should be to increase the ceiling height by 950mm, as all the OTHER CHOICES given in my question will NOT CONFORM with the Code.
Therefore, the correct answer is a. We should increase the ceiling height by 950mm.
Q#3: For safety purposes, you recommended to your client the use of a plastic skylight (instead of glass) to be installed as part of your client's requirement for her 54-sqm Dining Area. She wants to have four (4) large skylights of equal sizes arrange across the ceiling so as to save on electrical costs in that area. She wants the largest possible skylight possible without violating the National Building Code. As her architect, which among the following dimensions would you specify for each plastic skylight for this purpose?
a. 4 sets of 1.5m x 1.8m skylights
b. 4 sets of 2m x 4m skylights
c. 4 sets of 2.6m x 6.75m skylights
d. 4 sets of 2m x 5m skylights
According to the 2004 Implementing Rules and Regulations (IRR) of the National Building Code of the Philippines (PD 1096),
"The area of INDIVIDUAL plastic skylights shall NOT exceed 10.00 SQUARE METERS. The TOTAL AGGREGATE AREA of plastics used in skylights, monitors, and sawtooth glazing shall NOT exceed 20% of the FLOOR AREA of the room or occupancy sheltered."
(Rule XVI, Sec. 1604.3 of the 2004 IRR of PD 1096)
Since the skylight will serve a 54-sqm room (i.e., the Dining Area), the maximum TOTAL AGGREGATE AREA of the entire plastic skylight would be,
Maximum Total Aggregate Area (of the Entire Plastic Skylight) = [Floor Area of Room] x 20%
Maximum Total Aggregate Area (of the Entire Plastic Skylight) = [54 sqm] x 0.20
Maximum Total Aggregate Area (of the Entire Plastic Skylight) = 10.8 sqm
As your client requires four (4) skylights, then,
Area of Each Individual Plastic Skylight = [Maximum Total Aggregate Area of Skylight] / [Number of Required Individual Skylights]
Area of Each Individual Plastic Skylight = [10.8 sqm] / [4]
Area of Each Individual Plastic Skylight = 2.7 sqm
So, among the choices I gave,
Choice c.) [2.6m x 6.75m] can immediately be eliminated since the area of that skylight is [2.6m x 6.75m = 17.55 sqm], which EXCEEDS the maximum provision of the Code (i.e., 10 sqm area for an individual skylight).
Both Choice c.) [2m x 4m] and Choice d.) [2m x 5m] (or 8 sqm and 10 sqm, respectively) conforms to said provision in terms of individual skylight area, but, as the project would have 4 sets of such individual skylights, the total aggregate skylight areas (8 sqm x 4 = 32 sqm and 10 sqm x 4 = 40 sqm, respectively) will EXCEED the maximum provision of the Code (max. of 20% of the floor area sheltered, which is only 10.8 sqm total).
Only Choice a.) [1.5m x 1.8m] conforms with the Code as the area of individual skylight is [1.5m x 1.8m = 2.7 sqm] AND the total aggregate area is [2.7 sqm x 4 = 10.8 sqm], exactly the maximum aggregate area of skylight allowed by the Code based on the room area it serves.
Therefore, the correct answer is a. 4 sets of 1.5m x 1.8m skylights.
Q#4: If you are to design a multi-level parking building, irrespective of dimensions, what is the most recommended and ideal vehicular flow along driveways and ramps that you must employ among the following choices?
a. Two-way flow with 45-degree parking slots
b. One-way flow; counter-clockwise turns
c. One-way flow; no left turns
d. Two-way flow with parallel parking
One-way flow of vehicles inside a parking building limits the width of the driveway. It also forces vehicles to flow in one direction only thereby creating less "points of collisions". One-way flow also lessens possible accidents, such as when pedestrians are crossing the driveway or when a vehicle is backing up to the driveway, as one would only be primarily concerned to look at a single direction (where the traffic is coming from) instead of both directions.
A two-way parking layout versus a one-way parking layout.
Now, counter-clockwise turns (or left turns) is more ideal than clockwise turns (or right turns) because most, if not all, of the vehicles are left-hand-side drive (i.e., driver is located at the left). The turning would be more controlled when turning left than when turning right, especially on circular ramps.
Therefore, the correct answer is b. One-way flow; counter-clockwise turns.
Q#5: The audience-seating of the theater you designed has a central aisle width of 2.36m measured between the seats at the front row. The foyer, which serves as the exit, is located at the back of the audience seating area. If the total length of the aisle, from the front row to the back row is 15m, what then must be the width of this central aisle measured at the back row where such egress is located?
a. Width of 8.68 meters
b. Width of 3.20 meters
c. Width of 3.26 meters
d. Width of 2.36 meters
According to the 2004 Implementing Rules and Regulations (IRR) of the National Building Code of the Philippines (PD 1096),
"Every aisle shall be NOT less than 800 millimeters wide if serving only one side, and NOT less than 1.00 meter wide if serving both sides. Such minimum width shall be measured AT THE POINT FARTHEST FROM an EXIT, CROSS AISLE, or FOYER and shall be INCREASED by 30 MILLIMETERS for EVERY METER in LENGTH towards the exit, cross aisle or foyer."
(Rule XII, Sec. 1207.12.a of the 2004 IRR of PD 1096)
The solution for the width of the aisle located at the back row would then be,
Back Width = [Front Width] + 2(Length of Aisle - 1m)(0.03m)
The expression (Length of Aisle - 1m) means that the first meter of the aisle's length has NOT yet been increased by 0.03m (or 30 millimeters). The increase will start AFTER a 1 meter length (hence, we subtract 1m from the length of the aisle).
The "2" in 2(Length of Aisle - 1m)(0.03m) means that BOTH sides of that aisle will have a 30mm increase in width (for a total of 60mm) for every meter in length.
So, using the data I gave in my problem,
Back Width = [2.36m] + 2([15m] - 1m)(0.03m)
Back Width = [2.36m] + 2(14m)(0.03m)
Back Width = [2.36m] + 0.84m
Back Width = 3.20m
Therefore, the correct answer is b. Width of 3.20 meters.
Q#6: In an R-3 Zone, your client owns a rectangular inside lot with a 5-m frontage and a 17-m depth. Your client requires you to design a 2-storey residence with firewalls on both sides and suggested to put the said firewalls along the whole depth of his property for security reasons.
As an architect who knows the provisions of the National Building Code, what would then be your professional recommendation to your client?
The firewall must only be at one side, at a maximum of 13.6m in length.
Your client's suggested firewall length must be decreased by 2.70m on each side.
The firewalls can only be put at a maximum of 14.45m on each side.
You agree with the client as his suggestion conforms with the code.
According to the 2004 Implementing Rules and Regulations (IRR) of the National Building Code of the Philippines (PD 1096),
"For a R-3 USE or occupancy WITH A FIREWALL on TWO (2) SIDES, a firewall can be erected on a MAXIMUM of 85% of the TOTAL LENGTH of EACH SIDE property line; provided that ALL FIREWALL construction shall NOT exceed 65% of the TOTAL PERIMETER of the R-3 property, i.e., total length of all property lines, provided that firewalls in R-3 lots shall ONLY be allowed for a MAXIMUM TWO (2) STOREY component structure; and provided further that all the applicable stipulations of the Fire Code are strictly followed."
(Rule VII, Sec. 704.4.c.v.a of the 2004 IRR of PD 1096)
Given the data in my problem and the above Code provision on maximum firewall length, 85% of the side property lines (having a length of 17 meters on each side) is equal to,
Maximum Firewall Length (on Each Side) = [Length of Side Property Line] x 85%
Maximum Firewall Length (on Each Side) = [17m] x 0.85
Maximum Firewall Length (on Each Side) = 14.45m
However, it is important to note that the Code further specifically states that ALL firewall construction shall NOT exceed 65% of the TOTAL PERIMETER of a given lot.
In our problem, the lot perimeter would be,
Lot Perimeter = Summation of Lengths of all Property Lines
Lot Perimeter = [5m front] + [5m rear] + [17m left-side] + [17m right-side]
Lot Perimeter = 44m
So, the maximum length of all firewall construction at 65% of the lot perimeter is,
Total Maximum Length of All Firewalls = [Lot Perimeter] x 65%
Total Maximum Length of All Firewalls = [44m] x 0.65
Total Maximum Length of All Firewalls = 28.6m
So, IF we use our previously-computed maximum firewall length of [14.45m] (which was based on the 85% of the side property line), the total firewall construction will come out to be,
Total Length of All Firewalls = 14.45m (at left-side) + 14.45m (at right-side)
Total Length of All Firewalls = 28.9m ==> does NOT conform with the Code!
As we can see, the total length of all firewalls, have exceeded the maximum total length prescribed by the Code (computed at 28.6m max.).
So, if the maximum length of all firewalls must be [28.6m] as computed based on 85% of lot perimeter, and we have two side property lines of equal length to be provided with firewalls, we can get the maximum length of each firewall as,
Maximum Firewall Length (on Each Side) = [85% of Lot Perimeter] / 2 sides
Maximum Firewall Length (on Each Side) = [28.6m] / 2 sides
Maximum Firewall Length (on Each Side) = 14.3m ==> this length will now conform with the Code.
Thus, your client's suggestion to construct a 17-m firewall on each side (the whole length) is noncorfing to the Code. Based on our above computations, the maximum length of firewall should only be 14.3m per side. This would be a difference of [17.00m - 14.30m = 2.70m], thus, you must then recommend to your client to decrease his suggested firewall length by 2.70m on each side to conform with the Code.
Therefore, the correct answer is b. Your client's suggested firewall length must be decreased by 2.70m on each side.
Q#7: Your client, a movie theater owner, commissioned you to design her cinema's projection room. Most of the films they have are in the 35-millimeter format. Which of the following dimensions is the largest possible opening for the projection port you can specify for your client if such projection port will be covered with a pane of glass?
a. 23.5" x 29.75"
b. 325mm x 400mm
c. 0.25m x 0.30m
d. 50cm x 90cm
According to the 2004 Implementing Rules and Regulations (IRR) of the National Building Code of the Philippines (PD 1096),
"The provisions of this Section shall apply ONLY where ribbon-type motion picture films IN EXCESS of 22-MILLIMETER WIDTH and electric projection equipment are used.
(Rule XII, Sec. 1214.1 of the 2004 IRR of PD 1096)
Since a 35-mm film will be used in the movie theater, then, all the provisions of Section 1214 apply. Now, let's check the Code for provisions on the dimensions of the projection port opening,
"Ports Required - There shall be provided for each motion picture projector NOT MORE THAN ONE (1) PROJECTION PORT, which shall be LIMITED IN AREA to 750 SQ. CENTIMETERS, and not more than one (1) observation port, which shall be limited in area to 1,300 sq. centimeters. There shall be not more than three (3) combination ports, each of which shall not exceed 750 millimeters by 600 millimeters. Each port opening shall be COMPLETELY COVERED WITH A PANE OF GLASS; Except, that when acetate safety film is used, projection ports may be increased in size to an area not to exceed 4,500 sq. centimeters."
(Rule XII, Sec. 1214.4.a of the 2004 IRR of PD 1096)
So, it is clearly stated in the Code that projection ports shall be 750 sq.cm. (or equal to 0.075 sq.m.) maximum IF the opening is covered with GLASS.
Among the choices of opening sizes I have listed, the areas would be as follows,
Choice a.) 23.5" x 29.75" = 699.125 sq.in. or ~4,500 sq.cm. ==> size does NOT conform with the Code.
Choice b.) 325mm x 400mm = 130,000 sq.mm. or 1,300 sq.cm. ==> size does NOT conform with the Code.
Choice c.) 0.25m x 0.30m = 0.075sq.m. or 750 sq.cm. ==> size conforms with the Code.
Choice d.) 50cm x 90cm = 4,500 sq.cm. ==> size does NOT conform with the Code.
Therefore, the correct answer is c. 0.25m x 0.30m.
Q#8: You were hired by a very superstitious client who is fond with the number "8". She wants you to design and build her residence to be located at #8 Eight Street, Brgy. 888 in Project 8. Your client requires that her house will have 8 sides each measuring 8.88m in length. She then asks you, as her architect, what would be the area of her house? What will be your reply?
a. 223 sqm.
b. 380.74 sqm.
c. 200.96 sqm.
d. 888.88 sqm.
e. 415.43 sqm.
Since the shape is a polygon with equal sides, it is a REGULAR polygon. So we can use the formula,
A = [ (Side)2 x (# of Sides) ] / [ 4 x Tan(180 / # of Sides) ]
Solving the area of the house in my problem,,
A = [ (8.88m)2 x (8) ] / [ 4 x Tan (180 / 8) ]
A = 630.8352 sqm / (4 x Tan22.5)
A = 380.74 sqm
Therefore, the correct answer is b. 380.74 sqm.
Q#9: You have just designed a 49-storey medium-cost residential condominium. The podium component is 8 storeys high, with commercial spaces occupying the entire ground floor, the entire 7th floor and the entire 8th floor. All amenities are housed in the tower's 9th floor, while the top four (4) floors are occupied by seven (7) units of 115-sqm suites per floor. The other floors in the tower component will have the following units per floor: five (5) units of 21-sqm studio-type rooms; three (3) units of 34-sqm 1-Bedroom units; four (4) units of 51-sqm 2-Bedroom units; and two (2) units of 87-sqm 3-Bedroom units.
How many car parking slots should you provide for the residential units in this condominium building to conform with provisions of the National Building Code?
a. 92 car parking slots
b. 123 car parking slots
c. 314 car parking slots
d. 532 car parking slots
According to the 2004 Implementing Rules and Regulations (IRR) of the National Building Code of the Philippines (PD 1096), on Minimum Parking Requirements for R-5 Condominium Use/Occupancy,
"For 18-sqm to 22-sqm units -> Use 1 parking : 8 units
For Up to 50-sqm units -> Use 1 parking : 6 units
For Above 50-sqm to 100-sqm units -> Use 1 parking : 4 units
For More than 100-sqm -> Use 1 parking : 1 unit"
(Rule VII, Table VII.4, Item 1.2 for R-5 buildings of the 2004 IRR of PD 1096)
(Rule VII, Table VII.4, Item 1.2 for R-5 buildings of the 2004 IRR of PD 1096)
In my given problem, data shows that we have different types and sizes of residential units. Let us first analyze, breakdown, and organize all given information in my problem. Since the question only seeks the car parking slots for the Residential Units, we will only concentrate on the data provided for such Residential Units (disregarding the Commercial Spaces), for ease of solving.
Number of Floors with Residential Units = [Total Floors of the Building] - [All Floors with NO Residential Units]
Number of Floors with Residential Units = [49 Floors] - [8 Floors (Podium)] - [1 Floor (Amenity at 9/F)]
Number of Floors with Residential Units = 40 Floors (i.e., from 10/F up to 49/F)
For the "Top Four (4) Floors" which is from 46/F up to 49/F, we have,
115-sqm Suites: [7 Units of 115-sqm Suites per Floor] x [4 Floors] = 28 Total Units
For the Other Floors, i.e., from 10/F up to 45/F, which is a total of 36 Floors, we have,
87-sqm 3BR: [2 Units of 87-sqm 3BR Units per Floor] x [36 Floors] = 72 Total Units
51-sqm 2BR: [4 Units of 51-sqm 2BR Units per Floor] x [36 Floors] = 144 Total Units
34-sqm 1BR: [3 Units of 34-sqm 1BR Units per Floor] x [36 Floors] = 108 Total Units
21-sqm Studios: [5 Units of 21-sqm Studio Units per Floor] x [36 Floors] = 180 Total Units
So, based on the provision for the minimum parking requirements in the table from the Code, considering our building's residential unit sizes (we have 21-sqm, 34-sqm, 51-sqm, 87-sqm, and 115-sqm unit types),
For 18-22 sqm units @ 1:8 parking ratio ==> Applies to all 180 of our 21-sqm Studio Units
For up to 50-sqm units @ 1:6 parking ratio ==> Applies to all 108 of our 34-sqm 1BR Units
For 50-100 sqm units @ 1:4 parking ratio ==> Applies to all 144 of our 51-sqm 2BR and to all 72 of our 87-sqm 3BR Units
For more than 100-sqm units @ 1:1 parking ratio ==> Applies to all 28 of our 115-sqm Suites
Thus,
180 Total Studio Units x [1:8] = 22.5 or 23 car parking slots
108 Total 1BR Units x [1:6] = 18 car parking slots
216 Total 2BR and 3BR Units x [1:4] = 54 car parking slots
28 Total Suites x [1:1] = 28 car parking slots
Total # car parking slots for all residential units = 123 slots
Therefore the correct answer is b. 123 car parking slots.
Q#10: You are designing a straight-run staircase, with a riser height of 7-2/5", without any landings, connecting one floor to an upper floor. You want to find out the length of the space this stairs would occupy using the proportion formula RT=74. Checking the cross-section drawings, you saw that the floor-to-floor elevation is at 9'-3". What would be the total stair run (including the last tread/step)?
a. 12'-6"
b. Approx. 5.4545m (or 18')
c. 111.6"
d. 400cm
Ideally, if all data are given in a certain system, try to solve the problem within the same given system (as possible). The problem with conversions (especially when involving non-absolute/non-exact values) is that the answers will vary to some extent because some values of the measurements are "lost" due to the rounding off.
Moreover, take important note that the formula RT=74 (meaning, Riser x Tread = 74) is an English-based system of proportioning stairs. Using metric units with the same exact formula will NOT yield correct and exact results.
Let's first simplify all given values to same units (in inches):
Given Total Stair Rise = 9'-3" = (9' x 12") + 3" = 108" + 3" = 111"
Given Riser Height = 7-2/5" = 7" + (2/5)" = 7" + 0.4" = 7.4"
Now, we can also determine the number of steps/risers of such stair (which we need later on):
In question 5 I think its wrong. because "INCREASED by 30 MILLIMETERS for EVERY METER in LENGTH towards the exit, cross aisle or foyer.". the word EVERY METER means the whole length so no need to subtract 1...
In question 5 I think its wrong. because "INCREASED by 30 MILLIMETERS for EVERY METER in LENGTH towards the exit, cross aisle or foyer.". the word EVERY METER means the whole length so no need to subtract 1...
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