CTS-D Calculations Flashcards by Richard Ellis

An architectural drawing was created using the U.S. customary scale of 1/8 inch = 1 foot. You measure a distance of 4.25 inches on the drawing. What is the actual distance

The actual distance is 34 feet

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An architectural drawing was created using the SI scale of 1:50. You measure a distance of 52 mm on the drawing.
What is the actual distance in millimeters?

The actual distance is 2,600 mm

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An architectural drawing was created using the SI scale of 1:200.

You measure a distance of 13 mm on the drawing. What is the actual distance in millimeters?

The actual distance is 2,600 mm

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An architectural drawing was created using the U.S. customary scale of 1/4 inch = 1 foot. You measure a distance of 5 and 3/8 inches on the drawing. What is the actual distance in feet?

The actual distance is 21.5 feet

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A screen has an aspect ratio of 1.78:1 and the following height dimension: 165.3 inches (4199 mm). Determine the width and diagonal of the screen.

Rounded to the nearest tenth of an inch, the screen width is 294.2 inches.

Rounded to the nearest tenth of an inch, the screen’s diagonal is 337.5 inches.

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An existing screen is 216.5 inches (5499 mm) wide by 216.5 inches (5499 mm) high. A projected image with a 1.78:1 aspect ratio covers the entire width of the screen. What is the image’s height?

Rounded to the nearest tenth of an inch, the screen height is 121.6 inches.

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What is the aspect ratio of a screen with a width of 108 inches (2743) mm) and a diagonal of 135 inches (3429)?

Rounded to the nearest hundredth, the aspect ratio is 1.33:1, or 4:3

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You require a 16:9 screen with a height of 60 inches (1524 mm).

What will the screen’s diagonal be?

The screen diagonal is 122.4 inches.

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A 16:9 screen will be installed in a lecture hall. The screen’s diagonal is 72 inches (1829 mm). What is its width?

Rounded to the nearest tenth of an inch, the screen width is 63 inches.

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There are two listeners in a training room. Listener One is 31 feet (9449 mm) from a loudspeaker and Listener Two is 17 feet (5182 mm) away.

What is the expected change in decibels at Listener One’s position when compared to Listener Two’s position? Round your answer to the nearest decibel.

= -5.218

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An audio amplifier outputs 100 watts and then it is decreased to 50 watts.

What is the change in decibels?

-3.0

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An audio amplifier outputs 75 watts and then it is decreased to 50 watts. What is the change in decibels?

-1.7

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An audio amplifier outputs 10 watts and then it is decreased to 2 watts. What is the change in decibels?

-7

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A presenter is playing a music CD from her laptop for two people in a room. Listener #1 is 2 meters away from the presenter.

Listener #2 is 15 meters away from the presenter. What is the expected loss in SPL at the listener #2 position?

-17.5

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A presenter is speaking to a large audience. Listener #1 is 2 meters away from the presenter. Listener #2 is 15 meters away from the presenter. What is the expected gain in SPL at the listener #1 position?

17.5

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Using a coverage angle of 90 degrees, a mounted loudspeaker height of 12 feet (3.7 meters), and the listener is seated, calculate the diameter of coverage.

16 feet (4.9 meters)

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Using a coverage angle of 60 degrees, a mounted loudspeaker height of 12 feet (3.7 meters), and the listener is seated, calculate the diameter of coverage.

9 feet (2.7 meters)

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Using a coverage angle of 70 degrees, a mounted loudspeaker height of 15 feet (4.8 meters), and the listener is seated, calculate the diameter of coverage.

15.4 feet (4.7 meters)

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Calculate loudspeaker spacing for Edge to Edge, Partial Coverage and 50% Overlap if:
Loudspeaker height is 168 inches (426.72 cm)
Ear height is 48 inches (121.92 cm)
Coverage angle is 60 degrees

Edge to edge spacing = D = 138.56 inches or 351.96 cm

Partial coverage = D = 97.98 inches or 248.87 cm

50 percent overlap spacing = 69.28 inches or 175.98 cm

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Calculate loudspeaker spacing if:

Loudspeaker height is 144 inches (365.76 cm)
Ear height is 48 inches (121.92 cm)
Coverage angle is 68 degrees

Edge to edge spacing = D = 72.66 inches D = 184.54 cm

Partial coverage spacing = D = 51.38 inches D = 130.49 cm

50 percent overlap spacing = 36.33 inches D = 92.27 cm

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Calculate loudspeaker spacing for Edge to Edge, Partial Coverage and 50% Overlap if:

Loudspeaker height is 112 inches (284.48 cm)
Ear height is 62 inches (157.48 cm)
Coverage angle is 72 degrees

Edge to edge spacing = D = 72.66 inches D = 184.54 cm

Partial coverage spacing = D = 51.38 inches 130.49 cm

50 percent overlap spacing = D = 36.33 inches D = 92.27 cm

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Calculate loudspeaker spacing for Edge to Edge, Partial Coverage and 50% Overlap if:

Loudspeaker height is 168 inches (426.72 cm)
Ear height is 48 inches (121 .92 cm)
Coverage angle is 60 degrees

Edge to edge spacing = D = 138.56 inches D = 351.96 cm

Partial coverage = D = 97.98 inches

50 percent overlap spacing = D = 69.28 inches

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Calculate loudspeaker spacing for Edge to Edge, Partial Coverage and 50% Overlap if:

Loudspeaker height is 144 inches (365.76 cm)
Ear height is 48 inches (121 .92 cm)
Coverage angle is 68 degrees

Edge to edge spacing = D = 72.66 inches D = 184.54 cm

Partial coverage spacing = D = 51 .38 inches D =
130.49 cm

50 percent overlap spacing = D = 36.33 inches D =
92.27 cm

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Calculate loudspeaker spacing for Edge to Edge, Partial Coverage and 50% Overlap if:

Loudspeaker height is 112 inches (284.48 cm)
Ear height is 62 inches (157.48 cm)
Coverage angle is 72 degrees

Edge to edge spacing = D = 72.66 inches D =
184.54 cm

Partial coverage spacing = D = 51 .38 inches
130.49 cm

50 percent overlap spacing = D
= 36.33 inches D =
92.27 cm

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In a speaking facility, what is the required power to a loudspeaker if: The farthest listener is 4 m from the loudspeaker. The target sound pressure level at this position is 75 dB The loudspeaker’s sensitivity is listed as 89 dB 1 w /1 m

6.37 watts are required

In a speaking facility, what is the required power to a loudspeaker if: The farthest listener is 8 m from the loudspeaker The target sound pressure level at this position is 70 dB The loudspeaker’s sensitivity is listed as 89 dB 1 w /1 m

= 8.06 w

In a music performance facility, what is the required power to a loudspeaker if: * The farthest listener is 3 m from the loudspeaker. * The target sound pressure level at this position is 75 dB * The loudspeaker’s sensitivity is listed as 1 00 dB 1 w /1 m

= 2.85 w

In a speaking facility, what is the required power to a loudspeaker if: * The farthest listener is 6 m from the loudspeaker * The target sound pressure level at this position is 70 dB * The loudspeaker’s sensitivity is listed as 93 dB 1 w /1 m

= 1.80 w

If the efficiency rating of an amplifier is 65% and the amplifier draws 7 amperes (A) at 120 V of alternating current (VAC) what is the heat load it generates?

2850 Btu/hr

According to best practices what is the safe working load (SWL) requirement for overhead mounting of a 75 pound (33.75 kg) projector?

375 pounds (168.75 kg)

Considering a 2400 watt lighting dimmer is available, the design load for the system should NOT exceed

1920 watt

What is the recommended (highest) ambient lighting level illuminating a front projections screen within a video conferencing space?

5 footcandles (50 lux)

Why might you recommend to an AV client to set up a dynamic host configuration protocol (DHCP) server system?

Reduced possibility of address duplication

During the design phase of a project that has many network-enabled AV devices. What will the IT team need from the AV designer?

Bandwidth requirements

Say you are working in a conference room with boundary microphones. * The microphone is approximately 3 feet (.9 meters) from the people seated at the table. * The microphone is 7 feet (2.1 meters) from the loudspeakers. * The farthest listening distance (from the source to the listener) is 25 feet (7.6 meters) * The listener to the loudspeaker distance is 7 feet (2.1 meters) * The presenter can be heard without amplification from 5 feet (1 .5 m) away. * There is 1 microphone on an automixer. * The system is equalized. From the given information, calculate both NAG and PAG. Then, compare the two values. Is PAG greater than NAG?

Yes

Say you are working on the road with a 60's era rock band. * The singer is holding a handheld microphone approximately .1 feet (0.03 meters) from his mouth. * The microphones are 40 feet (12.2 meters) from the loudspeakers. * The farthest listening distance (from the source to the listener) is 400 feet (122 meters) * The listener to the loudspeaker distance is 400 feet (122 meters) * The singer can be heard without amplification from 15 feet (4.6 m) away. * There are 10 microphones on the stage. * The system is equalized. From the given information, calculate both NAG and PAG. Then, compare the two values. Is PAG greater than NAG?

Yes

When installing conduit it is important to remember that | the conduit will require:

Adequate mounting support to withstand cable pulling

The first step when determining conduit capacity for cable should be to determine the:

Allowable fill percentage based upon local codes and | regulations.

Why might you recommend to an AV client to set up a dynamic host configuration protocol (DHCP) server system?

Reduced possibility of address duplication

During the design phase of a project that has many network-enabled AV devices. What will the IT team need from the AV designer?

Bandwidth requirements

How can the AV designer address ambient noise issues caused by HVAC equipment and ductwork vibration?

Silencers, baffles, and insulation panels

Reflected sound could be

An asset if arriving within 50 ms after the direct sound

If your venue is located in an earthquake prone region what mounting precautions need to be taken to ensure the safety and security of persons and property?

Stabilizer arms, sway bars, and safety cables should be utilised

When designing an AV control system in a building that requires the fire control system to be integrated, what is a major concern if the fire alarm initiates?

The lights staying on in the room

If you have 3 loudspeakers wired in parallel, each rated at 8 ohms, the circuit’s impedance is 1.33 ohms.

2.67 ohms

Benchmarking

Benchmarking refers to the process of examining methods, techniques, and principles from peer organisations and facilities, which can be used as a basis for designing a new or renovated facility

Program Report

A program report is a document that describes the client’s specific needs, system purpose and functionality, and the designer’s best estimate of probable cost, in a non-technical format for review and approval by the owner. Also known as the AV narrative, or discovery phase report, return brief, or concept design report

Foot-Candle

Footcandle, abbreviated "fc," is an English unit of measurement expressing the intensity of light illuminating an object. The illumination from one candle falling on a surface of 1 square foot at a distance of 1 foot

Candela

A candela is the luminous intensity of a source that emits monochromatic radiation of frequency 540 x 1012 hertz, and has a radiant intensity in that direction of 1/683 watt per steradian. The candela has replaced the candlepower standard

Lumen

Lumen is a measure of the light quantity emitted from a constant light source across one square meter.

Lux

Lux is a contraction of the words luminance and flux; 10.7 lux is equal to 1 foot-candles.

If you have 3 loudspeakers wired in parallel, with the first rated at 4 ohms, the second rated at 8 ohms, and the third rated at 16 ohms, the circuit’s impedance is 2.29 ohms

Z = 1/((1/4)+(1/8)+(1/16))

Headroom

Headroom is the difference in dB SPL between peak and average level performance of an audio system. For a speech application, the recommended value is 10 dB and for program audio, the recommended value is 20 dB. ref

FSM

Feedback Stability Margin - refers to how close the system is to actual feedback. Any system on the edge of feedback will produce a ringing behavior prior to actual feedback. typically 6 dB

What size EMT (electrical metallic tubing) conduit is required when using a single RGBHV cable with an outer diameter of 0.792 inches (20. 12mm), a 53% fill percentage?

D > 1 .09 inches (27.6 mm

What size of EMT(electrical metallic tubing) conduit will be required when installing 2 miniature RGBHV cables with an outer diameter of 0.315 inches (8mm) and 3 composite video cables with an outer diameter of 0. 152 inches (3.86mm). The fill percentage is 40 percent

D > 0.818 inches (20.7 mm)

Three miniature RGBHV cables with an outer diameter of 0.474 inches (11.79 mm) are being installed in a conduit. Using electrical metallic tubing (EMT), what size conduit will be required and does a potential cable jam exist?

D > 1 .3 inches (33 mm)

Sampling Rate

Four Times the Frequency

If the calculation falls within this range, the next larger size conduit should be specified to avoid any potential jam

2.8 to 3.2

What size of EMT (electrical metallic tubing) conduit will be required when Installing 2 RGBHV cables with a circumference of 0.9896 inches and 3 composite video cables with a circumference of 0.8514.

ID=1.023, 1" EMT

What size of EMT (electrical metallic tubing) conduit will be required when Installing 8 audio cables with a circumference of 0.4084 inches, 8 CAT5 cables with a circumference of 0.5969 inched and 2 CAT5 Shielded cable with a circumference of 0.7948?

1-1/4" EMT

What is structured cabling?

A building or campus telecommunications cabling infrastructure that consists of a number of standardized smaller elements called subsystems.

RoHS

Lead-Free, stands for Restriction of Hazardous Substances also known as Directive 2002/95/EC, originated in the European Union and restricts the use of six hazardous materials found in electrical and electronic products.

Bandwidth

measure of the width of a range of frequencies, measured in hertz. The rate of data transfer, bit rate or throughput, measured in bits per second (bit/s)

Your customer wishes to stream CD (44.1 kHz ) quality music at 16 bits in stereo to his lobby area. How much bandwidth will he require?

1.4112 MBPS

You need to stream 10 channels of 96 kHz audio. You have 25 Mbps of bandwidth available. What is the highest bit depth you can use?

26.042 Bit Depth

You currently have the bandwidth capacity to stream 30 channels of 48 kHz, 24 bit audio. How many channels could you stream if you upgraded to 96 kHz, 24 bit audio?

15 Channels

What is the bit depth of an uncompressed digital black and white video at 30 fps with a resolution of 800 x 700 at 16 bit?

268.8 Mbps

What is the required bandwidth for 4:4:4 NTSC digital video (720 H X 480 V, 8 bits at 30 Hz)?

248.832 Mbps

What is the required bandwidth for a 4:4:4 computer image (1280 H X 1024 V, 8 bits at 80 Hz)?

2.5165824 Gbps

What is the required bandwidth for a 4:4:4:4 computer image (1280 H X 1024 V, 8 bits at 80 Hz)?

4.1418752 Gbps

What is the required bandwidth for 4:4:4:4 digital video (720 H X 480 V, 8 bits at 30 Hz)?

331.776 Mbps

What is the required bandwidth for 4:2:2 NTSC digital video (720 H X 480 V, 8 bits at 30 Hz)?

165.888 Mbps

What is the required bandwidth for 4:1:1 NTSC digital video (720 H X 480 V, 8 bits at 30 Hz)?

124.416 Mbps

A client requires a 4:4:4 computer image (1280 H X 1024 V, 8 bits at 80 Hz) with audio at ( 96 kHz ) at 16 bits in stereo and an overhead of 25% of the total used bandwidth. With 10 Gbps total bandwidth will you exceed recommended bandwidth capacity?

Yes ( 3.149568 Gbps)

A client requires 4:1:1 NTSC digital video (720 H X 480 V, 8 bits at 30 Hz) with audio at ( 96 kHz ) at 16 bits in stereo and an overhead of 25% of the total used bandwidth. With 10 gbps total bandwidth will you exceed bandwidth capacity?

No (1.5936 Gbps)

Lossless compression

Definition: allows a mathematically exact recovery of the original file.

Lossy Compression

Lossy compression methods, such as JPEG, discard information, which can result in image degradation via the appearance of compression artefacts or jaggies. Lossy methods are typically used for photographic images where loss of fidelity is acceptable.

Area of a Square

A=LW

Area of Rectangle

A=LW

Area of a circle

A = pi r2

Circumference of a Circle

c = pi x diameter

Inverse square law equation

(Intensity old/Intensity new) = ((Distance new)^2 / (Distance old)^2)

EAD (Equivalent Acoustic Distance)

is the farthest distance one can go from the source without the need for sound amplification or reinforcement to maintain good speech intelligibility. It is a design parameter dependent on the level of the presenter and the noise level in the room.

Bandwidth Capacity

Based on the estimates that only 70 percent of the rated capacity is considered available, and only 30 percent of the available capacity is available for streaming, consider 21 percent of a network's rated capacity available for streaming.

Assume a loudspeaker is generating 80 dB SPL at a distance of 22 feet (7m) away from the source outdoors. What would the level be at 88 feet (27m) away?

80 + (-12) = 68 dB SPL