I know the formula for calculating Space mean speed is pretty straight forward, but I am getting confused by the calculation shown in HCM.

If you look at page 7-3 in the end the SMS is supposed to be 38.298, I cannot understand how HCM has arrived at 25.5 mi/h.

I used nL/ (sum of travel times); n is the no. measurements, L is the distance. Can some one please review this for me.

Thanks.

# Space Mean Speed

Started by
winner9459
, Mar 17 2011 01:38 AM

5 replies to this topic

### #1

Posted 17 March 2011 - 01:38 AM

### #2

Posted 17 March 2011 - 02:39 AM

QUOTE (winner9459 @ Mar 16 2011, 09:38 PM) <{POST_SNAPBACK}>

I know the formula for calculating Space mean speed is pretty straight forward, but I am getting confused by the calculation shown in HCM.

If you look at page 7-3 in the end the SMS is supposed to be 38.298, I cannot understand how HCM has arrived at 25.5 mi/h.

I used nL/ (sum of travel times); n is the no. measurements, L is the distance. Can some one please review this for me.

Thanks.

If you look at page 7-3 in the end the SMS is supposed to be 38.298, I cannot understand how HCM has arrived at 25.5 mi/h.

I used nL/ (sum of travel times); n is the no. measurements, L is the distance. Can some one please review this for me.

Thanks.

Where in the HCM do you see the example? I see the formula at the end of page 7-3, but don't see a numerical example. I'm looking at HCM 2000

### #3

Posted 17 March 2011 - 04:14 PM

Probably this text would help you locate it.

"It is possible to calculate both time mean and space mean speeds from a sample of

individual vehicle speeds. For example, three vehicles are recorded with speeds of 30,

40, and 50 mi/h. The time to traverse 1 mi is 2.0 min, 1.5 min, and 1.2 min, respectively.

The time mean speed is 40 mi/h, calculated as (30 + 40 + 50)/3. The space mean speed is

38.3 mi/h, calculated as (60)[3 ÷ (2.0 + 1.5 + 1.2)]."

Thanks.

Where in the HCM do you see the example? I see the formula at the end of page 7-3, but don't see a numerical example. I'm looking at HCM 2000

"It is possible to calculate both time mean and space mean speeds from a sample of

individual vehicle speeds. For example, three vehicles are recorded with speeds of 30,

40, and 50 mi/h. The time to traverse 1 mi is 2.0 min, 1.5 min, and 1.2 min, respectively.

The time mean speed is 40 mi/h, calculated as (30 + 40 + 50)/3. The space mean speed is

38.3 mi/h, calculated as (60)[3 ÷ (2.0 + 1.5 + 1.2)]."

Thanks.

QUOTE (civilized_naah @ Mar 17 2011, 02:39 AM) <{POST_SNAPBACK}>

QUOTE (winner9459 @ Mar 16 2011, 09:38 PM) <{POST_SNAPBACK}>

I know the formula for calculating Space mean speed is pretty straight forward, but I am getting confused by the calculation shown in HCM.

If you look at page 7-3 in the end the SMS is supposed to be 38.298, I cannot understand how HCM has arrived at 25.5 mi/h.

I used nL/ (sum of travel times); n is the no. measurements, L is the distance. Can some one please review this for me.

Thanks.

If you look at page 7-3 in the end the SMS is supposed to be 38.298, I cannot understand how HCM has arrived at 25.5 mi/h.

I used nL/ (sum of travel times); n is the no. measurements, L is the distance. Can some one please review this for me.

Thanks.

Where in the HCM do you see the example? I see the formula at the end of page 7-3, but don't see a numerical example. I'm looking at HCM 2000

### #4

Posted 17 March 2011 - 04:32 PM

Space-man Speed sounds like a comic book character.

### #5

Posted 17 March 2011 - 06:16 PM

But, the text you extracted shows the SMS calculated as 38.3 (correct answer). Where is the 25.5 you are referring to?

### #6

Posted 17 March 2011 - 09:33 PM

that's interesting, the hard copy I am referring to has the wrong answer which made me go hmmmmm.....The text I posted is from the soft copy of HCM I have. Probably they updated it with the corrections.

Thanks for the help.

Thanks for the help.

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