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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