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Math Help How many seconds from my turn am I
Message
From
22/02/2013 12:13:47
 
 
To
22/02/2013 11:56:52
General information
Forum:
Visual FoxPro
Category:
Coding, syntax & commands
Environment versions
Visual FoxPro:
VFP 9 SP1
OS:
Windows XP SP2
Network:
Windows 2003 Server
Database:
MS SQL Server
Application:
Desktop
Miscellaneous
Thread ID:
01566768
Message ID:
01566770
Views:
93
This message has been marked as a message which has helped to the initial question of the thread.
>Any math experts that can validate this (Just like the old school questions)?
>
>A mile has 5,280 feet, and an hour has 3,600 seconds, so 60 miles per hour is:
>60 x 5,280 / 3,600 = 88 fps
>(60 mph = 88 fps )
>
>How to determine how many seconds from your destination you are (example driving in a car and you are 1500' from your next turn):
>
>1. Determine distance to travel (1500 feet)
>2. Determine speed
>3. Determine feet per second
>4. Determine how many seconds from the destination you are
>
>Let’s say you are 1500 feet from your turn and you are going 30 mph. You want to be notified 7 seconds from the turn.
> 30 x 5280 / 3600 = 44 feet per second
> 1500 feet / 44 feet per second = you are 34.090909 seconds from the turn
>
>I can process this every second until I am 7 seconds out. Is that the best method (assuming you could speed up or slow down it would need to be checked/computed every second)
>
>Looks right to me, but can anyone verify that is accurate?

You could store fps at 1mph (1.46666.....) to avoid repeating the 5280/3600. Then you'd just have Distance/1.46666/mph
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