+arndale Posted May 7, 2002 Share Posted May 7, 2002 Does anyone know if he speed shown on a GPS (mine's a Garmin E-map) is the horizontal ground speed or wether it takes into account any hills etc. The reason for the question is that a friend of mine wanted to borrow my GPS to see how fast he was skiing down a steep mountain side. Thanks Arndale Quote Link to comment
magellan315 Posted May 7, 2002 Share Posted May 7, 2002 Its the speed the GPS/you are moving at the that moment. If you are going uphill you will natuarlly be moving at a slower speed than going down hill. Quote Link to comment
+arndale Posted May 7, 2002 Author Share Posted May 7, 2002 Thanks for the answer but... I understand that if you are going up hill it is more likely you are going slower than downhill, but I would like to know if the speed is your actual speed or the speed in a horizontal direction. If you were going downhill at, say, a 60 degree angle you may actually travel 1 mile, but the distance on a map would show as less a distance. Hope that makes sense. Thanks Arndale Quote Link to comment
+Team Hoijong Posted May 7, 2002 Share Posted May 7, 2002 Well I dont know sure but i think it only counts horizontal speeds.. so when traveling vertical it will read zero. ehhh well.. i think i'm very wrong it will indicate also a speed when traveling vertical.. The GPS can compute elevation and can see from the satalietes postition how much you travel vertical and horizontal and will compute a real speed... Quote Link to comment
+Team Hoijong Posted May 7, 2002 Share Posted May 7, 2002 Well I dont know sure but i think it only counts horizontal speeds.. so when traveling vertical it will read zero. ehhh well.. i think i'm very wrong it will indicate also a speed when traveling vertical.. The GPS can compute elevation and can see from the satalietes postition how much you travel vertical and horizontal and will compute a real speed... Quote Link to comment
DisQuoi Posted May 7, 2002 Share Posted May 7, 2002 This Thread may answer your question. Quote Link to comment
Kerry. Posted May 7, 2002 Share Posted May 7, 2002 Horizontal only, no vertical taken into account and it's not going to show true downhill ski speed. Cheers, Kerry. I never get lost everybody keeps telling me where to go Quote Link to comment
Kerry. Posted May 7, 2002 Share Posted May 7, 2002 Horizontal only, no vertical taken into account and it's not going to show true downhill ski speed. Cheers, Kerry. I never get lost everybody keeps telling me where to go Quote Link to comment
Rocknroll Posted May 7, 2002 Share Posted May 7, 2002 isn't there some mathoholic that could take the drop in altitude and the speed and make this come out Quote Link to comment
Atilla the Pun Posted May 7, 2002 Share Posted May 7, 2002 First clear your backtrack, make your ski run and then save your backtrack to a route. Next you will calculate your speed between each pair of waypoints by the following method. Use the pythagorean theorem to obtain the actual distance traveled (or at least a number that's good enough for Gov't work) and keeping in mind that the VDOP is much greater than HDOP (vert. & horiz. Dilution of Precesion, so the altitudes the GPS records are less accurate than the horizontal positions, unless you have a calibrated barometric altimiter in your GPS) then figure the original time the GPS was calculating your speed at by multiplying your speed by the (horizontal) distance between waypoints. WPT1A = Waypoint1Altitude WPT2A = Waypoint2Altitude D = distance between waypoint 1 and 2 SQRT = take the square root of... SQRT((WPT1A-WPT2A) Quote Link to comment
Atilla the Pun Posted May 7, 2002 Share Posted May 7, 2002 Ok, this stupid board ate my reply, again. I'm going to a text editor to finish this and repost. AtP Quote Link to comment
Atilla the Pun Posted May 7, 2002 Share Posted May 7, 2002 Ok, this stupid board ate my reply, again. I'm going to a text editor to finish this and repost. AtP Quote Link to comment
Atilla the Pun Posted May 7, 2002 Share Posted May 7, 2002 to finish my physics dissertation..... The corrected formula that should have appeared above is: SQRT((WPT1A-WPT2A)^2+D^2 where ^2 means to square the value before it. This is the actual distance traveled between each pair of waypoints. Now, since speed=distance/time, and the GPS will give us the straight horizontal distance between each pair of waypoints, we can calculate the time it took you to travel between each pair of waypoints. You have horizontal distance and horizontal speed so the time to travel to each waypoint is HorizontalDistance*HorizontalSpeed. And now, finally, since you have the time it took you to travel from waypoint to waypoint and (from the first formula) the actual distance between wayopints, you can simply plug those values into this formula: Speed=Distance/Time and adjust for the units of time you want to use. It is recommended that you use a Magellan Meridian Platinum for this since the Magellans create more waypoints on a backtrack than Garmins do, and the MeriPlat is the only Maggy I know of that has a barometric altimiter. All this is off the top of my head, sorry if I rambled. Someone please check my assumptions and see if I swapped an operation or something. Thanks, AtP PS Clear as mud now? Quote Link to comment
Atilla the Pun Posted May 7, 2002 Share Posted May 7, 2002 to finish my physics dissertation..... The corrected formula that should have appeared above is: SQRT((WPT1A-WPT2A)^2+D^2 where ^2 means to square the value before it. This is the actual distance traveled between each pair of waypoints. Now, since speed=distance/time, and the GPS will give us the straight horizontal distance between each pair of waypoints, we can calculate the time it took you to travel between each pair of waypoints. You have horizontal distance and horizontal speed so the time to travel to each waypoint is HorizontalDistance*HorizontalSpeed. And now, finally, since you have the time it took you to travel from waypoint to waypoint and (from the first formula) the actual distance between wayopints, you can simply plug those values into this formula: Speed=Distance/Time and adjust for the units of time you want to use. It is recommended that you use a Magellan Meridian Platinum for this since the Magellans create more waypoints on a backtrack than Garmins do, and the MeriPlat is the only Maggy I know of that has a barometric altimiter. All this is off the top of my head, sorry if I rambled. Someone please check my assumptions and see if I swapped an operation or something. Thanks, AtP PS Clear as mud now? Quote Link to comment
peter Posted May 7, 2002 Share Posted May 7, 2002 quote:Originally posted by Atilla the Pun:The corrected formula that should have appeared above is: SQRT((WPT1A-WPT2A)^2+D^2 where ^2 means to square the value before it. This is the actual distance traveled between each pair of waypoints. Now, since speed=distance/time, and the GPS will give us the straight horizontal distance between each pair of waypoints, we can calculate the time it took you to travel between each pair of waypoints. You have horizontal distance and horizontal speed so the time to travel to each waypoint is HorizontalDistance*HorizontalSpeed. And now, finally, since you have the time it took you to travel from waypoint to waypoint and (from the first formula) the actual distance between wayopints, you can simply plug those values into this formula: Speed=Distance/Time and adjust for the units of time you want to use. It is recommended that you use a Magellan Meridian Platinum for this since the Magellans create more waypoints on a backtrack than Garmins do 1) Your equation is missing a parenthesis and should read: Actual Distance = SQRT((WPT1A-WPT2A)^2+D^2) 2) Don't use the backtrack function since doing this will eliminate the timestamps needed to calculate the speed. The tracklog gives you positions, altitudes, and times of successive points and is all you need. [Your procedure assumes you have a record of the horizontal speed between each set of backtrack points and I don't know of any GPS rcvr. that stores that data.] 3) Garmin and Magellan rcvrs. should work about the same. 4) In almost all cases you can skip all of this and just use the speeds shown by the GPS rcvr. directly without much error. A 30% grade is considered a steep ski run and on that grade the difference between horizontal difference and actual distance is less than 4.5%. And note that this assumes you're skiing straight down the run and not slalomming back and forth which would reduce the effective grade. If you know or can measure the grade then it's probably easiest to multiply the indicated horizontal speed by a correction factor (sqrt[1+grade^2]) to convert to actual speed. So for a 30% grade the correction factor is sqrt(1+.3^2) = sqrt(1.09) = 1.044. [This message was edited by peter on May 08, 2002 at 12:36 AM.] Quote Link to comment
Atilla the Pun Posted May 7, 2002 Share Posted May 7, 2002 quote:Originally posted by peter: 1) Your equation is missing a parenthesis and should read: Actual Distance = SQRT((WPT1A-WPT2A)^2+D^2) Thank you. I was in a hurry and missed that last close parenthesis when I recreated it. Also note that the D is actually the horizontal distance between the two waypoints. quote:2) Don't use the backtrack function since doing this will eliminate the timestamps needed to calculate the speed. The tracklog gives you positions, altitudes, and times of successive points and is all you need. [Your procedure assumes you have a record of the horizontal speed between each set of backtrack points and I don't know of any GPS rcvr. that stores that data.] Yeah, I was wondering about that as I was driving to work. But having the time of each waypoint makes it simpler. I also assume you mean "backtrack to route" function. quote:3) Garmin and Magellan rcvrs. should work about the same. Joe Mehaffey had a review with some screen shots showing that the Magellan receivers actually create more waypoints during a backtrack than the Garmins do. I would have posted a direct link but I couldn't find it right off. quote:4) In almost all cases you can skip all of this and just use the speeds shown by the GPS rcvr. directly without much error. A 30% grade is considered a steep ski run and on that grade the difference between horizontal difference and actual distance is less than 4.5%. And note that this assumes you're skiing straight down the run and not slalomming back and forth which would reduce the effective grade. If you know or can measure the grade then it's probably easiest to multiply the indicated horizontal speed by a correction factor (sqrt[(1+grade)^2]) to convert to actual speed. So for a 30% grade the correction factor is sqrt(1+.3^2) = sqrt(1.09) = 1.044. Now that's what I like, a simple answer. AtP Quote Link to comment
Atilla the Pun Posted May 7, 2002 Share Posted May 7, 2002 quote:Originally posted by peter: 1) Your equation is missing a parenthesis and should read: Actual Distance = SQRT((WPT1A-WPT2A)^2+D^2) Thank you. I was in a hurry and missed that last close parenthesis when I recreated it. Also note that the D is actually the horizontal distance between the two waypoints. quote:2) Don't use the backtrack function since doing this will eliminate the timestamps needed to calculate the speed. The tracklog gives you positions, altitudes, and times of successive points and is all you need. [Your procedure assumes you have a record of the horizontal speed between each set of backtrack points and I don't know of any GPS rcvr. that stores that data.] Yeah, I was wondering about that as I was driving to work. But having the time of each waypoint makes it simpler. I also assume you mean "backtrack to route" function. quote:3) Garmin and Magellan rcvrs. should work about the same. Joe Mehaffey had a review with some screen shots showing that the Magellan receivers actually create more waypoints during a backtrack than the Garmins do. I would have posted a direct link but I couldn't find it right off. quote:4) In almost all cases you can skip all of this and just use the speeds shown by the GPS rcvr. directly without much error. A 30% grade is considered a steep ski run and on that grade the difference between horizontal difference and actual distance is less than 4.5%. And note that this assumes you're skiing straight down the run and not slalomming back and forth which would reduce the effective grade. If you know or can measure the grade then it's probably easiest to multiply the indicated horizontal speed by a correction factor (sqrt[(1+grade)^2]) to convert to actual speed. So for a 30% grade the correction factor is sqrt(1+.3^2) = sqrt(1.09) = 1.044. Now that's what I like, a simple answer. AtP Quote Link to comment
Atilla the Pun Posted May 7, 2002 Share Posted May 7, 2002 Ok, it's not that the Garmin creates fewer waypoints for trackback, it just uses fewer of them in the actuall trackback. Here is the link to the screen shots made by Joe Mehaffey. AtP Quote Link to comment
peter Posted May 7, 2002 Share Posted May 7, 2002 quote:Originally posted by Atilla the Pun: Ok, it's not that the Garmin creates fewer waypoints for trackback, it just uses fewer of them in the actuall trackback. http://www.gpsinformation.net/mgoldreview/trackbacks.html is the link to the screen shots made by Joe Mehaffey. Yes, I'm familiar with that review and have already had an email discussion with Jack Yeazel about his screen shots since I feel that in actual hiking/biking the Garmin approach actually works much better. His screen shots are of a total hike of a few hundred feet and are not very realistic. We 'agreed to disagree.' However, the "trackback function" is irrelevant to the issue of finding speeds along the track since the original tracklog has all the information that's required. So neither the Magellan nor the Garmin would do any reduction of track points in this case. The active tracklog contains the following data for each point: time, date, lat., long., and altitude. That's all you need to calculate the horizontal, vertical, or slope speeds. Quote Link to comment
peter Posted May 7, 2002 Share Posted May 7, 2002 quote:Originally posted by mdmax371: It's your actual speed, I have checked my car speedometer with it and no matter what direction I'm traveling uphill or downhill the gps and my car show the exact same speed. Roads aren't usually steep enough for the horizontal and actual speeds to be significantly different. For example, the road up Kitt Peak in Arizona has a grade of 6% and is fairly typical for a mountain road. If you're driving up that road with a horizontal speed of 50.0 mph as shown on your GPS rcvr., then your actual speed along the slope will be 50.09 mph. That difference is too small for most of us to notice. OTOH, if you skydive and most of your velocity is vertical rather than horizontal then you'll quickly see that your GPS rcvr. is only reporting the horizontal component. Quote Link to comment
+niskibum Posted May 7, 2002 Share Posted May 7, 2002 "A 30% grade is considered a steep ski run" That depends on who you ask! If it ain't a double black diamond its a bunny hill. I have thought about checking my speed with my gps, but when you tuck it inside your jacket it loses lock and if you leave it hanging out it creates too much drag. In the words of Rudy in the movie Hot Dog "Yes, chinese downhill, it is the only way" (sounds a lot better with the German accent) Team Xscream Quote Link to comment
peter Posted May 8, 2002 Share Posted May 8, 2002 quote:Originally posted by niskibum: "A 30% grade is considered a steep ski run" That depends on who you ask! If it ain't a double black diamond its a bunny hill. The men's downhill course in the Utah Olympics had an average grade of 30.6% and the women's course had an average grade of 26%. Both had fairly short sections that were considerably steeper and where the horizontal vs. slope speed error would be higher, but the skier will generally carry his speed through to portions of the course with the average grade. I haven't been to those courses in person, but they didn't look like bunny hills to me. Quote Link to comment
+Team Hoijong Posted May 8, 2002 Share Posted May 8, 2002 My Magellan gives actual speed when tracking satalites.. It can count vertical speed and elevation so it computes with those two actual speeds.. Quote Link to comment
+pater47 Posted May 8, 2002 Share Posted May 8, 2002 If they make it to the bottom = fast enough If they stall midway = not fast enough If they fall and go sailing downhill out of control = too fast Ever notice everybody is willing to give THEIR 2 cents worth but only offer a penny for YOUR thoughts? Quote Link to comment
+niskibum Posted May 8, 2002 Share Posted May 8, 2002 I haven't been to those courses in person, but they didn't look like bunny hills to me. Like I said, it depends on who you ask. If you talk to Harry Egger, (top speed on skiis 248.104 kmh) who regularly skiis slopes of 57 degrees in a full tuck position, then yes, a 30 degree slope is a bunny hill. I am relativly sure that his speed was not calculated with a gps though--way too much drag! Quote Link to comment
irvingdog Posted May 8, 2002 Share Posted May 8, 2002 My kitty's breath smells like cat food......... my brain hurts.ö Quote Link to comment
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