SweatyYeti 2,101 Posted September 10, 2015 (edited) ^ Thanks, BH....that's interesting. Drew wrote: Can't we just calculate the speed, by using the horizontal measurement between TC-1 and TC-2? that is 12 feet. From RP's point of view Patty's path is about 10' How many frames does it take Patty to walk the 10' feet between TC-1 and TC-2? Actually, Drew....we should be able to calculate Patty's 'step length', by using that 12' length between trees TC-1 and TC-2. Just taking a look at the number of steps she takes covering that 12 feet.....(using MK Davis' stabilized version)....I get this, for a step length... Step Length : TC1 - TC2 = 12' 11.5' x 12 = 138" 138" / 3 steps = 46" per step. Patty walks almost parallel to the line of TC-1 - TC-2.....but not exactly....so I used a slightly shorter distance, of 11.5 feet. It looks like Patty took 3 steps between these two Frames....(my numbering of frames in MK's animation)... Those 3 steps...divided by 138"....gives us a step length of 46". And, that's without fully straightening her legs. Edited September 10, 2015 by SweatyYeti Share this post Link to post Share on other sites
Drew 499 Posted September 10, 2015 Remember the 12' is an actual field measurement, we are looking at Patty when the camera is not perpendicular to that measurement. So we have to account for that. Also, are you using centerline to centerline for the tree? heel to heel on the step? etc... It could be beetween 40-45" but still at least you are not basing it on a video of a film, with stops and starts in it to calculate your speed. Share this post Link to post Share on other sites
brianmk 2 Posted September 10, 2015 Sweaty Based on Patty's arm swing, could you say she took 2.5 steps between TC-1 TC-2? Share this post Link to post Share on other sites
SweatyYeti 2,101 Posted September 10, 2015 (edited) ^ Here are a few earlier frames that show Patty's arm position, along with her legs...for 2.5 steps... From her right arm being fully extended, to fully extended again is two steps. And then there's one more step...or a portion, thereof. Each step takes about 11.5 frames. But, if Patty covered that length in only 2.5 steps...then she was taking mighty long steps. Remember the 12' is an actual field measurement, we are looking at Patty when the camera is not perpendicular to that measurement. So we have to account for that. Also, are you using centerline to centerline for the tree? heel to heel on the step? etc... It could be beetween 40-45" but still at least you are not basing it on a video of a film, with stops and starts in it to calculate your speed. I'll reply to your post later today, Drew. Edited September 10, 2015 by SweatyYeti Share this post Link to post Share on other sites
Bigfoothunter 1,659 Posted September 10, 2015 (edited) Remember the 12' is an actual field measurement, we are looking at Patty when the camera is not perpendicular to that measurement. So we have to account for that. Also, are you using centerline to centerline for the tree? heel to heel on the step? etc... It could be beetween 40-45" but still at least you are not basing it on a video of a film, with stops and starts in it to calculate your speed. As I recall - there were only two start and stops in the film. None occurring once Roger closed the gap between he and the subject. Edited September 10, 2015 by Bigfoothunter Share this post Link to post Share on other sites
Guest Posted September 10, 2015 I had another look and the stoppages shouldn't affect the speed calculations at all. 52 seconds for 952 frames @ 11.2 frames per step, 41" per step is all the info that is needed. Had Roger not stopped the camera we would have the same number of frames on the reel and recorded the same number of steps. The only thing that changes is the total distance Patty walked across the creek bed from the moment Roger started the camera to when the reel ran out. Patty probably took closer to 100 steps instead of the 85 steps captured on film. But this doesn't matter because Patty's avg walking speed was based on the film not her trackway. ps Drew, don't tell me that more steps changes things and don't ask me a question I've already answered. Also, from the camera's POV TC-1 & TC-2 are NOT 12' apart and there are not enough sample steps to calc an average speed. But go ahead and redo the calculation to get an avg step length from TC-1 to TC-2. Then you might believe 41" is a good avg. Share this post Link to post Share on other sites
Drew 499 Posted September 10, 2015 Right, but the video has nothing to do with it. Share this post Link to post Share on other sites
brianmk 2 Posted September 10, 2015 Sweaty In the image you posted above (shown below), between the point where Patty is centered on TC-1 until the point where she is centered on TC-2 her right arm cycles 1.5 times. In other words, when she is centered on TC-1 her right arm is straight out and left foot is 1/2 a stride past centerline. By the time her right arm is straight forward again her left foot is planted on the centerline of TC-2. As she is stepping on the centerline of TC-2 her right arm then moves to a straight down position as she is vaulting (transitioning) over her left foot. Would this not be 2.5 steps? I wish I was able to do a GIF as she moves TC-1 to TC-2 Share this post Link to post Share on other sites
Bill 1,942 Posted September 10, 2015 Just for the record, the frames being displayed are the scans made by Rick Noll, not Mr. Davis. Credit should be given where due. Rick scanned a 2x zoom in print from the Green-Dahinden copy master, using a camera on a microscope and digitizing the frames at 3K resolution. We know this by the circular shadowing in the corners, which only the Noll frame scans have. 1 Share this post Link to post Share on other sites
Bigfoothunter 1,659 Posted September 10, 2015 Right, but the video has nothing to do with it. . You concern yourself with the number of steps taken on the film between two points seen on the film and then respond saying that the film has nothing to do with it. Explain yourself, Drew Share this post Link to post Share on other sites
Drew 499 Posted September 10, 2015 I didn't say the FILM, I said the video mentioned in the OP is irrelevant to the measurement. Share this post Link to post Share on other sites
SweatyYeti 2,101 Posted September 10, 2015 Sweaty In the image you posted above (shown below), between the point where Patty is centered on TC-1 until the point where she is centered on TC-2 her right arm cycles 1.5 times. In other words, when she is centered on TC-1 her right arm is straight out and left foot is 1/2 a stride past centerline. By the time her right arm is straight forward again her left foot is planted on the centerline of TC-2. As she is stepping on the centerline of TC-2 her right arm then moves to a straight down position as she is vaulting (transitioning) over her left foot. Would this not be 2.5 steps? I wish I was able to do a GIF as she moves TC-1 to TC-2 I'll look at that more closely later on tonight, Brian. And if I have time, I'll put together an animation of those frames. Bill wrote: Just for the record, the frames being displayed are the scans made by Rick Noll, not Mr. Davis. Credit should be given where due. Rick scanned a 2x zoom in print from the Green-Dahinden copy master, using a camera on a microscope and digitizing the frames at 3K resolution. We know this by the circular shadowing in the corners, which only the Noll frame scans have. Thanks for the info, Bill. I've only seen that animation referred to as MK's stabilized animation. In the future, I'll label 'still images' from it using Rick Noll's name, or initials. Share this post Link to post Share on other sites
Bill 1,942 Posted September 11, 2015 MK stabilized the Noll scans over MK's scan of F352 transparency to produce that stabilization. Share this post Link to post Share on other sites
Drew 499 Posted September 11, 2015 Bill, thanks for clearing that up. Share this post Link to post Share on other sites
Drew 499 Posted September 11, 2015 Bill, Sweaty, Gigantofootecus- What are your ideas on the angle patty is taking with regard to the walk between TC-1 and TC-2? 0 Deg line is perpendicular to the camera location, as shown on the sketch. I just threw a couple together, again, i'm not a photogrammetrist, but we could at least get close here with respect to the TC-1 and TC-2 tree measurement. I can throw some speed numbers out there too, but they are just based on this sketch, so have a look. Odinn posted a graphic in which he seemed to show that frame 389 was at tc1 and 426 was at tc2. If these are correct that is 37 frames or a hair over two seconds at 18 frames per second. So that means subject was traveling at 4.5 ft per second or about 3.1 mph. (3.4 if the body angle was 45 degrees). Share this post Link to post Share on other sites