Height of the hammer’s arc
Printed From: Nasgaweb
Category: Nasgaweb Forums
Forum Name: A.D.
Forum Discription: This forum is for Athletic Directors to discuss issues involving running a Heavy Events competition.
URL: http://www.nasgaweb.com/forum/forum_posts.asp?TID=7998
Printed Date: 3/26/26 at 3:57pm
Software Version: Web Wiz Forums 10.11 - http://www.webwizforums.com
Topic: Height of the hammer’s arc
Posted By: WALLY.OLECIK
Subject: Height of the hammer’s arc
Date Posted: 10/28/08 at 4:37pm
|
Now that the weather's going down the tubes, perhaps someone can help me out on a question l have. What would be the height of the arc of a 16 pound hammer throw that goes, say, 135 feet? This isn't just idle curiosity, we have a field that is rather short on one side and l'd like to know that the fence on that side is tall enuf to stop a long throw. lt would be very cool if one of you brain types out there could break the arc down into heights at ten foot intervals from 100 feet on up. Anyone?? lt would be very helpful and much appreciated. |
Replies:
Posted By: climber511
Date Posted: 10/29/08 at 2:55am
I am not the brain type - but here's a guess. The fence will be 1" to short to stop the hammer from going thru the windshield of the only Cadillac parked in the whole lot 
|
Posted By: WALLY.OLECIK
Date Posted: 10/31/08 at 12:22am
|
Wayne Hill? McSanta?? (lt's not everyday that l admit to not knowing something so give a guy a little help!!!) ------------- 16lb-hammer(at)sshga.org "Try not. Do or do not. There is no 'try!'" Yoda |
Posted By: Wayne Hill
Date Posted: 10/31/08 at 2:44am
|
OK, OK, I've been busy and sick and stuff. The issue here is the release angle. I wouldn't mind if someone checked me on this, but (neglecting shoulder height) I get Hmax = Distance * tan(theta) / 4 where theta is the release angle. In a nominally optimal ballistic trajectory, theta would be 45 degrees, so tan(theta) would be one, and the maximum height would be one fourth the distance of the throw (33' 9", for a 135' throw). In reality, because of the length of the implement, the release angle is almost always smaller, so the maximum height is less. For a 40 degree release angle, which is a pretty high release for a Scottish hammer, the max height would be close to one fifth the distance of the throw (27', for a 135' throw). Either of these is pretty darn high, so your fence probably isn't this high. However, depending on the real setupof the field, it may well be that the fence doesn't need to be nearly this high (so tell us more). ------------- "We may be small, but we're slow." - MIT Rugby |
Posted By: C. Smith
Date Posted: 10/31/08 at 3:40pm
|
I have hit a ceiling fan that was 35' or 40' with the 22lb. hammer at the Shamrock Games. Kevin Rogers can verifiy. ------------- |
Posted By: Wayne Hill
Date Posted: 11/01/08 at 2:53am
|
It helps to have a target.
------------- "We may be small, but we're slow." - MIT Rugby |
Posted By: WALLY.OLECIK
Date Posted: 11/02/08 at 9:43pm
The way we're set up, we throw from the S.W. corner of the field aiming at the N.E. corner. We have plenty of space except at the S.E. corner where the fence is just over 100' away. The bleachers in this area are set high enuf that the crowd is looking over the fence which is 10' tall (l think). So - what l'm reading is that this max height of 27' would be at the mid-point, or 67' 6" from the trig, down to 0' at 135". lf l'm thinking in the right direction, the hammer would be about 14' feet off of the ground at 100'. lf l'm close with the numbers, the hammer should not be more than about 10' off of the ground when it gets to the fence at between 100 and 120'. Am l close?? |
Wayne Hill wrote:Posted By: Wayne Hill
Date Posted: 11/03/08 at 12:22pm
|
The problem is that, at the top of the arc, the implement is traveling horizontally, so the path from there to touchdown starts flat and gradually steepens. At 100', the throw with the 40 degree release angle (i.e, the lower throw) is still at 20'9"(!). I'm not really sure how likely this is to cause a problem, though, because off-line throws usually aren't nearly as long as ones in the "sector". The problem is the word usually... ------------- "We may be small, but we're slow." - MIT Rugby |
Posted By: WALLY.OLECIK
Date Posted: 11/03/08 at 11:53pm
This could be a real problem because our field record is 133'+ and if someone made a similar throw down that sideline, we could be in deep doo-doo! Maybe l need to get someone (in better shape than myself) to make a few throws down the sideline with the 12# hammer to get some numbers so l can see how worried l should be. Thanks for working the numbers for me. l appreciate it. |
Posted By: Steve Conway
Date Posted: 11/04/08 at 2:33pm
Wouldn't the velocity of the hammer have something to do with the distance that it travels? I tend to agree with climber511; regardless of the height of the wall, the hammer will find a way to get over it  Always plan on the worst case scenario happening...
|
Posted By: McSanta
Date Posted: 11/05/08 at 5:47am
|
I can send you a spread sheet that will plot the trajectory of a hammer ignoring air resistance that is driven by input of : release angle and release velocity. However, that probably will not give you what you want I freely admit I am a math geek who avoided physics with utmost prejudice. However, if you want statistical analysis or want average time to death or other actuarial calculation ...
------------- Mark McVey "The work of science is to substitute facts for appearances and demonstrations for impressions." -John Ruskin |
Posted By: McSanta
Date Posted: 11/05/08 at 11:21am
|
Assuming NO drag --- formulas from: http://www.grc.nasa.gov/WWW/K-12/airplane/ballflght.html - http://www.grc.nasa.gov/WWW/K-12/airplane/ballflght.html t is time from release; x is horizontal distance; y is height I have ignored the height of the thrower which will increase the distance a bit as well as the height. Launch angle is 44 degrees which is close to optimal from what I can tell from the model -- biomechanical articles also indicate this Launch velocity is 66.05 ft / sec^2 ---
------------- Mark McVey "The work of science is to substitute facts for appearances and demonstrations for impressions." -John Ruskin |
Posted By: McSanta
Date Posted: 11/05/08 at 11:43am
|
I have the drag case worked out and will post later. I have in spreadsheet so if anybody want to double check the calculations, improve on them, .... my feelings will not be hurt. Since this is a safety issue, it would be proper and wise to have the formulas checked out as we do not want a safety incident anywhere let alone at one of the big games. I make no claims to accuracy and if the formulas are used to make safety judgement, you use them at your own risk.... in other words I do not want to be sued because I tried working on a math problem in an area where I have NO formal training! ------------- Mark McVey "The work of science is to substitute facts for appearances and demonstrations for impressions." -John Ruskin |
Posted By: McSanta
Date Posted: 11/05/08 at 6:54pm
|
If I got the formulas correct, the following graph is the path assuming
------------- Mark McVey "The work of science is to substitute facts for appearances and demonstrations for impressions." -John Ruskin |
Posted By: McSanta
Date Posted: 11/05/08 at 7:03pm
|
Two graphs the first shows the difference in distance when factoring drag and holding all else the same.
A more interesting result which maybe more helpful as the no drag model is simplier to work with. The plot of two paths in which the resulting distances are about 135' but one path consider drag, and the other does not. They paths lie almost on top of each other.
------------- Mark McVey "The work of science is to substitute facts for appearances and demonstrations for impressions." -John Ruskin |
Posted By: McSanta
Date Posted: 11/06/08 at 4:08am
|
last few caeveats: I did not consider the release height, which will slightly change the distance traveled as well as the trajectory. If my spreadsheet is correct, drag does not have a great impact on the path traveled and thus should not have a great impact on the problem you want to solve. The formulas at: http://en.wikipedia.org/wiki/Trajectory_of_a_projectile - http://en.wikipedia.org/wiki/Trajectory_of_a_projectile are good and with a spreadsheet others can plug and chug values in to answer some questions. ------------- Mark McVey "The work of science is to substitute facts for appearances and demonstrations for impressions." -John Ruskin |
Posted By: WALLY.OLECIK
Date Posted: 11/06/08 at 4:43pm
|
Wayne Hill and McSanta - Thanks so much!! Looks like l've got some changes to make to the lay-out of our field! ------------- 16lb-hammer(at)sshga.org "Try not. Do or do not. There is no 'try!'" Yoda |
Posted By: C. Smith
Date Posted: 11/25/08 at 10:47am
|
Those graphs are nice and all, but i'm telling you guys i have been 35'+ high with a 22lb hammer and it still landed at 101+. ------------- |
Posted By: McSanta
Date Posted: 11/25/08 at 11:54am
|
Not to fear, in the results sent to Wally, I included a "limiting" or worse case scenario. In this scenario, I assumed the thrower could not produce a release speed greater than that obtain on a wire-hammer by Olympic level thrower. (I found this value in a biomechanical journal and that was70 ft per second.) I am not an expert on sports medicine or bio mechanics but I believe this is a solid limiting assumption as the wire hammer is shorter in length and thrown with a movement that should generate more speed than the Scottish Hammer. Playing around with the launch angle, it appears that a launch angle of just under 60 degrees would produce the highest hammer throw (over 60 feet) and land about 135 out. At 100ft out, the hammer is 47' off the ground! If one assumed no drag, it is a simple mathematical solve to find the release angle that would produce the highest path at 100 ft out given a release speed. I did not get around to solving that for him. Never the less, Wally was given the limiting case which gives him a height of 47' to worry about. By his comments, I assume he is making approrpiate adjustments. ------------- Mark McVey "The work of science is to substitute facts for appearances and demonstrations for impressions." -John Ruskin |
Posted By: McSanta
Date Posted: 11/25/08 at 11:56am
|
From a practical point of view, how tall would a thrower have to be to launch the Scottish hammer at a 60 degree angle?
I think the thrower would have to be rather tall to launch the Scottish hammer on a 60 degree angle!!! Same calculation but a 45 degree angle, and the end of the hammer handle is just over 35 inches off the ground. So maybe 6'6 to 6'8" tall? ------------- Mark McVey "The work of science is to substitute facts for appearances and demonstrations for impressions." -John Ruskin |
Posted By: AncientOne
Date Posted: 11/26/08 at 5:22am
|
Hey- By the way, as I recall Craig Smith hit a ceiling fan mounted on steel cross beam in the Coliseum on the Illinois State Fairgrounds during the Shamrock Games. The cross beam is at 40' above the ground surface and the fan is a few feet lower. He spun the blades without breaking any and still threw the heavy hammer to 104'. -K ------------- Winners are remarkably adept at figuring out what's required to win. |
Posted By: McSanta
Date Posted: 11/26/08 at 6:49am
|
Release Angle: 52 degree Speed of hammer head upon release 57.5 ft/sec (82.5% of Oly level wire hammer thrower who is throwing Wire hammer is shorter in length and lighter implement) Drag coefficient = 0.72 Hammer head's diameter is 4.5" These values yield: Height just over 38' at just over 50' horizontal distance from launch point. Distance just over 104' ------------------------------- I am not a physicist or an engineer... However, the input (angle and velocity) to reproduce Craig's throw seems reasonable ... perhaps a bit steep on release angle but I don't never know. Given the coliseum earth (no turf to dig under, just well packed soil and hourse droppings?), how hammer boots don't always hold (excpecially later in the day ...but Super A's thrower earlier), early season form, .... maybe the angle is reasonable as the athletes maynot be as aggressive with the hammer throw in other conditions. I really don't know and the above maybe a plausible explanation of the release angle. --------------------- It is always cool watching the Shammies in the coliseum.
------------- Mark McVey "The work of science is to substitute facts for appearances and demonstrations for impressions." -John Ruskin |
Posted By: McSanta
Date Posted: 11/26/08 at 7:09am
|
In case anybody is interested: Some hammer flight paths that where generated for Wally's problem, the projected path of the highest hammer (at 100' out), the possible path for the world record throw, and a possible path of Craig Smith's imfamous or famous heavy hammer throw that hit the ceiling fan at the Shamrock Games.
------------- Mark McVey "The work of science is to substitute facts for appearances and demonstrations for impressions." -John Ruskin |
Posted By: WALLY.OLECIK
Date Posted: 1/05/09 at 2:42pm
|
Once again, thanks go out to Wayne Hill and McSanta for all of the assistance that they've given me on my safety issues. l've decided to restrict the hammer throws to a very narrow impact sector (40 degrees) to impress upon the throwers the importance of keeping the hammer away from the spectator seating area. This is the size of sector that will be used at the Master's World and slightly wider that used at many Games in Scotland. This won't change the length of the field on that side but if an athlete is going to throw that far, it seems that they will also be able to control the direction of their throws. |