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For someone who is 5'2(157cm) tall and weighing at 75kg(165lbs). How much force is required for that person to jump onto a box that is 66.9cm tall. Your help is appreciated
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Re: Force required for a vertical jump
The back of the envelope calculation is Force = Mass * Acceleration. -
Re: Force required for a vertical jump
Where's the "Like" button when you need it!Originally posted by ecramp48 View PostComment
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Re: Force required for a vertical jump
To jump onto the box you need a theoretical vertical velocity of 3.62 m/s [v=square root(2*g*box height)] with g - gravity acceleration 9.81 m/s²
Therefore, you need a vertical linear momentum of p=272 Ns [p=m*v] to accelerate the COM
Your maximum ground reaction force depends on the type (squat or countermovement) and duration of the jump and the shape of the force curve.
In case of a rectangular shape and a duration of 0.3s your vertical ground reaction force is 1642 N [F=p/s+m*g]
I have attached a file with the calculation and a typically force of a countermovement jump
ChristianComment
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Re: Force required for a vertical jump
Most instances of box jumping I have seen have the person landing in a crouched position, and then they stand erect, so to be more accurate, you need the height difference between their starting COM and the crouched-landing COM. The exact posture/COM could vary a lot even for subjects all 167 cm and 75 kg.
TerryComment
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Re: Force required for a vertical jump
Dear Mr. Jay,
The attached information may be useful in your research, however, the use of instrumentation will be required.
Exss 3850 9 summer linear kinetics
"Force plate in jumping" - slide 33
Attached file: Explanation of the bilateral deficit in human vertical squat jumping.pdf
Best Regards,
Wagner de Godoy
Biomech-L_Explanation of the bilateral deficit in human vertical squat jumping.pdfBiomech-L_Explanation of the bilateral deficit in human vertical squat jumping.pdfComment
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Re: Force required for a vertical jump
Jay,
An estimation of peak vertical jump force with assumptions or guesses marked with a (?) there are quite a few
We will call the displacement of the subject’s center of mass (CoM) in standing as zero. Let’s say that the displacement of the CoM at the point of take-off is raised 5cm (?) relative to standing. They will bend their legs when landing on the box, so let’s say this reduces required maximum CoM displacement by 25cm (?). The net vertical displacement of the CoM in flight is then:
66.9 – 5 – 25 = 36.9 cm
The vertical take-off velocity is: v = sqrt(2xgx0.369) = 2.69 m/s, and
Momentum at take-off is: P = 75x2.69 = 201.8 kg.m/s
The concentric (upwards) movement of a squat jump has gone from zero velocity to the take-off velocity. So the net impulse, the change in momentum, in this case is the final momentum = 201.8 N.s.
If this impulse was achieved in 0.5 sec (?) then the average force applied over this time period is: 403.6 N (impulse = force x time). This is not the peak force. Peak force may be twice (?) the average force in a squat jump (?). This does include the force due to gravity so we have to add on body weight (BW) to get total vertical force.
The peak force may therefore be around: [403.6 x 2] + BW = 807.2 + (75 x 9.81) = 1,543 N
For a counter movement jump (CMJ) the initial standing displacement of the subject CoM is still considered zero. There is an initial negative impulse relative to body weight followed by a positive impulse relative to body weight; the net impulse will still be equal to 201.8 N.s to achieve the same vertical CoM displacement of 36.9 cm. However, at the minimum body position force will be above body weight (it was equal to BW at the beginning of the squat jump), and with a constant impulse required (201.8 N.s) the difference between average force and peak force during the concentric phase will likely be less if the force is applied over the same time period (0.5 sec). It is likely that in the CMJ the time period is shortened.
If this same impulse was achieved in 0.3 sec (?) then the average force applied over this time period is: 672.7 N
The peak force may therefore be around: [672.7 x 1.5] + BW = 1009.0 + (75 x 9.81) = 1,745 N
It is all a bit of guess work to get average and peak force during the concentric phase. Knowing the duration of the concentric phase would help the estimation of average force; peak force however is still a fairly crude estimate and I am just guessing what it might be relative to average force during the concentric phase. I have not attempt to estimate the peak force during the eccentric phase of the CMJ.
AllanComment
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Re: Force required for a vertical jump
I think these responses illustrate the point that that simple questions usually have complex answers.
All of these points are good ones, and I highly recommend doing the math to estimate the answer. However all of these answers are still estimates because the only way to really know the answer is to measure it. You might think that a relatively simple setup with a force measuring device would give you an answer, but there are still many external factors to account for - many of these factors are "human" factors ... a subject who is confident will probably jump differently to someone who is not sure that they can do it, someone who has practiced will probably be very efficient and use less energy than someone trying it for the first time.
A very confident and agile subject might use a lot of force, jump and clear the box by half a metre before landing, while someone who is worried that the box might fail when they land on it, or that they might fall off the box as they land, will jump very cautiously with less force. If the box is quite low, a few centimeters, then the force required will be low - and we would expect that more force will be needed as the box height increases, but if subject is worried that the box height is close to the limit of their ability, they may drastically increase the force to make sure that they clear the top and land without tripping.
So even the simple act of measuring has complications.Comment
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