# Vertical Jump Power Calculators

Vertical Jump tests are usually just reported as the distance jumped in centimeters or inches. This does not always tell the full story. A heavier person jumping the same height as a lighter one has to do much more work to move a larger mass. Therefore it is sometimes useful to convert the score to units of power or work.

The mechanical work performed to accomplish a vertical jump can be determined by using the jump height distance that was measured (using Work = Force x Distance where Force = Mass x Acceleration). However, Power can not be calculated (Power = Work / time) since the time that force is acted on the body is unknown. Power can be directly measured using a force plate, though these are not readily available. Over time a few different formula have been developed that estimate power from vertical jump measurements. A few of these are presented below, with examples.

The examples below all use a hypothetical vertical jump of 60 cm (0.6 meters or 23.5 inches) by a person 75 kg and 180 cm. As you can see, there are widely different results achieved. This is partly due to it not always being clear if peak power or average power is being estimated.

## Lewis Formula

The Lewis formula or nomogram (Fox & Mathews, 1974) is a commonly used formula (found in many high school text books). This formula only estimates average power, and is based on a modified falling body equation. The original formula used the units of kg·m·sec.-1. To convert it to Watts, the standard unit for Power, the factor of 9.81 has been added.

Average Power (Watts) = √ 4.9 x body mass (kg) x √ jump-reach score (m) x 9.81

### Example

- Average Power = (square root of 4.9) x body mass(kg) x (square root of jump distance(m)) x 9.81
- Average Power = 2.2136 x 75 x 0.7746 x 9.81
- Average Power = 1261.6 Watts

## Harman Formula

To improve on the limitations of the Lewis formula, Harman et al. (1991) established equations for
both peak and average power through multiple regression procedures. The two equations
are listed below. In the first equation, it may actually be **-1,822** instead of **+ 1,822**, see comments below.

Peak power (W) = 61.9 · jump height (cm) + 36.0 · body mass (kg) + 1,822

Average power (W) = 21.2 · jump height (cm) + 23.0 · body mass (kg) – 1,393

#### Examples

- Peak power (W) = (61.9 x jump height (cm)) + (36 x body mass (kg)) + 1822
- Peak power (W) = (61.9 x 60) + (36 x 75) + 1822
- Peak power (W) = 3714 + 2700 + 1822
- Peak power (W) = 8236 Watts
- Average power (W) = (21.2 x jump height (cm)) + (23.0 x body mass (kg)) – 1393
- Average power (W) = (21.2 x 60) + (23 x 75) - 1393
- Average power (W) = 1272 + 1725 - 1393
- Average power (W) = 1604 Watts

## Johnson & Bahamonde Formula

Johnson and Bahamonde (1996) also developed a formula for the calculation of peak and average power from the vertical jump test, using the countermovement jump. These equation use the additional factor of body height.

Power-peak (W) = 78.6 · VJ (cm) + 60.3 · mass (kg) - 15.3 · height (cm) - 1,308

Power-avg (W) = 43.8 · VJ (cm) + 32.7 · mass (kg) - 16.8 · height (cm) + 431

#### Examples

- Peak power (W) = (78.6 x VJ (cm)) + (60.3 x mass (kg)) - (15.3 x height (cm)) -1308
- Peak power (W) = (78.6 x 60) + (60.3 x 75) - (15.3 x 180) - 1308
- Peak power (W) = 4716 + 4522.5 - 2754 - 1308
- Peak power (W) = 5176.5 Watts
- Average power (W) = (43.8 x VJ (cm)) + (32.7 x mass (kg)) - (16.8 x height (cm)) + 431
- Average power (W) = (43.8 x 60) + (32.7 x 75) - (16.8 x 180) + 431
- Average power (W) = 2628 + 2452.5 - 3024 + 431
- Average power (W) = 2487.5

## Sayers Formula

The Sayers Equation (Sayers et al. 1999) also estimates peak power output (Peak Anaerobic Power output or PAPw) from the vertical jump.

PAPw (Watts) = 60.7 · jump height(cm) + 45.3 · body mass(kg) - 2055

#### Example

- PAPw = (60.7 x jump height(cm)) + (45.3 x body mass(kg)) - 2055
- PAPw = (60.7 x 60) + (45.3 x 75) - 2055
- PAPw = 3642 + 3397.5 - 2055
- PAPw = 4984.5 Watts

## Bosco Formula

A formula for average power measurement from the Bosco Repeated Vertical Jump Test has been determined. The average power generated (W) is calculated from the test duration (Ts from 15 to 60 s), the number of jumps (n) total flight time (Ft ), and where g is the acceleration due to gravity, so that;

**W = (Ft Ts g2) / 4n (Ts - Ft)**

## Other

I have also come across this formula for calculating power from vertical jump height. This formula was used until recently at the NHL Combine.

**Power (ft-lb/sec) = 4 x weight (lb) x
√ jump height (ft)
**

## References

- Bosco C, Luhtanen P, Komi PV (1983)
*A simple method for measurement of mechanical power in jumping*.**European Journal of Applied Physiology**50:273-282. - Harman, E.A., Rosenstein, M.T., Frykman, P.N., Rosenstein, R.M., and Kraemer, W.J.

(1991).*Estimation of Human Power Output From Vertical Jump*.**Journal of Applied Sport Science Research,**5(3), 116-120. - Johnson, D.L., and Bahamonde, R. (1996).
*Power Output Estimate in University Athletes.***Journal of strength and Conditioning Research,**10(3), 161-166. - Keir, P.J., V.K. Jamnik, and N. Gledhill. (2003)
*Technical-methodological report: A nomogram for peak leg power output in the vertical jump*,**The Journal of Strength and Conditioning Research**Volume: 17 Issue: 4 Pages: 701-703. - Sayers, S., et al. (1999)
*Cross-validation of three jump power equations.***Med Sci Sports Exerc**. 31: 572.

## Related Pages

- Calculating power during running
- The Physics of the Vertical Jump
- Procedure for Vertical Jump Testing and using the Timing Mat method
- Vertical jump test results.
- Converting vertical jump units between cm and inches.
- Vertical jump devices in the fitness testing store.
- About force plates

*Got any comments, suggestions or corrections? Please let us know.*

## Old Comments

**Commenting is closed on this page, though you can read some previous comments below which may answer some of your questions.**

- Michelle Bivens • 7 years ago

Vertical jumping is a component of most sports. It is often taken for granted that an athlete instinctually knows how to jump vertically. Actually though, jumping vertically is a skill that can and should be taught to athletes

Research has proven that heavy lifting and plyometric training have effectively improved power output. A combination of both can result in gaining improvement in your vertical jump. - Joseph Barnes • a year ago

I'm not good at math so can someone tell me if my son has a 46 inch vertical how many feet does that convert into

- Cengiz ölmez • 2 years ago • edited

hi. thanks for everyrthink. but ı couldnt find lewis nomogram norms. can you help me?

- essay writers • 3 years ago

Thanks for sharing some great information and usage of this vertical jump calculator. It ideally gives a lot of different good learning and use for particular problems in mathematics that involves complex terms and equations.

- Mikhail • 7 years ago

Hi Rob.

I would like to know what will be the unit of power if 9,81 is not added? And can you help me with closed kinetic chain upper extremity test for upper extremity...in what unit does it give the power score?? - Colin • 7 years ago • edited

Are you sure the harman equation is +1822 and not -1822 - Rob Admin Colin • 7 years ago • edited

looking at the original paper it has listed +1822, even though it does seem to overestimate compared to the other equations listed.

- Colin Rob • 7 years ago

I know, i saw that too... but the Sayers paper (which includes Harman as one of the researchers) uses -1822... - I just bring it up because in my paper, the harman equation is giving very large peak power outputs compared to all other formulas... and i have found many research papers that also use -1822
- But maybe the Harman equation is off because of the small sample size (n=17)...anyways, thank you for the information

- Colin Rob • 7 years ago

Hi Rob, I could be wrong. I am just a student who is trying to write a research paper on vertical jump justing a vertec and force platforms... and the harman formula is completely throwing it off. -1822 would make the data fit closer with the other formulas and i am finding that +1822 completely overestimates... but still cannot find a definitive answer... any information you have would be greatly appreciated...

- J11 • 7 years ago

hi thanks for the info. i have a question for you: as you mentioned in the beginning, "A heavier person jumping the same height as a lighter one has to do much more work to move a larger mass. Therefore it is sometimes useful to convert the score to units of power or work." lets say 2 people both are 75kg/180cm with a vertical jump of 60 cm, however 1 person has 10% body fat and one has 30% body fat. if i would like to know the power they need jump up to 80cm, which formula works the best? thank you for your time!

- justjumping J11 • 7 years ago

The power for each person to jump 60cm would be the same, but with for the one with a higher body fat level (and therefore lower muscle mass) it would be a much more impressive effort. As for the best formula to use, I have no idea.