Thrush required to hover, take off and accelerate?

[Robert Rice]

thrush required to hover, take off and accelerate?

i need equations to figure out the following
how much force is required to accelerate a mass. ie knowing either the mass or pounds of force,
how to figure out acceleration curve,
basically I am trying to figure out how much force(thrust) is required to move and accelerate an object 300 lb. an also how much force(thrust) would be requires to maintain a steady hover.

 

[Galileo's Ghost]

Sounds like Newton's second Law should help you out...

 

[Redbelly98]

Robert, thanks for moving the question out of the blogs section.

It might help people answering to know some more information ... like is this a school homework question, a matter of personal curiosity, or relevant to a project you're working on?

Regards,

Mark

 

[Robert Rice]

jet pack! using micro-jets..

 

[Robert Rice]

jet pack.. ok let's say total weight of person, pack and fuel is 270 lbs and i have 300pounds of force available (1) at what rate would it accelerate from ground zero in vertical lift? (2) how much lift from a wing in required to achieve horizontal flight? (3) once horizontal flight is achieved, what would the acceleration curve be with the full 300 Lb/F applied.

 

[Redbelly98]

Okay, looks like this is not homework, which gives people more freedom in providing answers.

In order to hover, the thrust or lift force must equal the total weight.

To accelerate upward, the lift force must be greater than the weight. The more the lift force, the greater the acceleration (as long as lift is greater than the weight).

If L is lift force and W is the weight, then the acceleration "a" is:

$$a = g \frac{L-W}{W} \ \mbox{ or } \ g \cdot (\frac{L}{W}-1)$$

where g is the acceleration due to gravity of a free falling object: either 9.8 m/s² or 32 ft/s²

Using your example:
L = 300 lbs
W = 270 lbs

$$\begin{multline*} a = 32 \ \frac{ft}{s^2} \ \frac{300-270}{270}\\ = 32 \ \frac{ft}{s^2} \ \frac{30}{270}\\ = 32 \ \frac{ft}{s^2} \ \frac{1}{9} \\ = \frac{32}{9} \ \frac{ft}{s^2}\\ = 3.6 \ \frac{ft}{s^2} \end{multline*}$$