Rocket Thrust Question

Math and Physics are not my strongest suite, but I’m trying to figure something out.

I’m designing a Girandola, Imagine a ring with 4 rocket motors. If each motor produces 12oz of thrust and I set them vertically on the ring, I would expect 48oz of total Vertical Thrust (4x12oz).

But, if I set the motors at a 15 degree angle from vertical, how much of the 48oz of thrust is now going towards Vertical Thrust vs Rotational Thrust?

What is the equation to figure this out?

48 oz * cosine of 15 degrees

Rotational Thrust

I think this is caleld torque.

That gives me a 0.9659. What is this telling me?

Vector math problem.
You’re looking for the x and y values.

For a single rocket -
Fx = 12((cos(15)) = 11.580
Fy = 12((sin(15)) = 3.096

That’s the basics. There’s probably other factors that come in to play.

Note - depends on how you draw the triangle. Swap x and y values

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This is what you are describing, yes? (with more rockets, though)

edit: my pic has angle w.r.t. horizontal; you said vertical so my phi symbol should be on other side of that line.

So @artg_dms,
Are those values in oz of propulsion?
Because if you add them together you get 14.676oz which is more than the original value of 12oz. This is where this method is hurting my brain.

@mblatz, More like this but with 4 rockets.

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Yes - the units don’t change.
You don’t add the the x and y values. They’re sides / components of the vector triangle. You use the Pythagorean equation. It’s all in the page at the above link.


@nick, I see your pic. Makes sense now. @aanne got it right.

[full rocket thrust] * Cos(angle from vertical)

ex 1: no angle = zero degrees. Your FRT for one rocket = 12oz; Cos(0) = 1, so oriented straigt up gives you 12oz FRT * 1 = 12oz straight up. Trivial, but give you obviousl and undersdtably correct resiult, so makes sense.

ex 2: twist 90 degrees so the rocket is pointing exactly horizontally, i.e. no vertical thrust component.

Cos(90) = zero. So 12oz FRT * 0 = 0 oz trust straight up (its now all horizontal). Makes sense.

ex 3 (your situation): 12 * Cos(15) = 12 * .9659 = 11.59, i.e. almost all thrust is vertical…again this comports with what one would expect.

So total thrust of four 12-oz rockets at 15 degrees from vertical ~= 46.36 oz vertical thrust.

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Couldn’t resist


@mblatz, you did the part I’ve done a few time,

What is the horizontal force?

Do the Horizontal and Vertical forces Equal to the Force of the Rockets? Do they equal 48oz total?

@artg_dms thanks!

This works, though it is a ratio I don’t understand fully as it isn’t linear and I want it to be. :slight_smile:

It’s a trig thing. Not always intuitive. I had a good idea how to solve the problem, but did a google search to verify. Fond memories of Physics.

Substitute Sin() for Cos().

So when rocket is pointed fully up (angle is zero from vertical), the horizontal component of thrust is
12 * sin(0) = 0, as expected.

When rocket is pointed fully horizontal (angle is 90 from vertical), the horizontal component of thrust is
12 * sin(90) = 12, as expected.

When rocket is 45 degrees off of vertical , we’d expect thrust componentes to be equal along vertical and horizontal:

sin(45) = 0.7071
cos(45) = 0.7071

When rocket is 15 degrees off from vertical, then 12 * sin(15) = ~= 3.11 oz thrust. Multiply by four to get total ~= 12.42 oz thrust

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Maybe I’m not following the entire thread properly but what’s happening here is that when you tilt the motors 15 degrees, you’re losing 3.11 ounces (among all four motors and assuming the math above is correct) from the vertical component of thrust to the horizontal.

In this case the thrust of 12 ounces doesn’t change - the amount holding your object up does. Assuming the ring and motors weigh 48 ounces, your motors would need to provide 12.42 ounces of thrust each to maintain a hover (the vertical component of thrust would be 12 oz. per motor).

and for reference on what a Girandola is:

So you’ve answered both questions.

Vertical Thrust = 46.36 oz

Horizontal Thrust = 12.42 oz

Total Thrust of all Rockets can’t exceed 48oz, because of the conservation of energy. The rockets only produce 12oz of thrust each and there are 4 rockets.

So how can they produce 12.42 oz of Rotational Force and 46.36 oz of Lift Force? That is more than 48 oz of force. If energy can’t be created or destroyed, how are we creating an extra 10.78 oz of force? Unless in the real world we are not and these forces some how counter act themselves.

Meaning our equation is not telling us the actual force, or we are breaking the laws of physics and I just solved perpetual motion with ROCKETS!!! This is where all this breaks for me. If you put in 48oz of force, how can you get out 58.78 oz of force? You can’t, so then what is the correct amount of force?

@mdredmond this explains the mind fuck of it all.

Marshall (and Art, and Aditya) is correct. The confusion here is because you’re trying to add the two components of thrust together to get a total thrust value. You can’t just add them (well, you can if you want to confuse yourself, I guess) as if they were scalar (non-vector) numbers because you loose a critical piece of information, they’re not pointing in the same direction.

If you add them as vectors (because they are vectors), the math works just fine in reverse. 46.36oz thrust vertically plus 12.42oz thrust horizontally results in 48oz thrust at an angle of 15deg from vertical.

This sort of thing almost always requires pictures being drawn out before it makes sense. The closest thing I can think of that’s commonly studied is the forces of an object being pushed up an inclined plane. Look around for ‘free body diagram of an inclined plane’ and they’ll explain where the trig comes in.

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@michaelb, could we step out of pictures and do a real world test with 3 fish scales???

Would the example below work in the real world if the pull angle was 25 degrees?

Would the scales read like the picture?

The angle will change the exact numbers you measure, and the way you’ve drawn that picture isn’t what we understood from your description, but yes in principle you’ll see the same results we’re calculating if you set up the experiment with rope and fish scales.

In the picture you drew, if the 48oz force is 25deg below the horizontal, you’ll see 40.5oz horizontally and 20.2oz vertically.

If you want to maximize the confusion, pull at 45deg angle so a 48oz pull yields 34oz on each scale. The difference between vector and scalar addition is greatest that way.

There are also a couple of physics demonstrations on YouTube showing vector addition like this with strings and weights. ‘vector addition lab proceduce’ for example.

Please link the demonstration for me. I would love to see it.

I’m fairly certain this is just right triangle math.

Draw a line of length 12 at 15° from vertical. Now complete a right triangle with that line as the hypotenuse. Solve for the lengths of the other two sides and you have the thrust vectors in the vertical and horizontal.

Don’t let the fact that the triangle in this screenshot is not properly proportioned for the numbers we are talking about confuse things. I am just keeping it upright so we see what’s vertical and what is horizontal. I believe this is the solution.

Note that you don’t add the horizontal and vertical components of the thrust to get the 12 ounces of thrust. The only time that works is when the motor is completely horizontal or vertical.

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